Since we're on the topic of puzzles...

[QuikSand]

Okay, for those whose appetite has again been whet by the resurfacing of the Monty Hall puzzles, here's another puzzle that ends up with a similar final fork in the road. For those who weren't around to gather the forum's long-standing association with the number two-thirds... well, this puzzle was likely the one that solidified it (years ago) in the wake of the forum's first elaborate run-through of the Monty Hall puzzle.

Anyway...


You're a tricky sort of guy, and in order to take advantage of people with bar bets and the like, you carry around a special quarter with two identical "heads" sides. One day, you're taking a walk (presumably to a bar where you plan to fleece another innocent patron of a free drink) and it turns out you have exactly two quarters in your pocket -- the double-headed quarter, and another regular one.

As you pull your hand from your pocket, you realize that you just dropped one of the quarters. You look down, and all you can tell is that the side facing up is a heads (not tails). From your height, you can't tell by date or design whether the heads is from the double-headed coin or the regular one.

So, given this information... what is the likelihood that the coin sitting on the ground is the double-headed trick coin?

[Pumpy Tudors]

OK, to me, the simple answer seems to be that it's a 50% chance that the coin on the ground is the double-headed one. I mean, you had two quarters in your pocket, and the one on the ground is definitely one or the other. I can't think of a logical explanation for the face-down side of the coin to mean that it's anything other than 50-50.

Since this has something to do with 2/3, though, I'm clearly missing something here, which I'd be willing to accept.

[Fidatelo]

2/3

[Mustang]

My guess is that you originally have total of 3 heads and 1 tail (chicks dig you).
When looking at the dropped coin there are still a total of 2 heads and 1 tail available so, I'd assume that the other side has a 2 out of 3 chance of being a head

[Kodos]

50/50. The only way you'd know for sure is the 1 in 4 times that it ended up showing tails. The head being up tells you nothing, so you have even odds that it is the double-headed quarter or the regular.

But there's only a ten percent chance of that.

[BishopMVP]

50/50

[Huckleberry]

I find it amazing that people are answering 50/50 given the Monty Hall discussion but moreso given QuikSand's first paragraph. It appears that Pumpy is in possession of a spidey sense, though.

The answer is 2/3 just as in the boy/girl problem. You are looking at the heads side of a coin. There are three possible coin faces that you're looking at. If we call the two coins H1-H2 and H3-T, you could be looking at either H1, H2, or H3. All with equal likelihood. In two of those cases you are looking at the double-headed coin. In one of those cases you are looking at the "regulation" coin.

Therefore there is a 2/3 probability that you are looking at the double-headed coin.

[BrianD]

I find it amazing that people are answering 50/50 given the Monty Hall discussion but moreso given QuikSand's first paragraph. It appears that Pumpy is in possession of a spidey sense, though.

The answer is 2/3 just as in the boy/girl problem. You are looking at the heads side of a coin. There are three possible coin faces that you're looking at. If we call the two coins H1-H2 and H3-T, you could be looking at either H1, H2, or H3. All with equal likelihood. In two of those cases you are looking at the double-headed coin. In one of those cases you are looking at the "regulation" coin.

Therefore there is a 2/3 probability that you are looking at the double-headed coin.

This is probably the right answer. The odds are 50/50 before looking at the coin, but once you look and see the heads, everything changes.

[KWhit]

QS kind of gave the answer away in his first post.

[Maple Leafs]

Estimated time until somebody gives an answer based on semantics or human psychology that completely misses the point of the puzzle?

[Passacaglia]

P(coin is double-sided|head)=P(coin is double-sided and a head)/P(head) (from Bayes' Theorem)

P(coin is double-sided and a head) = 1*0.5 = 0.5
P(head) = 0.75

0.5/0.75 = 2/3

[KWhit]

Blah blah blah blah math.

:confused:

[Noop]

My question is why would you keep two quarters in your pocket if your going to scam someone. If this isn't his first time scamming someone he would already have his special coin away from other regular coins. So I question the intelligence of this person, ergo visa vi this question is not good due to the stupidity of the scammer.

.75% btw

[BrianD]

Estimated time until somebody gives an answer based on semantics or human psychology that completely misses the point of the puzzle?

Apparently 19 minutes? :)

[Maple Leafs]

I usually end up throwing this one out there in these sorts of threads. It's not quite as simple as some of the others in the setup, but I like it and maybe someone else will too.

You're playing a game with two other teammates. The three of you sit around a round table facing each other. Behind each of you are two lights, one red and one white. You can see the lights behind your teammates, but can not see the lights behind you.

At the start of each round, one light is turned on behind each player. Which light is turned on (red or white) is completely random, and independent of the other lights. All three may be the same color, or they can be some combination. Once the light is on, each contestant has to guess the color of their own light. They do this by pressing one of three buttons in front of them - "red", "white" and "pass".

If all three players pass, the round is a push. If at least one player guesses a color, and all players who guess are right, then your team wins $100. If at least one player guesses a color, and any players guess wrong, then your team loses $100.

You can not communicate with the other players, or signal each other in any way. You do not know what the other players have guessed before you make your own guess. You can not see your own light in any way (no reflections, etc).

However, you are allowed to discuss strategy with your teammates before the game starts. Can you come up with a strategy that will allow your team to win money at this game?

[st.cronin]

My first guess, without doing any math, is two players pass every time.

[Maple Leafs]

My first guess, without doing any math, is two players pass every time.
If the strategy is for two players to pass each time, then the remaining player will have a 50/50 chance of guessing his light. That means that your expected return would be zero -- you wouldn't lose (like you would with everyone guessing), but you wouldn't win either.

[st.cronin]

Yeah, I realized that after I posted. I'm not sure I can see a way to make the return greater than zero. The fact that you can see the other two lights doesn't mean anything, right? It doesn't make it more likely that you can guess the light behind you? In which case it's always at best 50/50 whether you guess right. And if more than one player is guessing, the odds of both being right go down to less than 50/50.

[Fidatelo]

When you say that if everyone passes the round is a push, does that mean the $100 gets stacked?

[KWhit]

My first guess, without doing any math, is two players pass every time.

That was my first thought too. It sounds like the best option (which obviously is just 50/50, but still seems better than the others.)

[Huckleberry]

If any player sees two lights of the same color behind their teammates they should guess the other color. If different lights, then pass.

Let's call the players A,B, and C. Here are the possibilities with their own light color listed respectively.

1. R,R,R
2. R,R,W
3. R,W,R
4. W,R,R
5. R,W,W
6. W,R,W
7. W,W,R
8. W,W,W

Now here are the results with the given strategy for each possibility:

1. All players guess White. Loss
2. Player C guesses White. A and B pass. Win
3. Player B guesses White. A and C pass. Win
4. Player A guesses White. B and C pass. Win
5. Player A guesses Red. B and C pass. Win
6. Player B guesses Red. A and C pass. Win
7. Player C guesses Red. A and B pass. Win
8. All players guess Red. Loss

Obviously you win 3 of every 4 times. Averages out to making $200 every 4 trials, or $50 every trial. Where do I sign up for this game?

[Fidatelo]

Huckleberry that's awesome!

[st.cronin]

Huckleberry is correct, I think. I would not have thought to look at it that way, with the color combinations as an organic whole.

[Passacaglia]

There are two envelopes on the table. You are told one has twice as much as the other. You open the first envelope, which contains $100, then are given the opportunity to switch. Do you switch envelopes?

[rkmsuf]

There are two envelopes on the table. You are told one has twice as much as the other. You open the first envelope, which contains $100, then are given the opportunity to switch. Do you switch envelopes?

maybe

[MJ4H]

Switch. EV of staying = 100, EV of switching = 125 [(200+50)/2]

[QuikSand]

There are two envelopes on the table. You are told one has twice as much as the other. You open the first envelope, which contains $100, then are given the opportunity to switch. Do you switch envelopes?

Oh boy.

[QuikSand]

Okay, I'll go ahead with the inevitable...

Switch. EV of staying = 100, EV of switching = 125 [(200+50)/2]

Okay, so now that you have switched from A to B and improved your expected value by doing so... should you now switch back? Doesn't the same algebra seem to apply?

[Huckleberry]

Egads. I can't instinctively solve that one. I'm going to lunch and I want to see an answer when I get back.

[MJ4H]

That was not part of the question.

(I'm well aware of this "paradox" but it is definitely an interesting question to me. The crux of the matter is that opening the envelope and discovering the amount can actually skew the probabilities a little bit since some amounts are more likely based on subjective criteria--for instance, can this "person" afford to have double that amount in the other envelope, what are the realistic chances of half a cent being used in this exercise, etc.--however, the paradox of switching back and forth still exists. But, I maintain it is irrelevant because the problem only allows one switch. I've had this discussion a number of times with many people, and I believe it is the correct answer. Note that this all assumes the chances of getting double and half are exactly 50% which, as I've alluded to, can change in extreme situations).

[st.cronin]

Okay, I'll go ahead with the inevitable...



Okay, so now that you have switched from A to B and improved your expected value by doing so... should you now switch back? Doesn't the same algebra seem to apply?

Knowns and unknowns are not thought about in the same way.

[QuikSand]

Knowns and unknowns are not thought about in the same way.

I'm just trying to coax along the "solution" here.

[Maple Leafs]

If any player sees two lights of the same color behind their teammates they should guess the other color. If different lights, then pass.
This is obviously the right answer.

The beauty of it is that everyone who guesses is still only ever a 50/50 chance to be right. But this strategy groups all the wrong guesses into two groups, and spreads all the right answers out over the other six.

(I like this puzzle so much because the first time I heard it, I thought about it for a half hour drive home and was utterly convinced at the end that there was no correct answer.)

[Pumpy Tudors]

(I like this puzzle so much because the first time I heard it, I thought about it for a half hour drive home and was utterly convinced at the end that there was no correct answer.)
Maybe if you had closed your eyes, you would've visualized the correct answer.

[albionmoonlight]

re: the lights puzzle.

The solution provided is elegant and correct. But I think that you need to add the caveat that the players have to go in a certain order. Otherwise, you could have this solution:

Abe, Bob, and Carl all sit at the table. Abe and Bob will always pass. If Carl's light is red, then Abe will pass first. If Carl's light is white, then Bob will pass first. Carl then wins 100% of the time.

[QuikSand]

Estimated time until somebody gives an answer based on semantics or human psychology that completely misses the point of the puzzle?

Well, we have an entrant here...

... since some amounts are more likely based on subjective criteria--for instance, can this "person" afford to have double that amount in the other envelope, what are the realistic chances of half a cent being used in this exercise, etc.

Well, I think it's pretty clear that introducing human psychology into the puzzle, and introducing evidence about what values are more or less likely to have been used is beyond the intended scope of this puzzle.

And if you set that aside... then there's clearly nothing special about the value $100. So what you're saying is that regardless of what's in the envelope that you open... whether it's one cent, $100, or a hundred trillion, it's obviously better to switch to the other envelope, period. That notion should be troubling to you... and is intended to shed some light onto the underlying difficulty of this whole setup.

Bottom line is that the pretty clear premise of the puzzle is that the values in the two envelopes are supposedly random values with no upper boundary...and while we are inclined to just skip past this frustrating concept and get quickly to the X and 2X shorthand (as you have above)... it is this premise that makes the puzzle itself essentially invalid to begin with, and thus a paradox.

[Huckleberry]

Hmm...while I was at lunch I thought maybe the key was considering X and 2X instead of 2X and X/2. This started with realizing the paradox when I considered that with the 2X and X/2 approach I would choose to switch even before seeing the amount. And then I'd choose to switch back. And then back again, ad infinitum.

But without looking at the amount, at least, it can be solved by understanding that I have a 50% chance of holding the X and a 50% chance of holding the 2X. So my EV of keeping the first envelope is 1.5X and my EV of switching is also 1.5X - makes me happy.

But knowing the amount kicks my ass, and I guess everyone else's if there's no solution and it is indeed a paradox. If I'm holding $100 this approach breaks down. No matter how I analyze it the EV of the second envelope is $125.

I guess the probability that the other envelope is the X and our envelope is the 2X isn't really 50%. Guess what it has to be to make the numbers work?

2/3

:D

[Maple Leafs]

Is it a $100 bill? Because if so, you could weigh both envelopes and see if they weighed the same. There is no such thing as a $200 bill, so if the other envelope weighed more then you'd know you should switch. If it didn't weigh more, it must be a $50 bill and you should stay.

Of course, if we're allowing for multiple bills and/or coins in each envelope then you get into using 2D linear equations to find the answer.

I'm assuming we're not allowing for the possibility of checks?

[MJ4H]

Well, we have an entrant here...





No, I was not introducing semantics or human psychology. I was merely addressing real aspects of the problem that is already a paradox. Neither of which had to do with semantics or human psychology. And neither of which were crucial to the answer.

Why don't you just explain the answer instead of leading people on and then belittling their attempts?

[MJ4H]

A little research shows that the solution basically cannot be explained to people without mathematical training, and even then is often "unsatisfying." Well great, color me stupid. I am not trained in mathematics, though I have a decent grasp of basic probabilities and most basic math. I do not understand the technical solution offered on the page I visited, so unless it can be summed up simply, I just won't get it. I had encountered this before and knew it was a paradox, but I certainly hadn't ever examined the math at that level.

So, in summary, I'm quite happy to just say "it's over my head."

http://soler7.com/IFAQ/two_envelope_paradox_solution.htm

[QuikSand]

Why don't you just explain the answer instead of leading people on and then belittling their attempts?

Actually, I already gave it my best shot.

Bottom line is that the pretty clear premise of the puzzle is that the values in the two envelopes are supposedly random values with no upper boundary...and while we are inclined to just skip past this frustrating concept and get quickly to the X and 2X shorthand (as you have above)... it is this premise that makes the puzzle itself essentially invalid to begin with, and thus a paradox.

It's not terribly satisfying, but basically the answer is "the entire puzzle is flawed." The notion of a value drawn completely at random from zero to infinity is one that removes this puzzle from the entire framework of basic, understandable probability. By skipping that fact, we allow ourselves to be drawn toward solutions based on the whole X and 2X (or X and 1/2 X, same thing) concept, which seem like they hold water, but they're simply built on a faulty foundation.

[MJ4H]

It's not terribly satisfying, but basically the answer is "the entire puzzle is flawed." The notion of a value drawn completely at random from zero to infinity is one that removes this puzzle from the entire framework of basic, understandable probability. By skipping that fact, we allow ourselves to be drawn toward solutions based on the whole X and 2X (or X and 1/2 X, same thing) concept, which seem like they hold water, but they're simply built on a faulty foundation.

Ah. That is actually a pretty good summary. The math side of it is a bit uglier, let's just say.

EDIT: It is so annoying to have to edit every single post I ever make because of typos, silly mistakes, my keyboard randomly skipping a letter I know I hit, etc. SIGH

[QuikSand]

So, in summary, I'm quite happy to just say "it's over my head."

That was essentially the cause for my initial reaction to this puzzle getting posted. I think the "solution" tends to be deeply anticlimactic. A bit like the old saw about three guys at a hotel and their $29. Setup is fine, explanation is weak... just not a crowd pleaser.

[MJ4H]

That was essentially the cause for my initial reaction to this puzzle getting posted. I think the "solution" tends to be deeply anticlimactic. A bit like the old saw about three guys at a hotel and their $29. Setup is fine, explanation is weak... just not a crowd pleaser.

Hehe, that is a good analogy. I hate that one.

[KWhit]

Actually, I already gave it my best shot.



It's not terribly satisfying, but basically the answer is "the entire puzzle is flawed." The notion of a value drawn completely at random from zero to infinity is one that removes this puzzle from the entire framework of basic, understandable probability. By skipping that fact, we allow ourselves to be drawn toward solutions based on the whole X and 2X (or X and 1/2 X, same thing) concept, which seem like they hold water, but they're simply built on a faulty foundation.

Serious question, as I don't get why this is a paradox yet.

Why does the 'zero to infinity' randomness enter into this? Would the answer be different if the question were this?


There are two envelopes on the table. You are told that they both contain amounts between $50 and $200 and that one has twice as much as the other. You open the first envelope, which contains $100, then are given the opportunity to switch. Do you switch envelopes?

[Huckleberry]

Serious question, as I don't get why this is a paradox yet.

Why does the 'zero to infinity' randomness enter into this? Would the answer be different if the question were this?

There are two envelopes on the table. You are told that they both contain amounts between $50 and $200 and that one has twice as much as the other. You open the first envelope, which contains $100, then are given the opportunity to switch. Do you switch envelopes?


Actually, it does make it different. Based on your example and assuming an equal distribution between $50 and $200, the probability that our $100 is the higher value is 2/3. I don't know why, but that sure seems right based on those numbers.

2/3 of the range is above our $100 and only 1/3 is below it. :eek:

[QuikSand]

Why does the 'zero to infinity' randomness enter into this? Would the answer be different if the question were this?

The difference is that in your construct, the value of $100 has special value, it's adding new material information to the puzzle. In the zero-to-infinity setup, no particular amount brings you any more information. $100 doesn't "teach" you anything, nor would $100 billion.

[KWhit]

The difference is that in your construct, the value of $100 has special value, it's adding new material information to the puzzle. In the zero-to-infinity setup, no particular amount brings you any more information. $100 doesn't "teach" you anything, nor would $100 billion.

But in the first (original) puzzle, there's still only two possibilities for what can be in that second envelope - $50 or $200. You are taught something by the original $100, right?

So that's why I'm confused about statements regarding infinity. Infinity doesn't matter, only $50 or $200.

(I feel quite sure I'm wrong on this one, but logically cannot wrap my head around the answer.)

[QuikSand]

But in the first (original) puzzle, there's still only two possibilities for what can be in that second envelope - $50 or $200. You are taught something by the original $100, right?

So that's why I'm confused about statements regarding infinity. Infinity doesn't matter, only $50 or $200.

(I feel quite sure I'm wrong on this one, but logically cannot wrap my head around the answer.)

I think the best way I can frame it is that:

YES, it's true that once you know the $100 value for Envelope A, the only possible values for Envelope B are $50 and $200

However, NO it's not true that the likelihood of those two values is 50/50, as we intuitively think it would be. What is the actual likelihood of the other value bing $50? That's where this puzzle becomes insoluble... there simply isn't any way to calculate that, because the distribution of the variables is one that has no grounding.

When you define the range of values (like you do when you establish the $50 and 200 bounds)... then the actual value $100 gives us an anchor, it's meaningful new information, that and lets us make a judgment on what are the remaining probabilities of the two remaining outcomes.

Now, this might not sound like compelling math to you, but off the top of my head, I'd frame it this way:

-there are two ways of coming up with two values, one twice the other, between 50 and 100.
-one is: pick a random number X from 50-100, and the other envelope gets 2x
-the other is: pick a random number Y from 100-200, and the other envelope gets 1/2 Y
-using that analogy, if we open envelope A and see $100, it either came from a process like X or Y above
-but the two processes are not equally likely to generate the number $100
-in fact, since process Y has twice the range, its likelihood of generating the exact value of $100 is exactly twice that of the other process
-(this admittedly involves the frustrating process of trying to compare the sizes of infinite ranges, but we can deal with that in only a relative sense)
-therefore, using Bayes' theorem, it's twice as likely that the $100 is the higher value than it is the lower value
-and therefore, the value of the other envelope = (2/3)(50)+(1/3)(200) = 100

How's that grab you?

[Maple Leafs]

Wait, what if they put a bill in one envelope and a check in the other? That totally screws my theory.

Crap, this is hard.

[st.cronin]

I think the problem is in stating the value of the second envelope as 125. The value is EITHER 50 or 200. Its value will never be anything else.

[QuikSand]

I think the problem is in stating the value of the second envelope as 125. The value is EITHER 50 or 200. Its value will never be anything else.

Well, if you can determine the likelihood of the two possibilities, there's nothing wrong with calculating an expected value. Of course it is never going to be that value, but the general concept of EV isn't out of reach here. What is out of reach is properly arriving at the likelihood of each of the two options.

[st.cronin]

If you did this as a simulation, though, the values of the envelopes should, over time, be equal. Right? The data would look something like:

x 2x
2x x
2x x
x 2x
x 2x
2x x

[QuikSand]

If you did this as a simulation, though, the values of the envelopes should, over time, be equal. Right?

Wrong, I'm afraid.

[Huckleberry]

If you did this as a simulation, though, the values of the envelopes should, over time, be equal. Right? The data would look something like:

x 2x
2x x
2x x
x 2x
x 2x
2x x

Sort of, but you'd have to set up the simulation different than how the problem is stated. You'd have to add a constraint whereby, for example, the lower value must be between $0.01 and $1,000 or whatever boundaries you want to set.

I don't see how you'd set up a simulation where the values in the envelopes could range from $0.01 all the way to $∞.

Better yet, your constraint would have to be that X = a constant and declared value. For example, one of the envelopes always contains $50 and the other envelope always contains $100. Basically, X doesn't change from trial to trial.

That's entirely different than the stated problem.

[st.cronin]

Wrong, I'm afraid.

Oh. Ok.

Sort of, but you'd have to set up the simulation different than how the problem is stated. You'd have to add a constraint whereby, for example, the lower value must be between $0.01 and $1,000 or whatever boundaries you want to set.

I don't see how you'd set up a simulation where the values in the envelopes could range from $0.01 all the way to $∞.

Better yet, your constraint would have to be that X = a constant and declared value. For example, one of the envelopes always contains $50 and the other envelope always contains $100. Basically, X doesn't change from trial to trial.

That's entirely different than the stated problem.

I agree that x has to be a declared value - but I don't see how that's different from the stated problem, as long as the subject doesn't know what x is.

[QuikSand]

The problem here is that even if you constrain the selection...the nature of the "one number is twice the other" setup makes the distribution nonlinear - the lower numbers in the range are more likely to be selected.

[Huckleberry]

That's the problem. In one case we don't know the value of X. In your simulation, we do.

That's like telling the guy with the original choice that the envelopes contain X = $100 and 2X = $200 and then asking him if he wants to switch after he opens the one with $100 in it. The available information is extremely important.

[Huckleberry]

The problem here is that even if you constrain the selection...the nature of the "one number is twice the other" setup makes the distribution nonlinear - the lower numbers in the range are more likely to be selected.

Right. Which is why in the earlier $50 - $200 range example if the selected envelope has $100 in it it is not a 1/2 probability that it is the larger value, it's actually a 2/3 probability.

Seriously, please tell me that's correct so that the numbers work for my rudimentary statistical brain. Staying EV = $100. Switching = (1/3)(200) + (2/3)(50) = $100.

[st.cronin]

I really have not come close to understanding about the last 5 posts by Quiksand.

Huckleberry, its not like that at all - if you know that $100 = x or 2x, then you know exactly the value of the other envelope. But if the subject doesn't know whether $100 = x or 2x, then he is faced with a decision, and the probability SHOULD be the same for the other envelope being either x or 2x.

[QuikSand]

Let me try another angle, again using the 50-to-200 setup, where we have an envelope that includes $100.

To momentarily simplify the problem, let's say that only the integers from 50 to 100 are possible values for X, and the other envelope will include 2X. But the problem there is that while all the integers from 50 to 100 are eligible to get selected, only even integers from 102 to 200 are possible. In essence, the set of integers from 102 to 200 has only half the chance (on average) of being selected as the numbers 100 and below. (This gets a little silly because of the specific boundaries at work here, I understand)

Point is... no matter how finely you space out your intervals... you still end up with this paradox. The number of values between 50 and 100 is *always* going to be the same as the number of values between 100 and 200 -- since they map into X and 2X pairings perfectly. And the space from 100 to 200 is *always* going to be twice s big as the space between 50 and 100, we should all agree. So, whether it's integers, gaps of 1/1000, or any other incrementally tiny measurement... you'd always have exactly the same number of "hits" below and above 100... meaning the "density" of values being used above 100 is half that of the ones being used below 100. If you can handle that logic... that's what leads you to take it to its logical extreme, and essentially assume that there is "no" interval being used, and we are dividing the 50-100 range into an infinite number of possibilities... however, we are by necessity applying the same size of infinity (I know, I know) across the twice-as-large range from 100 to 200. And that still means half the density. Meaning that, in essence, any given number above 100 is half as likely as any number below 100.

In this case, it means that the other envelope is only 1/3 likely to be $200, and is 2/3 likely to be $50... which is where we were before.

[QuikSand]

Right. Which is why in the earlier $50 - $200 range example if the selected envelope has $100 in it it is not a 1/2 probability that it is the larger value, it's actually a 2/3 probability.

Seriously, please tell me that's correct so that the numbers work for my rudimentary statistical brain. Staying EV = $100. Switching = (1/3)(200) + (2/3)(50) = $100.

That's correct, though this is not rudimentary stuff.

[st.cronin]

There are two envelopes on the table. You are told one has twice as much as the other. You open the first envelope, which contains $100, then are given the opportunity to switch. Do you switch envelopes?

Let me try another angle, again using the 50-to-200 setup, where we have an envelope that includes $100.

Ok, I actually did understand that post, but it seems like the solution to a different problem. The original problem has no 50-200 setup. It is EITHER 50/100, or 100/200. There are no other possibilities.

[Huckleberry]

Thank you for the explanation. I knew the answer had to be a 2/3 - 1/3 split on the probability (the math has to work out) but wouldn't have been able to post a rigorous proof to save my life.

[QuikSand]

Ok, I actually did understand that post, but it seems like the solution to a different problem. The original problem has no 50-200 setup. It is EITHER 50/100, or 100/200. There are no other possibilities.

Hey, either I am writing this post while sitting at the North Pole, or I'm not. Just because there are only two possibilities doesn't mean they are equally likely.

The framework of the possible values is absolutely relevant here. In the 50-to-200 variation thats been tossed out, it actually allows us to calculate the expected value of the other envelope. The fact that they ($50 and $200) are not equally likely is the main point to draw from it, I think. The other envelope containing $50 is more likely than it containing $200, period.

But it ought to also shine a light on why the original setup is incorrect. To just assume that the two other possible values for the envelope, $50 and $200, are equally likely is a complete fallacy, and it's what underlies the whole puzzle and its resultant paradox.

[Huckleberry]

Ok, I actually did understand that post, but it seems like the solution to a different problem. The original problem has no 50-200 setup. It is EITHER 50/100, or 100/200. There are no other possibilities.

It is definitely tricky to get your head around it.

I don't think it can be explained better than he just did, though. Speaking very generally, lower numbers are twice as likely to be picked as higher numbers for a given range. Of course that gets us into a very counterintuitive (sorry about using that word) situation where no matter which envelope we pick, once we have picked it and seen its value we know that we have a 2/3 probability of having picked the higher amount. :eek:

Or not. I guess it all depends on the value picked and the range prescribed.

Oh forget it. I'll settle for half understanding it. It seems I was right before the "Or not." statement above. It is twice as likely that the other envelope is less than that the other envelope is more than the one selected.

[QuikSand]

Of course that gets us into a very counterintuitive (sorry about using that word) situation where no matter which envelope we pick, once we have picked it and seen its value we know that we have a 2/3 probability of having picked the higher amount. :eek:

Not really. The fact of the matter is that the value $100, in that carefully-constructed setup, is the *only* value that could be either low or high.

If we went through the same construct with a range of something like 1 to $1,000, we'd see a more graduated distribution between 2 and 500... and the higher your number in envelope A in that range, the more likely it is to have been the higher of the two.

[KWhit]

I appreciate the efforts going into dumbing this down for me (us), but I'm still not there yet.

This must just be getting down to semantics of the wording of the problem. Because the way I see it, you have a 50/50 chance of initially selecting the envelope with the more money in it. But your possible gain ($100 to $200) is greater than the possible loss ($100 to $50), so you switch. It is a good gamble to make. (This is basically the expected value argument).

The fact that you KNOW there's $100 in one of the envelopes has to be taken into account. There aren't an infinite number of choices. We know that there are only two possibilities left - $50 or $200.

[st.cronin]

What strikes me as incorrect about this solution is that we don't have a RANGE of values in the problem - we have either/or. Its not as if 100 was randomly generated - the random generation was 2x or x, one of which equals 100.

[QuikSand]

I appreciate the efforts going into dumbing this down for me (us), but I'm still not there yet.

This must just be getting down to semantics of the wording of the problem. Because the way I see it, you have a 50/50 chance of initially selecting the envelope with the more money in it. But your possible gain ($100 to $200) is greater than the possible loss ($100 to $50), so you switch. It is a good gamble to make. (This is basically the expected value argument).

The fact that you KNOW there's $100 in one of the envelopes has to be taken into account. There aren't an infinite number of choices. We know that there are only two possibilities left - $50 or $200.

I think I am running out of ways to explain why this (underlined above) is not so...but while the initial chance of drawing the higher envelope is obviously 50/50, once you have the value, and you have the range from which the values were drawn, you can now use that new information to recalculate the correct likelihood, as I have done above with your 50-to-200 example.

And in the original puzzle, this is not even possible, since the entire concept of a random distribution from zero to infinity is essentially a meaningless concept.

So, I understand your common sense *wants* the odds to be 50/50, but they simply are not. We know what they are in your setup, and in the original setup they don't really even exist.

[QuikSand]

What strikes me as incorrect about this solution is that we don't have a RANGE of values in the problem - we have either/or. Its not as if 100 was randomly generated - the random generation was 2x or x, one of which equals 100.

The underlying issue here is the concept of sizes of infinity, which is just terribly tough to grasp. I've probably done my best to walk through it, but there may be other discussion of this problem that do a better job. I suspect that my discussion above, even if it doesn't sway you, is the best I can do.

But honestly... surely we all agree that just because there are only two remaining options doesn't mean that the chances must be 50/50? Right?

[Huckleberry]

Not really. The fact of the matter is that the value $100, in that carefully-constructed setup, is the *only* value that could be either low or high.

If we went through the same construct with a range of something like 1 to $1,000, we'd see a more graduated distribution between 2 and 500... and the higher your number in envelope A in that range, the more likely it is to have been the higher of the two.

Okay, now this I don't understand simply because the 2/3-1/3 probabilities can't change and still have our EVs stay the same for staying and switching. Not if we're staying with the twice as much or half as much in the envelope example.

Let's say we have a range for both envelopes of $1 to $1000 dollars, integers only. So basically the lower value can be from $1 to $500 and the upper value can be from $2 to $1000.

Now let's say we open an envelope and it has $100 in it. In order for the numbers to work, $50 has to be twice as likely as $200 in the other envelope. Based on the density explanation above, that makes sense.

What if we opened it and there were $372 in it? Well, $186 has to be twice as likely as $774 or the numbers don't work. Right?

How does the fact that if we open the envelope and see an odd amount we know for a fact it's the lower value work into this? :)

(Just kidding, I know we're approximating the range with integers to make it easier to discuss)

[st.cronin]

The underlying issue here is the concept of sizes of infinity, which is just terribly tough to grasp. I've probably done my best to walk through it, but there may be other discussion of this problem that do a better job. I suspect that my discussion above, even if it doesn't sway you, is the best I can do.

But honestly... surely we all agree that just because there are only two remaining options doesn't mean that the chances must be 50/50? Right?

Yes, sure. I'm fine with that. Probability in general is problematic. And I suspect there's some underlying semantic problem as well, although I don't actually see it now.

As I see the original problem, you get a value that is either x or 2x, with no more probability of one than the other. If that's true, then it should follow that, over many trials, the values of the first envelope should be the same as the values of the second envelope. If that's not true, then there is something about the original problem, not math, that I don't understand.

[KWhit]

But honestly... surely we all agree that just because there are only two remaining options doesn't mean that the chances must be 50/50? Right?

Sure.

But another way to state this question essentially is this:

There are two envelopes in front of you with different amounts of money in them. You pick one up. What are the odds that you have chosen the envelope with the higher amount of money in it?

50/50.

I'm sure you'll tell me that this is a totally different scenario, but I'll probably not understand why.

:)

[QuikSand]

Let's say we have a range for both envelopes of $1 to $1000 dollars, integers only.

Well, in the simplification to integers only, we skew things a bit, don't we? By simply removing half of the integers over 500 from any consideration, that makes this a very imperfect simplification. And that's where the troubles start.

[QuikSand]

I'm sure you'll tell me that this is a totally different scenario, but I'll probably not understand why.

So, we can just skip it and move on, then. Done.

[st.cronin]

Sure.

But another way to state this question essentially is this:

There are two envelopes in front of you with different amounts of money in them. You pick one up. What are the odds that you have chosen the envelope with the higher amount of money in it?

50/50.

I'm sure you'll tell me that this is a totally different scenario, but I'll probably not understand why.

:)

Yes, yes. I agree with KWhit.

[Huckleberry]

Well, in the simplification to integers only, we skew things a bit, don't we? By simply removing half of the integers over 500 from any consideration, that makes this a very imperfect simplification. And that's where the troubles start.

Sure. So remove the integer constraint. Still has to be twice as likely that the other envelope is the lesser amount than it is that it's the higher amount, no?

Sure.

But another way to state this question essentially is this:

There are two envelopes in front of you with different amounts of money in them. You pick one up. What are the odds that you have chosen the envelope with the higher amount of money in it?

50/50.

I'm sure you'll tell me that this is a totally different scenario, but I'll probably not understand why.

:)

It's a totally different scenario because you didn't have the constraint of one is twice as much as the other. As QS explained, this changes the probability density function above and below the selected value. If the constraint were that one of the envelopes contains four times as much money as the other then the probability that the other envelope contains less than the one you selected becomes 4/5 instead of 2/3.

[gkb]

I'm too dumb to understand this thread. I think I need to order some of the basic math and statistics classes from The Teaching Company.

[QuikSand]

Okay, I will try one more angle. We all understand the idea of a random number between (low bound) and (high bound) and what that means, how it's calculated, etc. Perfectly smooth distribution, any number in the range has an equal chance of being selected, we get it.

In this puzzle, though, the selection is **much, much** more complex. It might not seem that way, but it really is. Picking two numbers in the range such that one is twice the other is a much more complex function, and (believe it or not) it leads to some numbers being more likely to be selected than others. Yes, there are numbers with in any bound range that have no partner in one direction... but there's more to it than that. It is a complex distribution that skews toward lower values.

So, the pair of 50/100 is more likely to get picked than the pair of 100/200. if you don't agree with that, then you are most likely applying the logic of that simple, smooth distruibution (the one that comes naturally to all of us) rather than the actual, more complex, distribution necessitated by the two components and their relationship.


Getting anywhere?

[KWhit]

Yes, yes. I agree with KWhit.

Well there goes that theory!

[st.cronin]

The idea of a random number is what has been introduced, I think. That's where this solution doesn't match up with my understanding of the problem. There is NO indication in the problem that 50, 100, or 200 are determined by a random function. The only thing that is randomly determined is WHICH envelope the subject opens.

[QuikSand]

The idea of a random number is what has been introduced, I think. That's where this solution doesn't match up with my understanding of the problem. There is NO indication in the problem that 50, 100, or 200 are determined by a random function. The only thing that is randomly determined is WHICH envelope the subject opens.

And I would argue that it's *you* who is introducing something here, by exerting through sheer force of will, that the two options are equally likely.

[KWhit]

Okay, I will try one more angle. We all understand the idea of a random number between (low bound) and (high bound) and what that means, how it's calculated, etc. Perfectly smooth distribution, any number in the range has an equal chance of being selected, we get it.

In this puzzle, though, the selection is **much, much** more complex. It might not seem that way, but it really is. Picking two numbers in the range such that one is twice the other is a much more complex function, and (believe it or not) it leads to some numbers being more likely to be selected than others. Yes, there are numbers with in any bound range that have no partner in one direction... but there's more to it than that. It is a complex distribution that skews toward lower values.

So, the pair of 50/100 is more likely to get picked than the pair of 100/200. if you don't agree with that, then you are most likely applying the logic of that simple, smooth distruibution (the one that comes naturally to all of us) rather than the actual, more complex, distribution necessitated by the two components and their relationship.


Getting anywhere?

Yeah. I get that. I think I'm where st.cronin is in questioning the randomness concept. My thinking is that one person picked a random number and then performed a calculation against it to derive the other number.

[st.cronin]

And I would argue that it's *you* who is introducing something here, by exerting through sheer force of will, that the two options are equally likely.

Well, do you believe that if I put two envelopes in front of you, with different amounts of money, that you would actually be more likely to pick one over the other? If not, then I think we agree.

[Huckleberry]

I'm trying to come up with a somewhat graphical explanation for visual learners.

[Maple Leafs]

Hey, either I am writing this post while sitting at the North Pole, or I'm not.
Are there any bears outside your window, and if so what color are they?

[Huckleberry]

Dammit I wish I could copy and paste an image. Not working.

[QuikSand]

My thinking is that one person picked a random number and then performed a calculation against it to derive the other number.

That's fine... but along a confined range, doing so yields a non-uniform distribution of possible choices, which is really the essential point in getting away from the insistence that this remains a 50/50 proposition.

[Huckleberry]

Here is the image showing what most everyone is thinking of as representing the probability distribution for any value in our range being in one of the envelopes:

http://i286.photobucket.com/albums/ll105/Huck_L_Berry/evendist.png

However, the one envelope having twice as much as the other changes the actual distribution to looking like this:

http://i286.photobucket.com/albums/ll105/Huck_L_Berry/2envelopesdist.png

As you can see, once we select any value in the distribution as being in one of the envelopes, the probability that half that value is in one of the envelopes is twice the probability that twice that value is in one of the envelopes.

Just a graphical attempt at an explanation. Please do not think the y-axis values are being presented as accurate. Just know that I used an even distribution for the first one and a 1/SQRT(X) distribution for the second. That distribution ensures that y(x) = 2*y(4x) which is the situation we're dealing with.

[MJ4H]

The way I'm thinking of it is lower numbers are easier to hit, like a target that is nearer to you, than higher numbers, like a target that is farther from you. Think of the formula that derives at the number (in this case, simply 2x) as the aiming mechanism, maybe? If I take my 2x gun and shoot, it is easier to hit lower numbers than higher numbers. If I take a more complicated gun, maybe something like 5x^3+16 (or something?? just picked something random without much thought there) then it might be even harder to hit the high numbers--you would have to be much more precise with your aiming. I wonder if that makes any sense, or if those of you with your brains fully wrapped around this can see if my brain is starting to at least curl towards it?

Man the more I look at even that, which was clear to me a second ago, the more nuts I think I am.

In summary: ????????????

[st.cronin]

The distribution part of the math makes perfect sense to me, MJ4H. I just don't see how it applies to the problem. Again, I ask: If I put two envelopes in front of you, one with 100 dollars, one with 200 dollars, which one would you be more likely to pick? If you say its a coin flip, then you see the problem the same way I do. If you actually think you're more likely to pick the one with 100 dollars, then... you're either a giant pessimist, or you understand the problem in a totally different way than I do.

[MJ4H]

But, that isn't the problem as stated.

edit to add: but, for the love of God, I feel your pain. I just know that changing the problem as you did avoids the entire issue. It may seem like a superficial change, but it isn't. Also, ouch this still hurts.

[st.cronin]

But, that isn't the problem as stated.

The way I read it, it is exactly. Which, as I suspected, makes this whole thing semantical.

X is not variable! It is simply UNKNOWN to the subject. Unknown is not the same thing as variable.

[Fidatelo]

Holy shit there are fucking graphs in here.

*backs away slowly*

[MJ4H]

You don't like graphs? Well yeah, then.

http://i7.photobucket.com/albums/y257/Gorgonian14/1104al2.gif

[gstelmack]

So, the pair of 50/100 is more likely to get picked than the pair of 100/200.

This was the point of the thread where I went "Okay, now I get it".

[Huckleberry]

My quick explanation at a proof before heading out with my daughter:

1.) A range of values must be defined

An infinite range of possible values makes this problem meaningless. It is impossible to have an even distribution over an infinite range. Just consider it for a moment. What is the constant y-value of each point's probability if the x-values can go to infinity? If it's zero then the total probability of selecting any number as a part of the pair is zero. That doesn't work. If it's greater than zero than over the entire set of x-values the total probability will go to infinity as well. Also doesn't work.

2.) Given a defined range, we can define the probability function

Based on the example QS gave above. Select any interval of possibilities for the lower value. Then select the same interval for a set of possible higher values. There are twice as many possible lower values in the lower interval than there are possible higher values in the higher interval. For example: Choose 1-100 for the lower values and limit the possibilities to every integer. Choose 2-101 for the higher interval. There are 100 possible lower values in the lower interval but only 50 possible higher values (even numbers) in the higher interval. If you'd like, expand the possible values to ever 1/2. Now there are 200 possible lower values in the lower interval and 100 possible higher values (every whole number).

Now shrink the interval. Make it 1-10 and 2-11. Then 2.5-3.5 and 5.5-6.5 and the same holds true. Eventually as the interval shrinks you will see that for every possible value there is double the amount of probability for a lower value as there is for a higher value with the same interval.

And yes, I know I explained that poorly.

3.) Given this knowledge I believe I can describe the probability function

We now that y(x) = 2*y(4x). We also know that I've made a giant leap to that assumption, but tough.

This suggests an equation of the form C / X^2 where C is a constant. We will look for the formula describing the probability that a given value in the known range will be selected as a member of the pair. We know that the integral of this formula from the lower boundary to the upper boundary must sum to 2 (two members of the pair). Therefore we can determine the formula by setting 2C*SQRT(UB) - 2C*SQRT(LB) = 2. C = 1/(SQRT(UB)-SQRT(LB))

So for the range of 100-10000 for both values, the probability function for any value as a member of the pair = (1/90) / SQRT(X)

For example, the probability that 400 is selected as a member of the pair is, I believe, .000556 or so. Here I'm not sure if I should instead be integrating the function between 399.5 and 400.5, but if I should then the probability of 400 is also .000556 which is nice. :D

4.) Knowing all of this we can essentially use the formula as a cheat sheet

Let's say 400 is in our envelope. We can use the formula to determine that P(800) = 0.000393 and P(200) = 0.000786. Fortunately for me this is 2*P(800). So we can calculate our expected value using relative probabilities. EV staying = $400. EV switching is 200*P(200)/(P(200)+P(800)) + 800*P(800)/(P(200)+P(800)) = $400.

This also works for all values because let's say that we select $800 in our first envelope. Well, based on the upper and lower boundaries we know that P(1600) = 0. So using the above formula but plugging in the new values gives EV staying = 800 and EV switching = 400.

Got to run. I'm sure I'll see the errors pointed out later. :)

[BrianD]

For those that are still unsure of the problem, I would suggest trying to come up with a way to model the situation and run a few trials to see what you come up with. The only way you can run trials is to put constraints around the problem that aren't originally there.

[st.cronin]

For those that are still unsure of the problem, I would suggest trying to come up with a way to model the situation and run a few trials to see what you come up with. The only way you can run trials is to put constraints around the problem that aren't originally there.

Post 53 I thought I had put forward how the problem should be modeled, but I've been told its wrong. Maybe you can explain why?

[BrianD]

Although here is a different thought. In most of these puzzles, the initial easy probability is changed when some new information is given. The odds of the 2-headed quarter being on the ground was 50/50 before looking down and seeing the head. In this latest example, couldn't we assume that the new information of seeing the $100 doesn't actually give us any new information since we don't know any of the constraints? Couldn't we say that the initial 50/50 chance of having the higher amount in our envelope is still the same since we don't know any differently after looking inside? Does that change anything at all?

[Julio Riddols]

All I know is that if I had two envelopes to choose from and one had twice as much as the other, I would switch simply because the amount gained would be greater than the amount lost. It would be worth the gamble to try and turn that 100 into 200, since the worst that could happen is it becomes 50.

It would be like someone flipping a regular coin, and if it lands on heads you give them 50, but if it is tails they give you 100. Worth the gamble.

[QuikSand]

Okay, let's dial all the way back to the original two envelopes puzzle. Even before we open the envelope. We know it has some value inside, let's call it X, while the other one contains Y. If the entire universe of possible values is fair game, then it really shouldn't matter what number X is, correct? Nothing special about $100.

So, the other envelope either had X/2 or 2X in it, we'd agree. If we accept the argument (being made here) that this is now a 50/50 proposition, then we naturally conclude that the EV of envelope Y is 1.25X. Or, in essence, we *know* that it's better to switch from envelope X to envelope Y.

Now, we hold envelope Y. And we now know that whatever Y is, we know that X is either Y/2 or 2Y. And if you again assume this is a 50/50 proposition, then you are again at the clear conclusion that the EV of X is greater than Y, so you again *know* that it's better to switch from envelope Y to envelope X.

This puts you right back to the previous paragraph. Obviously, you are stuck square in the paradox. It simply cannot be that X is better than Y and Y is better than X. This is your classic reductio ad absurdum - if we make assumptions that lead us directly to an impossible result (like this one) then we need to go back and refine our original assumptions.

And... of course... the flawed assumption is the 50/50 probability. And it traces from the unbounded random distribution, a function which cannot be rationally analyzed, nor simulated in any meaningful way.

[QuikSand]

It would be like someone flipping a regular coin...

You are assuming your own desired conclusion here. If you start with the assumption that this is a 50/50 proposition, then of course your logic makes sense.

What if the option was instead to roll a fair die, and on a roll of 1-4 you got $50, but on a 5 or 6 you got the $200. Still a no brainer? Is this the sort of illustration that gets us anywhere closer to dropping the assumption that this is a 50/50 proposition?

[st.cronin]

QS, if you can, please explain why post 53 is incorrect.

Edit: I am close to positive that the problem is that QS is treating x as a variable, instead of as an unknown constant.

[BrianD]

People are assuming after seeing the $100 that the range of possible values in the other envelope is $50-$200. This isn't the case. The range at the beginning of the problem was either $50-$100 or $100-$200. We don't know what the range was so we can't assign probabilities. Unknown is not the same as 50/50.

[Julio Riddols]

Why would it be anything other than a 50/50 shot? The other envelope has to be either twice as much or half as much. Either way, the possible gain would outweigh the possible loss because there are only two envelopes. Either you got the big one or the little one. How can it be any other way?

If we were talking about 3 envelopes, I see your point. But we're only talking about 2.

[Julio Riddols]

People are assuming after seeing the $100 that the range of possible values in the other envelope is $50-$200. This isn't the case. The range at the beginning of the problem was either $50-$100 or $100-$200. We don't know what the range was so we can't assign probabilities. Unknown is not the same as 50/50.

But you are told that the other envelope is either twice as much or half as much.. Not "somewhere between" twice as much and half as much. There is no range, only two possible values in that other envelope. Given that the first envelope contains 100, the second MUST contain either 50 OR 200.

[QuikSand]

If you did this as a simulation, though, the values of the envelopes should, over time, be equal. Right? The data would look something like:

x 2x
2x x
2x x
x 2x
x 2x
2x x

Okay, you're hammering on this post again and again. I don't even know what you're saying here.

Is this tied to the original problem? The one where we really can't even discuss it because it is entirely founded on a meaningless concept of a random distribution between zero and infinity?

If so... then there's simply no way we can say anything meaningful about the probabilities at work with this problem. It's not 50/50, it's not 2/3-to-1/3, it's just a meaningless undefined probabilistic distribution, a paradox.

I know this isn't the answer you are looking for. But it's the explanation for why your side of the puzzle ends up arguing in a complete circle with itself.

[QuikSand]

Given that the first envelope contains 100, the second MUST contain either 50 OR 200.

That is true. But they are not necessarily equally probable, as you and others are assuming/insisting.

[Buccaneer]

I can hear QS banging his head against the wall all the way here in Colorado.

[st.cronin]

Okay, you're hammering on this post again and again. I don't even know what you're saying here.

Is this tied to the original problem? The one where we really can't even discuss it because it is entirely founded on a meaningless concept of a random distribution between zero and infinity?

If so... then there's simply no way we can say anything meaningful about the probabilities at work with this problem. It's not 50/50, it's not 2/3-to-1/3, it's just a meaningless undefined probabilistic distribution, a paradox.

I know this isn't the answer you are looking for. But it's the explanation for why your side of the puzzle ends up arguing in a complete circle with itself.

When does randomness enter into the problem, is my question? The values in the envelopes are fixed, not random.

[QuikSand]

When does randomness enter into the problem, is my question? The values in the envelopes are fixed, not random.

But by definition, the values were selected to fit the terms stated. And there is some probability that the other envelope contains half as much, and some probability that it contains twice as much. And those two probabilities add to 1, they are exhaustive and exclusive. We have to agree there.

The only place we disagree is whether, as you argue, they are necessarily 50% each. What you have is your voice to say that they are. I have mathematics to say that either they are not (in the bounded version of this problem) of that the math is so screwed up that the probabilities are essentially meaningless (in the case of the original, unbounded version).

[BrianD]

But you are told that the other envelope is either twice as much or half as much.. Not "somewhere between" twice as much and half as much. There is no range, only two possible values in that other envelope. Given that the first envelope contains 100, the second MUST contain either 50 OR 200.

That is actually untrue. The other envelope isn't either twice as much or half as much. One envelope is twice the other. If I give you one envelope and then tell you that the other is "either twice as much or half as much", that assumes equal probability between the two outcomes and you should switch. By saying that one envelope is twice the other, your options are that either you picked the higher one or you picked the lower one. Knowing what is in your envelope doesn't change that fact. By assuming that knowledge of you envelope gives two possibilities for the other envelope, you are adding false information to the problem. There is still only one other possibility, you just don't know what that possibility is.

[BrianD]

What if we look at this from the other side. I put $50 in one envelope and $100 in the other envelope. QuikSand grabs and opens up the $100 envelope. From here he says "great, the other envelope must be either $50 or $200. We know there isn't a 50% of there being $200. The fact that he doesn't know this isn't a possibility doesn't give it a 50% chance.

[Buccaneer]

That is actually untrue. The other envelope isn't either twice as much or half as much. One envelope is twice the other. If I give you one envelope and then tell you that the other is "either twice as much or half as much", that assumes equal probability between the two outcomes and you should switch. By saying that one envelope is twice the other, your options are that either you picked the higher one or you picked the lower one. Knowing what is in your envelope doesn't change that fact. By assuming that knowledge of you envelope gives two possibilities for the other envelope, you are adding false information to the problem. There is still only one other possibility, you just don't know what that possibility is.

Very well said (to back up QS). You pick up env1 and the other could be twice. You changed your mind and pick up env2 and the other could be twice. Both can't be right but you have no way of knowing.

[Julio Riddols]

That is true. But they are not necessarily equally probable, as you and others are assuming/insisting.

Why? 1/2= 50%. There are two choices, and you pick one of them. How is that not equally probable? Is one being dangled from high above and you must be able to jump a certain height to get it?

I don't see how there is any logic required. Its a simple case of picking one, then being told that the second could be double that or half that. I'll take that bet every time.

Now if you don't get the higher one, yeah, you lose 50 bucks in the given example.. But if you get the higher one, you gain 100. There is simply no other way to explain this without using variables that were not a part of the original question.

[st.cronin]

But by definition, the values were selected to fit the terms stated. And there is some probability that the other envelope contains half as much, and some probability that it contains twice as much. And those two probabilities add to 1, they are exhaustive and exclusive. We have to agree there.

Yes.

The only place we disagree is whether, as you argue, they are necessarily 50% each. What you have is your voice to say that they are. I have mathematics to say that either they are not (in the bounded version of this problem) of that the math is so screwed up that the probabilities are essentially meaningless (in the case of the original, unbounded version).

Bounded, unbounded - are those the only options? What if, instead of monetary values, one has a picture of Prince Charles, and the other has a picture of Jennifer Lopez? How do you analyze that?

[Julio Riddols]

What if we look at this from the other side. I put $50 in one envelope and $100 in the other envelope. QuikSand grabs and opens up the $100 envelope. From here he says "great, the other envelope must be either $50 or $200. We know there isn't a 50% of there being $200. The fact that he doesn't know this isn't a possibility doesn't give it a 50% chance.

Yeah, but he would be right to take the gamble is all I am saying. Because if you took $100 and $200 and he picked the 100 first, he gains 100. Sure, he has no idea whether he has picked the higher or lower envelope first, but if thing a is twice thing b, that means thing b has to be half of thing a.

If envelope a= 2, then envelope b has to be either twice that or half that. Either way, one is still twice the other.

[MJ4H]

Why? 1/2= 50%. There are two choices, and you pick one of them. How is that not equally probable?

Because in the set of all possible numbers that can exist in the envelopes, there are more lower numbers than high numbers. This makes it more likely if you switch that you will be switching to a low number. To demonstrate, what if there were many envelopes with these numbers:

1, 2, 3, 4, 5, 6, 7, 8, 9 ,10, 11, 12, 13, 14, 15, 20, 25, 30, 45, 50, 100, 200, 400, 500, 1000

If you pick an envelope and it has the number 200 in it, would you be wise to switch it with another envelope? There are 800 or so possible higher numbers, and 199 or so possible lower numbers, so your chances are better that the envelope you switch to is higher. Except that it obviously isn't. Because the distribution is not linear.

And the distribution for this example is not linear either. There is not an equal chance of 5 and 500,000 being in the envelopes. There is also not an equal chance of 50 and 200 being in the other envelope. There is not an equal chance of 50 and 100 being in the two envelopes and 100 and 200 being in the envelopes. This is because the distribution is not linear. This non-linearness was discussed a bit earlier in the thread. That is the tough part to grasp, frankly. Once you understand that, I think it is pretty easy to see why the options aren't weighted equally by probability.

[Julio Riddols]

There are 4 tables.

There are two envelopes on each table. You are told one has twice as much as the other. You open the first envelopes on each table, each of which contains $100, then are given the opportunity to switch. Do you switch envelopes?

Possible losses = 50 dollars at each table. 4 x -50 = -200.

Possible gains = 100 dollars at each table. 4 X 100 = 400

if you take the gamble on each table, win twice and lose twice, you still gain $100.

[MJ4H]

We understand the basic expected value calculations, yes. They are based on the idea that it is 50/50 that the other envelope will be the higher number. It is not. The distributions are not linear.

[Julio Riddols]

Because in the set of all possible numbers that can exist in the envelopes, there are more lower numbers than high numbers. This makes it more likely if you switch that you will be switching to a low number. To demonstrate, what if there were many envelopes with these numbers:

1, 2, 3, 4, 5, 6, 7, 8, 9 ,10, 11, 12, 13, 14, 15, 20, 25, 30, 45, 50, 100, 200, 400, 500, 1000

If you pick an envelope and it has the number 200 in it, would you be wise to switch it with another envelope? There are 800 or so possible higher numbers, and 199 or so possible lower numbers, so your chances are better that the envelope you switch to is higher. Except that it obviously isn't. Because the distribution is not linear.

And the distribution for this example is not linear either. There is not an equal chance of 5 and 500,000 being in the envelopes. There is also not an equal chance of 50 and 200 being in the other envelope. There is not an equal chance of 50 and 100 being in the two envelopes and 100 and 200 being in the envelopes. This is because the distribution is not linear. This non-linearness was discussed a bit earlier in the thread. That is the tough part to grasp, frankly. Once you understand that, I think it is pretty easy to see why the options aren't weighted equally by probability.

Yes, but the original question tells you that the number in the envelope is 100. So you already know that one envelope is 2x the other, meaning there are only 3 possible values that can be distributed between the two envelopes. if one is 100, it could be twice envelope 2, or envelope 2 can be twice it. so, therefore, envelope 2 must be either 50 or 200, not 1,2,3,4,5,6,7,8, etc.

[BrianD]

One is still twice the other, but knowing the value of one doesn't give you two possibilities for the other one. There is only one option for the other envelope. The fact that you don't know what that value is doesn't magically create another possibility.

Since you don't know what the potential range is, you could make your envelope swap choice without opening the first one. Seeing what is in your envelope doesn't change the initial range, is just gives you the mistaken belief that the range has changed.

[Huckleberry]

You're still assuming equal probability. Every example you post assumes it. The problem is that the assumption is wrong.

[MJ4H]

Yes, but the original question tells you that the number in the envelope is 100. So you already know that one envelope is 2x the other, meaning there are only 3 possible values that can be distributed between the two envelopes. if one is 100, it could be twice envelope 2, or envelope 2 can be twice it. so, therefore, envelope 2 must be either 50 or 200, not 1,2,3,4,5,6,7,8, etc.

Yes, the other envelope is either 50 or 200. But it is not 50/50 between them.

[Fidatelo]

I just want to play a game where i get to open envelopes full of money and keep some of it.

[BrianD]

Yes, but the original question tells you that the number in the envelope is 100. So you already know that one envelope is 2x the other, meaning there are only 3 possible values that can be distributed between the two envelopes. if one is 100, it could be twice envelope 2, or envelope 2 can be twice it. so, therefore, envelope 2 must be either 50 or 200, not 1,2,3,4,5,6,7,8, etc.

Let's try it this way...pick two values for the envelopes that fit this problem and give me the values.

[QuikSand]

Bounded, unbounded - are those the only options? What if, instead of monetary values, one has a picture of Prince Charles, and the other has a picture of Jennifer Lopez? How do you analyze that?

Wow, our semantic problems are greater than I had imagined.

[Buccaneer]

Let's try it this way...pick two values for the envelopes that fit this problem and give me the values.

That's just mean. :)

[Julio Riddols]

I understand you're saying there are not 3 numbers involved.. I get that. But it is worth a one time gamble to see if the second envelope is twice the first. If the first envelope is twice the second, you're only cutting the value of the first envelope by half rather than doubling it.

Thats why it is worth a shot. You might come out with half of what you could have had, but you also might double that if you happened to pick the lower number first.

Overall, if I needed the money, I would just take one envelope and run, period.

[Julio Riddols]

ok, here are two envelopes.. A and B. I know the values in them, and one is twice the other. Pick one.

[Julio Riddols]

FYI, I am really trying to see the other side of this that you guys are arguing.. I am. I just don't see how you come to the conclusion you do. If i have 2 items and one is double the other, one is also half the other.

[Buccaneer]

FYI, I am really trying to see the other side of this that you guys are arguing.. I am. I just don't see how you come to the conclusion you do. If i have 2 items and one is double the other, one is also half the other.

Which one? The one with X amount?

[BrianD]

ok, here are two envelopes.. A and B. I know the values in them, and one is twice the other. Pick one.

Fine, we'll do it your way. I'll pick A.

[Huckleberry]

There are only two possibilities.

Either A is X and B is 2X or A is 2X and B is X.

My EV from picking one and staying with it is my chance of picking X multiplied by its value plus my chance of picking 2X multiplied by its value. 0.5*X + 0.5*2X = 1.5X

My EV from picking one and switching to the other is my chance of picking X multiplied by the value of the other envelope (I switched) plus my chance of picking 2X multiplied by the value of its corresponding other envelope (I switched). 0.5*2X + 0.5*X = 1.5X

[Julio Riddols]

But there is no x amount in the original question.. once you are given the value of one, the other can only be twice that or half that.. I don't see how there is any other way to look at it.

[Julio Riddols]

ok, a is 4, do you want to see if b is the lower or the higher of the two? dont forget if it is lower you only lose 2, but if it is higher, you gain 4.

[Buccaneer]

other can only be twice that or half that..

And which envelope would that be?

[KWhit]

B?

[Buccaneer]

B?

Or A?

[QuikSand]

That is true. But they are not necessarily equally probable, as you and others are assuming/insisting.

Why? 1/2= 50%. There are two choices, and you pick one of them. How is that not equally probable?

Okay. I'm honestly asking this just to see how far apart we are. You're really set on this 50/50 thing... so I'm wondering if you are even open to the concept that a two-option system could be anything but. Forgive me if you think this is trivial... but after your last several increasingly insistent examples, I'm not so sure.


Let's say that you have won $100 by playing a previous puzzle. And now I'm offering you a Monty Hall style deal. I have taken a 52-card deck, and written values on each card -- on the ace of spades, I wrote $200, and on every other card, I wrote $50. I then drew one card from that deck at random, and placed it into the envelope. So, my offer to you is this: if you give me the $100, you get the prize indicated by the card inside the envelope. Only two outcomes are possible here, $200 or $50.

Are you taking that bet?

Why would it be anything other than a 50/50 shot? The other envelope has to be either twice as much or half as much.

I think the guy who wrote that would take this bet. After all, there are two possible outcomes here, $200 (winning $100) and $50 (losing $50)... and losing $50 is less than gaining $100. Right?

...or are you at least conceivably open to the possibility that "two options" doesn't necessary mean "equal probabilities" yet?

[KWhit]

Okay, let's dial all the way back to the original two envelopes puzzle. Even before we open the envelope. We know it has some value inside, let's call it X, while the other one contains Y. If the entire universe of possible values is fair game, then it really shouldn't matter what number X is, correct? Nothing special about $100.

So, the other envelope either had X/2 or 2X in it, we'd agree. If we accept the argument (being made here) that this is now a 50/50 proposition, then we naturally conclude that the EV of envelope Y is 1.25X. Or, in essence, we *know* that it's better to switch from envelope X to envelope Y.

Now, we hold envelope Y. And we now know that whatever Y is, we know that X is either Y/2 or 2Y. And if you again assume this is a 50/50 proposition, then you are again at the clear conclusion that the EV of X is greater than Y, so you again *know* that it's better to switch from envelope Y to envelope X.

This puts you right back to the previous paragraph. Obviously, you are stuck square in the paradox. It simply cannot be that X is better than Y and Y is better than X. This is your classic reductio ad absurdum - if we make assumptions that lead us directly to an impossible result (like this one) then we need to go back and refine our original assumptions.

Now this I get.

[Huckleberry]

But there is no x amount in the original question.. once you are given the value of one, the other can only be twice that or half that.. I don't see how there is any other way to look at it.


The envelopes are filled with a certain amount before I pick. That amount doesn't change. That amount isn't dependent on which envelope I pick. What starts in A stays in A and what starts in B stays in B. Is all of this correct?

If so, then yes there is an X amount and a 2X amount. Just because you don't say the words "there is an X amount" doesn't mean it's not the case. There are two values in the envelope. One is twice the other. So one is X and one is 2X.

You are suggesting a scenario where I pick envelope A and then you flip a coin to determine how much you're going to put into envelope B.

In which case I would switch.

[Buccaneer]

It simply cannot be that X is better than Y and Y is better than X.

This is the answer.

[Julio Riddols]

QS, I see what youre saying with the monty hall deal, but there are basically 51 envelopes in that bet, with only one being worth 200. No, I wouldn't even think about that gamble.

But if you told me that there are two cards and one has 50 on it and one has 200 on it, I would.

[BrianD]

ok, a is 4, do you want to see if b is the lower or the higher of the two? dont forget if it is lower you only lose 2, but if it is higher, you gain 4.

Let me stop you there. I might be tempted to think that I've got a 50% chance of B being 2 and a 50% chance of B being 8. Take a look at B. Do I really have a 50% chance? Does the fact that I don't know if we are playing a 2-4 game or a 4-8 game mean that there is an equal chance of both? No, we are playing one or the other. The fact that I don't know which doesn't mean that we could be playing the other one.

[Buccaneer]

But if you told me that there are two cards and one has 50 on it and one has 200 on it, I would.

Of course but that's more information than you were given in the original puzzle.

[QuikSand]

QS, I see what youre saying with the monty hall deal, but there are basically 51 envelopes in that bet, with only one being worth 200. No, I wouldn't even think about that gamble.

But if you told me that there are two cards and one has 50 on it and one has 200 on it, I would.

Okay, so we seem to have cleared the hurdle that the only possible odds on two options is 50/50. We're making progress, I think.

Now... if we can just get you to stop building in the 50/50 assumption in to every argument you make. I honestly don't think there is anyone in this thread who can't follow the argument that a 50% chance at $200 is worth suffering a 50% chance of $50. The argument is not that you're wrong there... but that you are completely wrong in assuming that it is a 50/50 probability. It's been said multiple times, I'm just trying another phrasing.

[BrianD]

Let me stop you there. I might be tempted to think that I've got a 50% chance of B being 2 and a 50% chance of B being 8. Take a look at B. Do I really have a 50% chance? Does the fact that I don't know if we are playing a 2-4 game or a 4-8 game mean that there is an equal chance of both? No, we are playing one or the other. The fact that I don't know which doesn't mean that we could be playing the other one.

And just to move things along, I'll stick with envelope A.

[KWhit]

So what's really being said is that the $100 result in the original envelope is more likely to be the result of $50 x 2 as opposed to $200 / 2.

Not sure why that's the case, but that's the basic idea, right?

[Buccaneer]

And just to move things along, I'll stick with envelope A.

And I'll pick B. How is it that we are playing the same game?

[Julio Riddols]

And just to move things along, I'll stick with envelope A.

Envelope b was 8.

[Julio Riddols]

So what's really being said is that the $100 result in the original envelope is more likely to be the result of $50 x 2 as opposed to $200 / 2.

Not sure why that's the case, but that's the basic idea, right?

This is where I am sticking.

[BrianD]

Envelope b was 8.

OK, so was there ever a chance that envelope b was 2?

[QuikSand]

So what's really being said is that the $100 result in the original envelope is more likely to be the result of $50 x 2 as opposed to $200 / 2.

Not sure why that's the case, but that's the basic idea, right?

Yes, that is the basic idea.

But again... the real problem is that in the original version of this problem, the whole foundation of the puzzle lies on completely faulty math. The temptation (obviously) is to just skip over that fact and keep on working with it, ignoring that fact. But that's what leads, in substantial part, to all the wackiness. Infinities are tricky little bastards to deal with.


There's a familiar old math trick where you start out with something like A=B and from there go on to prove, step by step, that 1=2. And it's a tightly constructed proof, it follows perfectly rational rules throughout, any 9th grade algebra student (well, an A student at least) could tell you so. So, why doesn't it prove to us that 1 does indeed equal 2? Because it has a cleverly hidden division-by-zero inside, where something is being divided by (a-b). Division by zero is undefined, it doesn't exist, it renders the rest of the discussion pointless. It doesn't matter that the rest of the proof is nicely construted and follows rules perfectly -- it has an illegal function in it, it is not valid, period.

This is, in many ways, the same thing. We are tempted to smooth over the impossible distribution function described and continue on with our "analysis" and we end up with all manner of conclusions. But the real answer is simply that "this doesn't work, because the underlying math doesn't work." Once you put bounds on the values involved, it becomes a bit simpler to think about, but it's still (clearly) tough for some folks to shake the familiar concept of nice, even distributions of random numbers...which this absolutely is not.

[BrianD]

dola,

or to put it another way, what were my odds before I picked an envelope?

[Julio Riddols]

OK, so was there ever a chance that envelope b was 2?

no, and I see that.. But if one had to be double the other, I would think it would be a good bet to gamble on that.

[Julio Riddols]

dola,

or to put it another way, what were my odds before I picked an envelope?

I was thinking this earlier, trying to get what you guys are saying.. I mean, honestly, I trust that what you are saying is right, because you wouldn't argue the point so hard if it wasn't.. I am only trying to understand how you are right.

The chances of you picking the lower envelope first are 50/50 out of 2 envelopes, and it is also 50/50 that you would pick the higher one of the two.. That means the second envelope is just as likely to be double as it is to be half.

[Huckleberry]

No. It can never be half. There was a 4 in one envelope and an 8 in the other.

It can never be 2.

[QuikSand]

The chances of you picking the lower envelope first are 50/50 out of 2 envelopes, and it is also 50/50 that you would pick the higher one of the two.. That means the second envelope is just as likely to be double as it is to be half.

So, (since you never introduced a known dollar value here, Ii won't either) how do you ever escape your loop, then? If the other envelope is a 50/50 shot to be twice as big... then you definitely want to switch, right?

Except that once you switch, it's still true that the other envelope (your original one) has a 50% chance of being twice as big, by your logic. So, do you gain again by switching back to your original envelope? Your logic insists this is true, since a 50/50 chance means no matter what envelope you hold, you are better off switching.

[Julio Riddols]

No. It can never be half. There was a 4 in one envelope and an 8 in the other.

It can never be 2.

What I am saying though, is that as the one picking the envelopes, you don't know if you got the higher one first or not.

After picking envelope A, not knowing whether you got the higher or lower envelope, a chance at +4 is still a good gamble to take if the worst you could do is -2.

[MJ4H]

Unless, say, you get the +4 1 in 5,000 times and the rest of the time you get -2. Just for instance. Saying it is a good gamble does'nt make it one. It isn't in this case. In the envelopes case, it is a break even one.

[QuikSand]

I think it's likely time to simply declare an impasse. There's information about this puzzle on the web in various places, and while a lot of it gets into too-deep math, not all of it does. For anyone who wants to really dig more into this, have at it. I've done my best to try to put things into English in different ways, but this is just too slippery a puzzle to feel like the end result is going to be a whole thread full of contented and satisfied solvers. Just not gonna happen.

[BrianD]

I was thinking this earlier, trying to get what you guys are saying.. I mean, honestly, I trust that what you are saying is right, because you wouldn't argue the point so hard if it wasn't.. I am only trying to understand how you are right.

The chances of you picking the lower envelope first are 50/50 out of 2 envelopes, and it is also 50/50 that you would pick the higher one of the two.. That means the second envelope is just as likely to be double as it is to be half.

Right, but the point is that I have a 50% chance of picking the 4 and a 50% chance of picking the 8. I have a 0% chance of picking the 2 or the 16 because they don't exist. Once I picked the 4, that didn't change my 0% chance of picking the 2 or the 16. They were never options even thought I didn't know that.

[Julio Riddols]

So, (since you never introduced a known dollar value here, Ii won't either) how do you ever escape your loop, then? If the other envelope is a 50/50 shot to be twice as big... then you definitely want to switch, right?

Except that once you switch, it's still true that the other envelope (your original one) has a 50% chance of being twice as big, by your logic. So, do you gain again by switching back to your original envelope? Your logic insists this is true, since a 50/50 chance means no matter what envelope you hold, you are better off switching.

Yes, true. But as a one time gamble, knowing the value of the first envelope could either be halved (the first envelope was higher) or doubled (the first envelope was lower) the possible gain would be worth the gamble against the possible loss.

I understand that there are only two numbers involved here, but as the one who chose the envelope, you have no idea which of those two numbers you got.

[QuikSand]

Yes, true. But as a one time gamble, knowing the value of the first envelope could either be halved (the first envelope was higher) or doubled (the first envelope was lower) the possible gain would be worth the gamble against the possible loss.

I understand that there are only two numbers involved here, but as the one who chose the envelope, you have no idea which of those two numbers you got.

Okay, I'll try again.

So, you switch. You give back envelope X, and take instead envelope Y. It's now envelope Y that is in your hot little hands.

Now you have Y instead of X. And now I offer you a chance to switch back to X. By the math you continue to insist is true, isn't it *still* a 50/50 chance that X is either twice Y or half Y... meaning that you have to switch again?

Are you following my question here?

[BrianD]

Oh how about this. Let's use your 4-8 example. We'll play 100 times. 50 times you get the 4 and 50 times you get the 8. By your rules you should switch every time. This means you still get the 4 50 times and the 8 50 times. The end result is that you average 6 each time.

[Julio Riddols]

I don't mean to disappoint anyone here.. I have actually been really enjoying this. I like challenging my brain..

I can see where you're saying it is impossible to have wound up with 2 since 2 wasn't part of the possible number set in the first place. I see that. That we can surely agree on. I'm not saying that it is a bad idea to keep either envelope. Either one could be the higher one. All I am saying is that you are given the information that one envelope is double the other, nothing more. you get to see the value of the first envelope, and then you know the second envelope will either cut that in half or double it. My logic here is, since you aren't told which envelope it is that you picked first, the second is just as likely to be double that as it is to be half, in the eyes of the one who is choosing. Where I find it a good gamble is one way you lose 50%, the other you gain 100%.

Now as the one giving the choice, yeah, you know which envelope the chooser has. You know what the second envelope holds. Of course you know how futile it is for the other person to gamble if you know the second envelope is the lower one. But he doesn't know that, he wasn't given that info. He has to assume he has one or the other, and it is worth a shot to find out.

[QuikSand]

Julio, I'll try the same question another way.

Say you're playing this game. You are in a room, and you choose envelope X. And we give the other envelope Y to Riddick Fonseca in the other room.

Since you know that it's a 50/50 shot to either gain 100% or lose 50%, you say you'd rather switch -- that the expected value of his envelope is more than that of your envelope. So, you're better off with Y than with X.

Meanwhile, The Riddick has just done the same calculation, and reached the same conclusion -- he's completely convinced that *your* envelope is the better one, since its expected value is 1.25X that of his envelope. So he wants to switch, too. so, he's better off with X than with Y.

Are you feeling at all uncomfortable with this?

[Julio Riddols]

Oh how about this. Let's use your 4-8 example. We'll play 100 times. 50 times you get the 4 and 50 times you get the 8. By your rules you should switch every time. This means you still get the 4 50 times and the 8 50 times. The end result is that you average 6 each time.

That goes a long way to explaining your side.. I get that. Over a series of equal chances between 4 and 8, you're probably just going to break even. But in a one time gamble, which is all I have been trying to say, it is worth a shot to possibly double your envelope against the possibility that you halve it.

I really think we're seeing eye to eye here, just that you're thinking in the broad terms of doing this over and over, in which case it is futile to keep switching envelopes. But if you do it just once, gaining 100% is something I will put against losing 50%.

I honestly wouldn't be surprised if the last page of this thread was pointless and that I understood your side the whole time.

[QuikSand]

I honestly wouldn't be surprised if the last page of this thread was pointless and that I understood your side the whole time.

In my view this side conversation about 2,4, and 8 has not advanced anyone's understanding of the puzzle at all.

[Buccaneer]

How can X be greater than Y and Y be greater than X?

[BrianD]

There is the fallacy though. The second one isn't as likely to be double as it is half. You don't know what range you are playing in so you assume it can be either way. This is wrong. The proper way to look at it is that after your choice you have a 50/50 chance of holding the bigger envelope. After the switch you have a 50/50 chance of holding the bigger envelope. Knowing what is in one envelope doesn't change the possibilities. The trick is that by not giving you enough information to properly assign probability, you guess at the probability.

Always swapping will make you feel like you are giving yourself the best odds, but still never get better than the 50/50 which means you'll end up averaging between the high and low amount.

[Buccaneer]

Just advancing the A or B choice.

[BrianD]

In my view this side conversation about 2,4, and 8 has not advanced anyone's understanding of the puzzle at all.

I was trying to go at your division by 0 explanation from another direction. In the initial setup, the odds of me picking the envelope worth 2 was 0%. Picking the 4 envelope first doesn't magically create a potential for the other envelope to be a 2. 0% isn't just a little bit less than 1%. 0% means it is an impossibility. It seemed worth a shot.

[Julio Riddols]

Im not saying it gives you the best odds at all, its clearly a win or lose.. but if you win, the gain outweighs the loss. if I pick 4 and go up to 8, I gain.. WAIT.

Light came on. I see. I gain 4 or lose 4 either way, since there are only 2 possible values. I get it now. I really apologize for making it that damn hard to explain, but I think that gets it for me. Is that where you guys are at?

[QuikSand]

Is that where you guys are at?

Nope.

[Julio Riddols]

Shit.

[Julio Riddols]

I can see how you could go for infinity choosing whether to trade envelopes or not trade them and it is pointless to think that going one way would be better than the other..

[Julio Riddols]

I hope no one is losing sleep over this.. I'm very worried about trying to gameplan for a meeting against any of you guys in FOF MP though. :)

[gkb]

Let's say that you have won $100 by playing a previous puzzle. And now I'm offering you a Monty Hall style deal. I have taken a 52-card deck, and written values on each card -- on the ace of spades, I wrote $200, and on every other card, I wrote $50. I then drew one card from that deck at random, and placed it into the envelope. So, my offer to you is this: if you give me the $100, you get the prize indicated by the card inside the envelope. Only two outcomes are possible here, $200 or $50.

Are you taking that bet?

...or are you at least conceivably open to the possibility that "two options" doesn't necessary mean "equal probabilities" yet?

I thought this was a really good example. I'm not claiming to fully understand this yet, but I believe this got me much closer.

[Shkspr]

I can hear QS banging his head against the wall all the way here in Colorado.

:)

Actually, I already gave it my best shot.

I think the best way I can frame it is that:

Wrong, I'm afraid.

Let me try another angle,

I think I am running out of ways to explain why this (underlined above) is not so...

And that's where the troubles start.

So, we can just skip it and move on, then. Done.

Okay, I will try one more angle.

Getting anywhere?

That's fine... but

Okay, let's dial all the way back to the original two envelopes puzzle.

Okay, you're hammering on this post again and again. I don't even know what you're saying here.

Wow, our semantic problems are greater than I had imagined.

Okay. I'm honestly asking this just to see how far apart we are.

We're making progress, I think.

I think it's likely time to simply declare an impasse.

I've done my best to try to put things into English in different ways,

Just not gonna happen.

Okay, I'll try again.

I'll try the same question another way.

Are you feeling at all uncomfortable with this?

In my view this side conversation about 2,4, and 8 has not advanced anyone's understanding of the puzzle at all.

Nope.

[st.cronin]

Okay, I'll try again.

So, you switch. You give back envelope X, and take instead envelope Y. It's now envelope Y that is in your hot little hands.

Now you have Y instead of X. And now I offer you a chance to switch back to X. By the math you continue to insist is true, isn't it *still* a 50/50 chance that X is either twice Y or half Y... meaning that you have to switch again?

Are you following my question here?

I reject this analysis, because once you open Y, both X and Y are KNOWN. There is no decision to be made. Honestly, I think people are intentionally understanding the problem in a way that permits this distribution tangent, because there is just no way that I can see it applying to this problem. I still think the way I tried to model the problem is the correct way:

There are 2 envelopes, x and 2x. I give you one of them. This is the only point in the problem where probability needs to be considered - it is a fact that I have given you either the higher one or the lower one, but there is no reason to believe that either is more or less likely. If you assume an equal chance, there is no reason to switch, because in a thousand trials you will end up with the same dollar value in the envelope I gave you, as in the envelope I held back. In a thousand trials the envelope I gave you will have something close to 500(x) + 500(2x), and the envelope I held back will have something close to 500(2x) + 500(x). I still don't understand what's wrong with that way of looking at it.

[QuikSand]

st.cronin, the diminishing returns of this debate are adding up. I get it, you don't buy the analysis. Really, that's fine.

The selection you quote above, deliberately and obviously using X and Y rather than dollar values, was discussing the setup where you don't open the envelopes and their value remains unknown. Give me a break... do you honestly believe that I'm envisioning a "debate" about whether an opened envelope with $100 and an opened envelope with $200 are up for grabs here? Spare me. Anyone who has suffered through all the torture of this thread so far has earned at least the respect of not being a complete idiot. Please grant me that, and use a few context clues.

By your analysis, if you leave the envelopes *unopened* you are in an absurd situation of *always* improving your situation by switching from X to Y, and then back to X, and then back to Y, which is simply evidence of a flaw in your logic. And even when you open one envelope, by your still-faulty analysis, you still have the absurd situation of preferring the other envelope regardless of what dollar value was in the one you opened. This is all a function of the mathematical thin ice that the puzzle is based on, but if you insist on just skipping past that fact, then you're never going to concede anything, and you will presumably comfortably rest in your infinite loop of logical absurdity forever. And I can live with that, to be honest.

[gstelmack]

I reject this analysis, because once you open Y, both X and Y are KNOWN. There is no decision to be made. Honestly, I think people are intentionally understanding the problem in a way that permits this distribution tangent, because there is just no way that I can see it applying to this problem. I still think the way I tried to model the problem is the correct way:

There are 2 envelopes, x and 2x. I give you one of them. This is the only point in the problem where probability needs to be considered - it is a fact that I have given you either the higher one or the lower one, but there is no reason to believe that either is more or less likely. If you assume an equal chance, there is no reason to switch, because in a thousand trials you will end up with the same dollar value in the envelope I gave you, as in the envelope I held back. In a thousand trials the envelope I gave you will have something close to 500(x) + 500(2x), and the envelope I held back will have something close to 500(2x) + 500(x). I still don't understand what's wrong with that way of looking at it.

The problem you have is that x is more likely to be a smaller amount than a larger one. This all comes back to:

50/100 is a more likely pairing then 100/200. Don't focus on the after opening bit, focus on this. Once you understand it or accept it, then it's easy to make the leap to understanding that once you know you have 100, it is more likely that the other envelope holds 50 than that it holds 200.

[st.cronin]

50/100 is a more likely pairing then 100/200.

This is true if you are using a random function to generate the values. Why in the world anybody would think that's the case is beyond me.

[QuikSand]

This is true if you are using a random function to generate the values. Why in the world anybody would think that's the case is beyond me.

Hey, there might be a glimmer of progress there.

What do you think is the case? What supports your ironclad assertion that the underlying distrubution method results in a 50/50 chance of the two remaining options? Other than your just preferring it that way, that is.

[st.cronin]

By your analysis, if you leave the envelopes *unopened* you are in an absurd situation of *always* improving your situation by switching from X to Y, and then back to X, and then back to Y, which is simply evidence of a flaw in your logic. And even when you open one envelope, by your still-faulty analysis, you still have the absurd situation of preferring the other envelope regardless of what dollar value was in the one you opened. This is all a function of the mathematical thin ice that the puzzle is based on, but if you insist on just skipping past that fact, then you're never going to concede anything, and you will presumably comfortably rest in your infinite loop of logical absurdity forever. And I can live with that, to be honest.

I thought you had explained why the EV analysis was wrong. Are you really saying this is incorrect math?

If you assume an equal chance, there is no reason to switch, because in a thousand trials the envelope I gave you will have something close to 500(x) + 500(2x), and the envelope I held back will have something close to 500(2x) + 500(x). I still don't understand what's wrong with that way of looking at it.

[QuikSand]

Are you really saying this is incorrect math?

Your math is fine, if you are correct that the two options are equally likely, and that in 1000 trials you'd expect 500 of each.

About 30 times in this thread, it has been pointed out (by me and several others) that this is a false underlying assumption.

And since you prefer not to hear that argument, you don't.

As nearly as I can tell, that's where we are.

[st.cronin]

Hey, there might be a glimmer of progress there.

What do you think is the case? What supports your ironclad assertion that the underlying distrubution method results in a 50/50 chance of the two remaining options? Other than your just preferring it that way, that is.

I have never insisted that there is a 50/50 chance. What I am saying is that if you assume a 50/50 chance, there is no reason to switch or not switch. There is nothing in the puzzle to assume any probabilistic function for whether the opened envelope is x or 2x. It could be that the puzzle master always hands out 2x. It could be that he hands out 2x to women, x to men.

[QuikSand]

Blessedly, I'm leaving soon for a meeting (heh, meeting with some actuaries, no joke) and will be away a while. I think if I continue to have the urge to participate in this thread, I will instead commit some ban-worthy offense here and get myself booted from the forum.

[BrianD]

There are 2 envelopes, x and 2x. I give you one of them. This is the only point in the problem where probability needs to be considered - it is a fact that I have given you either the higher one or the lower one, but there is no reason to believe that either is more or less likely. If you assume an equal chance, there is no reason to switch, because in a thousand trials you will end up with the same dollar value in the envelope I gave you, as in the envelope I held back. In a thousand trials the envelope I gave you will have something close to 500(x) + 500(2x), and the envelope I held back will have something close to 500(2x) + 500(x). I still don't understand what's wrong with that way of looking at it.

Is this the point you were trying to make last night? If so, I didn't understand what you were going for.

I think this has to be the right way to look at it. The end result of the envelope game is that it doesn't matter what you do when given the option to switch. You have a 50% chance of getting the higher amount initially and a 50% chance of having it after a switch. There are only two options, the higher amount or the lower amount. Assuming two options which don't include the known open envelope is where people have been making mistakes and why they think changing is better.

Unless I am missing something, I think your way of looking at it is fine...or at least gives what I believe to be the right result.

[Maple Leafs]

Over a series of equal chances between 4 and 8, you're probably just going to break even. But in a one time gamble, which is all I have been trying to say, it is worth a shot to possibly double your envelope against the possibility that you halve it.
I think this is where the problem is.

In terms of probability, there is no difference between the odds over a series of chance and the odds for a "one time" gamble. They're indistinguishable.

Let's say I offer a bet. We roll one die. If it comes up a 6, I give you $100. If it comes up 1 - 5, you give me $25.

Obviously, it's a bad bet for you. The longer we play, the more I'll win. However, as a one-time gamble you stand to win more than you lose. That doesn't make it a good bet, though.

If a gamble is a long-term bad idea, then it's a bad idea in the sort term too.

[st.cronin]

Blessedly, I'm leaving soon for a meeting (heh, meeting with some actuaries, no joke) and will be away a while. I think if I continue to have the urge to participate in this thread, I will instead commit some ban-worthy offense here and get myself booted from the forum.

I confess, this has actually been my goal all along. :D

[QuikSand]

There are 2 envelopes, x and 2x. I give you one of them. This is the only point in the problem where probability needs to be considered - it is a fact that I have given you either the higher one or the lower one, but there is no reason to believe that either is more or less likely. If you assume an equal chance, there is no reason to switch, because in a thousand trials you will end up with the same dollar value in the envelope I gave you, as in the envelope I held back. In a thousand trials the envelope I gave you will have something close to 500(x) + 500(2x), and the envelope I held back will have something close to 500(2x) + 500(x). I still don't understand what's wrong with that way of looking at it.

Actually, I think you are correct in your approach here. Sorry for the short shrift earlier, I was misreading. I perhaps am suffering form melding together the many different confused approaches being tossed around in this thread.

We have had a ton of 50/50 enthusiasm here, and I accidentally mistook your post above for more of the same. But your argument here actually looks fine, sorry for my misreading.

[Julio Riddols]

I've given up on the notion that an envelope is more likely to hold 50 dollars than it is to hold 200. I have no idea how that can be true.

I've realized it is futile to switch envelopes after picking the first one. I say take the money and run, and lets be done with it. What is trying to be proven here is trivial at best and will never matter in my life. (or at least I hope not. I'll just leave both envelopes and walk away if that is the case.)

Its clear to me that no amount of explaining how one envelope is more likely to contain 50 than it is to contain 200 is going to help me understand how that is even remotely true, given the question that was originally asked. So, to save the trouble for you guys, I am going to let go here. I'd rather not wind up angry over something so minute.

[st.cronin]

QS, in all seriousness, thank you for your participation in this thread. Its been rewarding for me, at least.

[rkmsuf]

I've given up on the notion that an envelope is more likely to hold 50 dollars than it is to hold 200. I have no idea how that can be true.

I've realized it is futile to switch envelopes after picking the first one. I say take the money and run, and lets be done with it. What is trying to be proven here is trivial at best and will never matter in my life. (or at least I hope not. I'll just leave both envelopes and walk away if that is the case.)

Its clear to me that no amount of explaining how one envelope is more likely to contain 50 than it is to contain 200 is going to help me understand how that is even remotely true, given the question that was originally asked. So, to save the trouble for you guys, I am going to let go here. I'd rather not wind up angry over something so minute.



"The only winning move is not to play."

[Julio Riddols]

Actually, I think you are correct in your approach here. Sorry for the short shrift earlier, I was misreading. I perhaps am suffering form melding together the many different confused approaches being tossed around in this thread.

We have had a ton of 50/50 enthusiasm here, and I accidentally mistook your post above for more of the same. But your argument here actually looks fine, sorry for my misreading.

This is exactly what I was trying to explain toward the end of last night, but obviously didn't have the words to do it. I see why it would be pointless to switch envelopes, and the rest of what St. Cronin said is where I have been the whole time. My ability to explain myself was obviously crippling to my argument.

[Julio Riddols]

QS, in all seriousness, thank you for your participation in this thread. Its been rewarding for me, at least.

Agreed. Despite the frustration of it all, I think some progress was made.

[BrianD]

The problem you have is that x is more likely to be a smaller amount than a larger one. This all comes back to:

50/100 is a more likely pairing then 100/200. Don't focus on the after opening bit, focus on this. Once you understand it or accept it, then it's easy to make the leap to understanding that once you know you have 100, it is more likely that the other envelope holds 50 than that it holds 200.

I hate to do this since we seem to have reached a consensus, but can I challenge this assertion? If we are not given the bounds of the problem, can we really say what any probability is to say what is more likely?

[BrianD]

dola

And since this is turning into a love-fest...I've missed these problems. I like a chance to exercise the brain on math/logic problems.

[Huckleberry]

I actually confused myself last night trying to come up with examples.

So if it doesn't bother anyone, I'd like some help working through this example.

I set up a game based on this problem where the lower value can be anywhere from $0-$1,000. I apply a random distribution to this range. So all values in that range are equally likely to be selected as the lower value. The higher value, obviously, is then set to be two times the lower value. So the range of higher values is $0-$2,000 with half the probability density of the lower values but the same number of values.

Now let's play the game.

The lower value is randomly selected, the higher value is calculated, and the two envelopes are filled. The identical envelopes are placed on a table in front of me and I pick one and open it. Now here are the two different possible results that need to be analyzed as I see it:

1.) I open the envelope and find any value greater than $1,000.

Well clearly I have to stay with my original selection. The value I have selected is not within the boundaries defining the possible lower values. Therefore it is necessarily the higher of the two values I'm dealing with. In EV terms, EV staying = X, EV switching is P(2X)*(2X) + P(X/2)*(X/2) = 0*2X + 1*X/2 = X/2.

2.) I open the envelope and find any value equal to or less than $1,000.

I think we can all agree this is the option that we are talking about. Assuming, of course, that you have come to understand that we have to define a range of possible values or else the discussion is meaningless both logically and mathematically.

So my question is what is the proof that my value of X less than or equal to $1,000 is twice as likely to be a lower value than a higher value? I intuitively (there's that word again) understand that the density of lower values is twice the density of higher values which implies this is true. But is there somewhere I can slog through equations that prove this?

The reason I ask is that I used Excel (I know, don't say it) to run this simulation. I'm using Excel 2007 so I just dragged the RAND function, CEILING(RAND*1000,1), and 2*CEILING(RAND*1000,1) down to the bottom of the sheet. Deleted the bottom row so I could have a title row, so the bottom line is that I ran the simulation 1,048,575 times.

For my example I selected $500 as the value less than or equal to $1,000. In my simulation the value of $500 appeared 2,088 times. It appeared 1,039 times as the lower value and 1,049 times as the higher value.

This doesn't seem right.

Someone explain.

It appears that once we define boundaries then if we open an envelope containing a value greater than the upper limit on the lower value we should stay (obviously and with 100% certainty our EV of staying is double or EV of switching) while if we open an envelope with less than or equal to the upper limit on the lower value we should switch.

I got 524,731 total values greater than $1,000 (lower and higher values combined). I got 1,572,419 total values less than or equal to $1,000 (once again combined). I'm sure that works into this somehow.

Hmm...talking myself through it it looks like I need to combine the probability that I'm 3 times more likely to select a value less than or equal to the upper limit on the lower value than I am to select one greater.

So my overall EV of switching looking at the problem before the value is known is probably going to work out to be the same as staying, no?

[Julio Riddols]

dola

And since this is turning into a love-fest...I've missed these problems. I like a chance to exercise the brain on math/logic problems.

I'm thinking next time I will take some brain-gain 4000 before stepping into one of these, so I can get this brain buff as hell.. Start smacking mental home runs at a record setting pace.. Get questioned about possible brain 'roids.. :p

[BrianD]

For my example I selected $500 as the value less than or equal to $1,000. In my simulation the value of $500 appeared 2,088 times. It appeared 1,039 times as the lower value and 1,049 times as the higher value.

This doesn't seem right.

Someone explain.


I'm going to pull just this piece out and say that it should seem right. If you are generating random numbers between 0 and 1000, every number has an equal probability of coming up. This means that 250 and 500 have an equal probability. Because every x=250 comes with a 2x=500, you should have a nearly identical occurrence of 500 being x and 2x.

[BrianD]

I'm thinking next time I will take some brain-gain 4000 before stepping into one of these, so I can get this brain buff as hell.. Start smacking mental home runs at a record setting pace.. Get questioned about possible brain 'roids.. :p

I need to do the same as I tend to get my ass kicked in these threads.

[Maple Leafs]

Assuming, of course, that you have come to understand that we have to define a range of possible values or else the discussion is meaningless both logically and mathematically.
I think this is the key.

The original puzzle has no range of values. That's why it turns into an apparent paradox.

[Huckleberry]

Okay, I'm going to approach this from a slightly different angle. I almost hate to do this, but not enough to stop me from doing it anyway.

1.) First of all, it is still true that the original problem is unbounded and nonsensical in a mathematical sense. That hasn't changed. However...

2.) I am going to ignore that and discuss the probabilities only from the single case example perspective that st.cronin and Julio Riddols are focusing on. Their contention, and the seemingly obvious answer, is that the EV formula is based on the other envelope being 50% likely to be 2X and 50% likely to be X/2 while our known value is locked in as the X value. This is the problem - the assumption that our known value is X. 50/50 probabilities still apply, however the assignment of the variables is incorrect.

Our known value is not locked in as X. Our known value is only known to be either the X or the 2X in the envelopes. Our other envelope is known to be the 2X if our known is X and the X if our known is 2X. Opening one of the envelopes and seeing its value in this case actually adds no new useful information. The EV formulas are accurately written as follows:

EV(H) - EV of Holding
EV(S) - EV of Switching
V(I) - Value of Initial choice
V(O) - Value of Other envelope
P(IX) - Probability that Initial choice is X
P(I2X) - Probability that Initial choice is 2X
P(OX) - Probability that Other envelope is X
P(O2X) - Probability that Other envelope is 2X

EV of Holding your initial choice

EV(H) = V(I)

V(I) in terms of X

V(I) = P(IX)*X + P(I2X)*2X = .5*X + .5*2X = 1.5X

EV of Switching

EV(S) = P(OX)*X + P(O2X)*2X = .5*X + .5*2X = 1.5X

Therefore EV(S) = EV(H)

Q.E.f'nD.

[st.cronin]

I almost hate to do this.

Clearly. :D

[MJ4H]

Hmm. Huckleberry, that looks pretty darn good.

[MikeVic]

This thread is confusing.

[Fidatelo]

I think the odds of it confusing someone are 50/50, but I'm pretty certain QuickSand will disagree.

[Huckleberry]

Hmm. Huckleberry, that looks pretty darn good.

I know, that's what scares me. The proof that 1 = 2 looks pretty good at first, too, if you're not careful. :D

I don't know why, but running through some simulations made me think of this approach. I looked at random values in the two envelopes, and as soon as you assign actual values in the envelopes it becomes obvious.

Let's say the first trial there's $100 in one envelope and $200 in the other envelope. Well, EV(H) = P(100)*V(100) + P(200)*V(200) = .5*$100 + .5*$200 = $150. Nobody should dispute this. And EV(S) = P(100)*V(200) + P(200)*V(100) = .5*$200 + .5*$100 = $150. Again, this shouldn't be disputed. So EV(H) = EV(S).

Anyway, I laid out a boatload of random trials in Excel then added a column for each outcome (picking A and holding, picking A and switching, picking B and holding, picking B and switching). Obviously the sum of all values realized by holding equaled the sum of all values realized by switching. At that point I was just going to add the table to this thread to show that the answer obviously has to be that switching holds no benefit. Then I was looking at the table and realized that the fact that we don't know whether the selected value is X or 2X could be used to restate the EV formulas.

[Huckleberry]

Apparently I like my solution more than anyone else but there's another good thing about it. Based on the above EV calculations, we define our EV of 150 as 1.5X - this seems odd at first because we are working in a situation where 150 is either X or 2X.

But defining 150, our known value, as 1.5X equates the situation where we have opened an envelope and seen $150 with the situation where $100 and $200 are in the envelopes and we haven't selected one yet. And this is true. Our EV in that situation is the same as our EV after selecting $150. So my phrasing earlier about the selection giving us no new information is not really true. The situation where we have opened an envelope with known value X is equivalent to the situation where we have two envelopes with known values 2X/3 and
4X/3.

Weird but true.

I think.