Probability puzzle

[BrianD]

We've talked about the Monty Hall problem before, but I am unsure if the problem I have in mind is along the same lines.

My problem deals with the game Deal or No Deal and getting down to the final 2 cases. For those who are unfamiliar with the game, there are 25 dollar values ranging from $.01 to $1,000,000 placed in 25 suitcases. A contestant picks one case to hold on to and then proceeds to open the remaining cases to see what dollar amounts are in them. At various points in the game the contestant is given a money offer based on the dollar amount still undiscovered to walk away and end the game. If the contestant refuses all offers and gets down to 2 cases, the contestant is given the choice of keeping their case and winning the dollar amount inside it, or trading for the last remaining case and winning the dollar amount inside it.

My question is this. After picking 1 of the 25 cases to hold on to and then opening 23 other cases, there are two dollar amounts still in play. The $.01 and the $1,000,000. You have already decided not to take any bribes along the way, so you have to decide to stay with your original case, or trade for the only other unseen case. What should you do, and why?

Note: for the purposes of this puzzle, it doesn't really matter how much is in the smaller case. You want the $1,000,000, so what is the best way to go after it?

[QuikSand]

Pretty sure this is correct:

The difference between this setup and the classic Monty hall setup is that Monty opens doors with knowledge, and in your setup, the briefcases are being opened randomly. With your setup, the chances of the $1m prize being in either of the remaining two cases is 50%. So, do whatever you want.

[BrianD]

I came up with a different answer, but I don't know if it is correct. I'll see if anyone other discussion develops before I share my thoughts.

[Celeval]

I agree with QS - to look at it a different way, you're choosing two briefcases... the one you originally selected, and the last one open. There's a 4% chance of ithe $1m being in the one you selected, but there's only a 4% chance that the $1million is in the other, and you've already hit one of those. So there's a 50/50 chance of it being in either one.

[Marc Vaughan]

I agree with the right honourable QS.

[Passacaglia]

I'm with QS, too. Just post it, Brian, so we can start the flaming!

[gstelmack]

Bah, by that point you should already have walked away with the $300K or so offer you got a few suitcases back...

[BrianD]

OK, I am going to nominate my thread for crappiest puzzle thread of all time. Before posting the thread I did a bunch of math to see if the answer would be 50/50. Since it didn't come out to be 50/50 I posted the puzzle. After QS gave his answer, I went back and checked my math and discovered a mistake. Upon correcting the mistake, I see that it is 50/50. Or more accurately, there is a 4% chance of picking the right case initially with a following 100% chance of opening 23 cases without finding the $1,000,000. There is a 96% chance of not picking the right case initially with a following 4.16% chance of opening 23 cases without finding the $1,000,000. Multiply the two probabilities and you get a 4% chance of getting to that situation without having the right case.

[albionmoonlight]

Well, we can't let this die this quickly. Let's see if someone will bite on this:

You should not trade. There were two possibilites to start--you either had the million dollars or you did not. If you did not have it, then there is a great chance that it would have turned up when he was opening the other cases. But it did not turn up. It is almost impossible to open 23 of 24 cases and not turn up $1,000,000 if the $1,000,000 was on the board.

Therefore, it is very, very likely that you have the $1,000,000 (otherwise it would have turned up already). So you should not trade unless you are really really dumb.

How's that?

[albionmoonlight]

dola--

maybe the real puzzle will be to predict who will bite on my answer and start defending it.

[Anthony]

i think the better puzzle question would be:

at what point does it make sense to take a deal for (for example) $300K earlier on than to actually try to win the $1million and walk away with potentially $0?

i know i wouldn't ever get as far as the last 2 cases, i would take the guaranteed lesser amount and forego a non-guaranteed larger amount.

[Alan T]

i think the better puzzle question would be:

at what point does it make sense to take a deal for (for example) $300K earlier on than to actually try to win the $1million and walk away with potentially $0?

i know i wouldn't ever get as far as the last 2 cases, i would take the guaranteed lesser amount and forego a non-guaranteed larger amount.


I am guessing there is some mathematical proof for figuring this out, and I assume the show creator uses that to determine how much to actually offer each time as well.

I don't really watch this show, but the few times I did, my personal thought was that you should decide to yourself at what point does the money mean a better life for you or make things substantially easier? For me, I would assume at any point I got an offer close enough to paying off my mortgage, I should jump at it. If you go beyond that, you're being greedy and reckless I would think.

[Anthony]

i would walk away if they gave me a deal of $250K or more. anything over $200K and you walk away with at least $100K aftet taxes.

[QuikSand]

I am guessing there is some mathematical proof for figuring this out, and I assume the show creator uses that to determine how much to actually offer each time as well.

There's another factor here -- how much does one value an additional dollar? If you think it through, there's honestly a diminishing return even for something that we tend to think of as having pretty constant value, like money. If your *first* dollar is going toward absolute essentials, it's worth more to you than the *millionth* dollar that may be going toward fuel for the speedboat.

Anyhow... there has been an awful lot of research in to trying to estimate this sort of function -- but since this sort of calculation has much more to do with personal preferences (risk acceptance, amount of declining value of added dollars) there's never going to be one correct answer to a question like this.

[Anthony]

i would also assume that for casting purposes they try to land people who claim would actually want to play for the million dollars. do you think the show would be much fun (i've watched a couple episodes of it, it gets rather repetitive after a bit for me to watch it with any regularity) if every contestant who went on decided to walk away at the 1st good deal? that wouldn't make for good programming. i'm thinking they look for a certain profile - have a mortgage, has children, etc - something that would entice someone to be greedy in an instance where they normally wouldn't be greedy. i know i would walk away rather early cuz of the old "bird in the hand" logic, but someone weaker willed than i would do the math in their head and say "i've already won enough to pay off the mortgage - if i stick around a bit more i can afford one kid's college tuition" and after that it's "i can afford college for *all* my kids if i play this right, i'm so fucking close right now". you gotta think they want to cast people who really got something to play for rather than myself who's looking at this as an opportunity to make some quick money.

[Maple Leafs]

From a pure math standpoint, the offers from the bank will almost always be poor. It's in the best interest of the producers to have the contestant keep playing as long as possible. Even when the number of cases gets low enough that the math is trivial (like last night's show, that got down to just two cases left) you can see that the offers are low.

Psychologically, people tend to assign more value to a loss than to a gain. In short, a dollar lost hurts more than a dollar gained feels good. On the other hand, the producers have the opportunity to screen contestants to find people who are less likely to take this approach.

Bottom line: You will almost always be getting bad value by accepting the bank's offer, although there may be times when it would preferable to do so to avoid enormous risk.

[BrianD]

From the few shows I've seen, it seems like the bank offer is always higher than the average of the remaining cases. The times when the bank offer is low seems to be when mostly high amounts are left to be picked almost ensuring that any case selected will be higher than the offered amount.

[Maple Leafs]

From the few shows I've seen, it seems like the bank offer is always higher than the average of the remaining cases.
I don't think so. It will usually be higher than the median (i.e. middle) value since the amounts are top heavy, but not the mean (i.e. what we usually call the average).

[Marc Vaughan]

i think the better puzzle question would be:
at what point does it make sense to take a deal for (for example) $300K earlier on than to actually try to win the $1million and walk away with potentially $0?

I think thats down to personal preference and how much you need the money.

Some people are more inclined to game, some less so - personally I'm happy gambling with lesser amounts of money (ie. fruit machines, pocket change sort of stuff) but with the figures in that game I'd be looking for an 'out' as soon as possible taking whatever amount I felt was reasonable before I lost it all ...

i know i wouldn't ever get as far as the last 2 cases, i would take the guaranteed lesser amount and forego a non-guaranteed larger amount.

I'd have 'drawn' until the offer went over around $100k ... the only time I'd go to near the end is if I got really unlucky and hit the vast majority of the high boxes in the first few picks (in which case it'd probably be worth it to see it through because the potential 'loss' is lessened to a huge extent).

[Marc Vaughan]

From a pure math standpoint, the offers from the bank will almost always be poor. It's in the best interest of the producers to have the contestant keep playing as long as possible. Even when the number of cases gets low enough that the math is trivial (like last night's show, that got down to just two cases left) you can see that the offers are low.

If you watch the show you'll find the offer starts very low, then around the 'mid-point' it'll gust with the streak a player is on - if they hit a lot of low numbers in a slot then it'll go a little high (ie. statistically near a decent offer based on the odds), if they lose a big figure it'll gun low (ie. lower than they're statistically likely to recieve).

This is psychologically to encourage a dilemma in the player and make the game more interesting.

[Commo_Soldier]

I am not sure about anyone else but when I watch or play a game of deal or no deal, I select cases and try to remember where the top 8 dollar amounts are.

My reason for doing this is if two or more cases in close proximity to either case has had one of the top 8 amounts it is more unlikely that the case near those amounts would also have a high amount. Therefore I always go with the case that has had the fewest high amounts near it. I have noticed this works a majority of the time when I play a game or watch the show on TV.

Additional note: there are actually 26 cases not 25.

[MJ4H]

I would be blown away if that worked because it would mean that the money is not distributed into the cases randomly.

[EagleFan]

A lot of the decision depends not only on the offer you get but what your lowest possible outcome is. If I am on there an by some incredible luck eliminate the ower values quickly I would be more willing to gamble. This is one of those games that would be easy to second guess yourself over. It would be fun to try it once for real though.

[Commo_Soldier]

I would be blown away if that worked because it would mean that the money is not distributed into the cases randomly.

I am not great at statistics, but I believe this would be the better play. Even with randomness I think the odds would be better that the million is not right next to high amounts. I mean what are the odds that of 5 cases in a row 3 will hold a super high number? Not as great compared to just two of the 5 and the third at another location on the board.

[Pumpy Tudors]

I am not great at statistics, but I believe this would be the better play. Even with randomness I think the odds would be better that the million is not right next to high amounts. I mean what are the odds that of 5 cases in a row 3 will hold a super high number? Not as great compared to just two of the 5 and the third at another location on the board.
Something seems wrong here.

[MJ4H]

I am not great at statistics, but I believe this would be the better play. Even with randomness I think the odds would be better that the million is not right next to high amounts. I mean what are the odds that of 5 cases in a row 3 will hold a super high number? Not as great compared to just two of the 5 and the third at another location on the board.

No, randomness doesn't understand location. The odds of the million being next to $500,000 are the same as the odds of the million being next to $0.01.

[Commo_Soldier]

I am not sure how well I can explain it since I am not great at stats and probability. However I will try to explain my reasoning with this analogy. Say you have a drawer with 26 socks. 8 of which are black and 18 are white. If someone were to draw socks at random, the first being case one and last being case 26. If you laid them out next to each other and looked at them in groups odds are better that you will not find to many black socks bunched up in a small group of 5 or so in a row. That is my reasoning with the cases. Although completely random the odds of the select few cases that are worth 6 figures or more are just as likely not to be right next to each other. I could be wrong, but I think this is an accurate.

[MJ4H]

[$500,000] [ ? ] [$750,000] [$1] [ ? ] [$0.01] [ ? ] [ ? ] [$5]

Which of the 4 cases most likely has the $1 million in it would you say? Is there one you would say is less likely than the others?

[Commo_Soldier]

No, randomness doesn't understand location. The odds of the million being next to $500,000 are the same as the odds of the million being next to $0.01.

Yes, but what are the odds that the first case starts at a penny and then increments so the 26th case is $1,000,000? Now what are the odds that the cases are out of order? I mean it is possible that cases are grouped together or are in order. However it is more likely that they are not grouped together because of the amount of cases involved. I am not saying it is a super high ratio, but do not think it is out of the question that the ration is something like 60-40 that they are not grouped up.

[MJ4H]

The odds of them being in incremental order are exactly the same as any other of the millions of possible orders. (likely many many orders of magnitude more than millions of possible orders, that is)

[Pumpy Tudors]

I am not sure how well I can explain it since I am not great at stats and probability. However I will try to explain my reasoning with this analogy. Say you have a drawer with 26 socks. 8 of which are black and 18 are white. If someone were to draw socks at random, the first being case one and last being case 26. If you laid them out next to each other and looked at them in groups odds are better that you will not find to many black socks bunched up in a small group of 5 or so in a row. That is my reasoning with the cases. Although completely random the odds of the select few cases that are worth 6 figures or more are just as likely not to be right next to each other. I could be wrong, but I think this is an accurate.
If you're just trying to pick out one black sock in your example (I'm assuming the black socks represent the highest dollar amounts), it doesn't matter where you pick. Your odds of picking a black sock are exactly the same no matter what was next to it. If they're indeed randomly placed, then the location of one black sock or high dollar amount is not related to the location of any of the others. There's just no connection.

[MJ4H]

This has some martingale potential, I think.

[Commo_Soldier]

[$500,000] [ ? ] [$750,000] [$1] [ ? ] [$0.01] [ ? ] [ ? ] [$5]

Which of the 4 cases most likely has the $1 million in it would you say? Is there one you would say is less likely than the others?

Yes, I would say the first question mark is more likely to have a low number compared to the other three. Like I said I could be wrong on all of this, but I just think the odds are that three black socks will not be drawn in a row when you factor in that only 27% of all the socks are black.

[BrianD]

My originally crappy thread is starting to develop legs. :)

[Commo_Soldier]

The odds of them being in incremental order are exactly the same as any other of the millions of possible orders. (likely many many orders of magnitude more than millions of possible orders, that is)

There is one chance that they will be in exact order out of all the possibilities. All the remaining possibilities put them out of order. This is where I am going. There is a chance that they are clumped together, however when you look at all the possibilities and the low amount of valuable cases compared to all of them it is unlikely. It would be different if say 1/2 of the cases had over $100,000, but only 27% do. With this in mind it is much more likely that they will not be in a small group when you look at all possible solutions than they will.

[Pumpy Tudors]

Yes, I would say the first question mark is more likely to have a low number compared to the other three. Like I said I could be wrong on all of this, but I just think the odds are that three black socks will not be drawn in a row when you factor in that only 27% of all the socks are black.
The odds of three black socks being in positions 1-2-3 are exactly the same as three black socks being in positions 1-12-22 or 3-6-7 or 10-19-25 or whatever. You're not making your situation any better (or worse) by selecting positions that are right around each other. The socks (or cases of money) are completely independent of each other.

[MJ4H]

I'm pretty sure this is some weird manifestation of the gambler's fallacy we are looking at, here.

[Pumpy Tudors]

I'm pretty sure this is some weird manifestation of the gambler's fallacy we are looking at, here.
Yeah, I think so, too. Commo_Soldier seems to be arguing his case in good faith, so I don't want to look like I'm making fun of the guy. I think he's missing something fundamental here, but many, many people do that when it comes to things like this.

[Commo_Soldier]

Yeah, I think so, too. Commo_Soldier seems to be arguing his case in good faith, so I don't want to look like I'm making fun of the guy. I think he's missing something fundamental here, but many, many people do that when it comes to things like this.

I see what you guys are saying on this, but for some reason I just think it not to be true. I guess I will have to create a simple computer program to compute all the possible solutions on a smaller but similar scale to see if I am just dreaming, since it has happened before.

[Barkeep49]

Commo: If I flip a coin and it turns up heads, the next time I flip it, it's still 50/50 that it turns up heads. If I've flipped a coin 99 times and it's come up heads each time, it's still 50/50 that it comes up head the 100th time. The odds of a coin doing that are very very very small, but each individual flip is still 50/50.

The same is true of the briefcases, assuming that they truly are random.

[Commo_Soldier]

Flipping coins is a completely different beast.

Now I have done a quick example but plan on looking into this further. Say you have 5 cases with possibilities of 1,2,3,4,5 randomly in each case. This creates a situation where there are 120 different possibilities.

Now out of the 120 combination's 4 and 5 appear next to each other in only 48 out of the possible combination's compared to the 72 times they do not appear next to each other. With this if I was left with [?] [2] [4] [?] [3] for my cases I would want Case #1 because the odds are slightly in favor of 4 and 5 not being next to each other. Granted they are not certain, but better then 50/50.

As I mentioned I am going to look at this better later, but it appears to me that my idea may be true.

[graygoose12]

But wouldn't 1 and 4 appear next to each other just as often as 4 and 5. So there is just as good a chance as the 1 being next to the 4 as the 5 being next to the 4.

[Commo_Soldier]

Yes, they would. I was thinking the same thing while posting. This is probably going to lead to me being wrong, but I have to look just a little deeper since I am frequently bored at work with nothing better to do.

[graygoose12]

To add, the 2, 4, and 3 can only be in those positions in 2 out of the 120 combinations. So there is a 50 % chance that it is this combination: 1.2.4.5.3 and a 50 % chance it is this combination: 5.2.4.1.3

So there is a 50/50 chance of the 5 being first or fourth.

[Pumpy Tudors]

Flipping coins is a completely different beast.
Except it isn't.

[Surtt]

Yes, they would. I was thinking the same thing while posting. This is probably going to lead to me being wrong, but I have to look just a little deeper since I am frequently bored at work with nothing better to do.

If you flip a coin a hundred times, most of the time you will not get a nice head-tail-head-tail pattern. They will seem to come in groups (some times large groups), they do not look random, but they are. In the end if you count them up you will get 50/50 (or close to it)

[Commo_Soldier]

Ok, so after thinking it through more and more I realize that I was indeed wrong. Maybe it just appeared to be correct because I have played deal or no deal quite a few times and more often then not I have done extremely well by picking cases as I described above. I guess I have just been getting lucky though.

[Vince]

To add, the 2, 4, and 3 can only be in those positions in 2 out of the 120 combinations. So there is a 50 % chance that it is this combination: 1.2.4.5.3 and a 50 % chance it is this combination: 5.2.4.1.3

So there is a 50/50 chance of the 5 being first or fourth.

This.