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]

Originally Posted by 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.

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]

Spoiler

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]

Originally Posted by Passacaglia

Blah blah blah blah math.

[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]

Originally Posted by Maple Leafs

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]

Originally Posted by st.cronin

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]

Originally Posted by st.cronin

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]

Originally Posted by 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?

maybe

[MJ4H]

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

[QuikSand]

Originally Posted by 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?

Oh boy.

[QuikSand]

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

Originally Posted by MattJones4Heisman

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]

Originally Posted by QuikSand

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]

Originally Posted by st.cronin

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

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

[Maple Leafs]

Originally Posted by 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.

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]

Originally Posted by Maple Leafs

(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]

Originally Posted by Maple Leafs

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...

Originally Posted by MattJones4Heisman

... 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

[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]

Originally Posted by QuikSand

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_...x_solution.htm

[QuikSand]

Originally Posted by MattJones4Heisman

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.

Originally Posted by QuikSand

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]

Originally Posted by QuikSand

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]

Originally Posted by MattJones4Heisman

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]

Originally Posted by QuikSand

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]

Originally Posted by QuikSand

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]

Originally Posted by KWhit

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?

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.

[QuikSand]

Originally Posted by KWhit

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]

Originally Posted by QuikSand

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]

Originally Posted by KWhit

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.