rsl12

This paradox bothered me for a long time, but I think I've figured it out. Would be interested in hearing other people argue their point.

Two cups are presented to you. You are told that one cup contains twice as much money as the other cup. Without knowing which cup contains what, you can choose to keep the contents of one of the cups. Which do you choose?

Let's suppose that you are allowed to look in one of the cups before choosing. You find that Cup A contains X amount of dollars. Therefore, you can reason that there is a 50% chance that cup B contains half that amount, and a 50% chance that it contains double the amount. Therefore, the expected value of the amount of money you will get, if you choose cup B is

(0.5*X)+(2X) divided by 2, or 1.25X.

Therefore, if you choose cup A, you will definitely get X dollars, but odds are that choosing cup B gives you the better deal.

So it seems that if you choose to look under cup A, it would be better to choose cup B. But if you had instead chose to look under cup B, it would have been better to choose cup A! How is that possible?? Why bother looking at the money in one cup in the first place? Let's suppose you weren't allowed to look in the cups, but followed the same logic as above. You would say, "Let's assume that cup A has X dollars. Therefore, the expected amount in cup B is 1.25*X dollars, and it's better for me to choose cup B." Or else you would say, "Let's assume that cup B has Y dollars. Therefore, the expected amount in cup A is 1.25*Y dollars, and it's better for me to choose cup A." Something is wrong.

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Pragma

Nope, nothing is wrong with your math there. That is the actual sequence of events. You would be unable to make a decision, because the expected value of one is always higher than the other, regardless of which one you pick. We did this problem in my Stat class a month or two ago.

You're better off just flipping a coin, heads for the one on the left, tails for the one on the right, than trying to statistically figure out which will give you more money - you'll go insane trying to do that, as the math is cyclic. "If I choose A, then B gets me more money. So I choose B, but now A gets me more money - repeat"

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CSflim

Ah! the good ol' "grass is always greener syndrome".

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rsl12

I *don't* think the math is right. If you look at the problem *without* assuming that there is $X in cup A, then you could argue something like this:

There's two possible scenarios: cup A has $X and cup B has $2X, or else cup A has $2X and cup B has $X. The expected value in both the cups is $1.5X, so it doesn't matter which one I choose.

That makes TOTAL sense, right? So where's the mathmatical flaw in the original argument?

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Pragma

The math is correct - we did this in my stat class.

Assume that you have X dollars in cup A, and cup B has either 1/2 as much or twice as much.
Expected value from choosing cup B is (1/2)(1/2) * (2)(1/2) = 1.25

Therefore, repeating the same math for choosing cup A is also 1.25

So no matter which one you choose, you stand to gain more value by choosing the other.

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rsl12

If you did it in stats class and they said it was right, then I say that your Entire stats class is wrong! If you calculate it that way the expected values are X and 1.25X, respectively. If you calculate it the way i did 2 messages ago, the expected values are 1.5Y and 1.5Y. How can the math be right in both cases, and you have expected values equal to each other in one case, and not equal in another?

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KnifeMissile

Both evaluations of expectation are correct. I think the problem is simply that you're taking expectation too seriously. First, I will try to show how the two evaluations are correct, since there seems to be some confusion about this, and then I'll try to show why they are not contradictory.

First, suppose we play a game. I will give you some amount of money and show you a box. In the box is either half the amount of money I've just given you or double it. Given a choice between the two, which should you take? Expectation says you should always pick the other box and, in this case, it will be correct! If you were to do this several times and always pick the sure thing then, on average, you will find yourself with less money than if you had always picked the other box. This is reflected by the expectation being 1.25 * X, where X is the amount of money you already have.
By the way, this is precisely the game you play after picking the first cup of this thread's game. In this case, X is the amount of money revealed. Please note that you do not know weather X is the greater or lesser money.

Secondly, lets play this thread's game. It would seem that it doesn't matter which cup you choose since you don't know any difference between the cups. If X amount of money is in one cup and 2 * X is in the other then the expectation is 1.5 * X. Please note that, in this case, you know that X is the lesser money.

X is not the same value in both cases so you can't simply compare them. The evaluations of expectation are correct, it's just that the expectation doesn't suggest what you think it does because they're both different situations.


I've always liked this story when trying to explain probability. I'm going to the corner store and I'm definitely buying a tub of ice cream. They only carry the two flavours chocolate chocolate chip and coockie dough. What are the chances that I'm buying the cookie dough?

Well, if this is all you know, then the chances are 50%.

But, if you happen to know that chocolate chocolate chip out-sells cookie dough, say by 9:1, then you can say that the chances I'll buy the cookie dough are 10%.

But, if you happen to know that my favourite flavour is cookie dough and I buy it more often than chocolate chocolate chip, say by 9:1, then you can say the chances I'll buy the cookie dough are 90%.

Which probability is correct? They all are! Probability isn't an absolute, it depends on what you know. Even when you're talking about seemingly perfectly random events, like dice, you still must make the assumption that all the sides of the dice are equally as likely to occur. The dice could have been weighted, or irregular, or anything! Whenever calculating probability, you are always making assumptions because you have insufficient knowledge to be certain...

rsl12

knifemissile, thanks for your reply. I feel i will have to argue with you, however.

and what i will argue is one single point: that the two games, (the original game and your game of being shown a box and then given a choice to change for double or 1/2 the amount) are NOT the same.

Let's start with the original game, and then modify it in mid-game to make it exactly like the 2nd game. Cup A contains $1, cup B contains $2. You can look at only one of them. There are 2 possibilities.

A. You look at cup A. Now, we can modify the game to be the same as yours by flipping a coin and deciding whether to leave the $2 in cup B or replace it with $0.50. Now you know the expected value in cup B is $1.25, and of course you should pick cup B.

B You look at cup B. Now we modify cup A by flipping a coin and deciding to leave $1 in it or change it to $4. The expected value is $2.50, and of course you should pick cup A.

There is a difference!!! in these two cases, it is OBVIOUS that switching gives you the higher return. But if you had left the game alone as it was without looking inside either cup, then it is obvious it should make NO DIFFERENCE which cup you choose--you have a 50/50 chance no matter what. And the fact that you get to look inside one cup does not, in any way, change your ability to get the larger amount of money. I'll show you--there are 2 possibilities:

A. You look under cup A, and see there is $1. Then you switch to cup B. You get $2.

B. You look under cup B, and see there is $2. You switch to cup A, and get $1.


And if you don't bother switching,

A. You look under cup A, and get $1.

B. You look under cup B, and get $2.


In both cases, your expected winnings are $1.50, so switching made absolutely no difference! Common sense should tell you this is the right way to look at the problem, and that the other way is wrong.

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rsl12

or to make it more clear...

You pick up cup A and don't look inside. Now you know that you have the cup that contains either $2X or $X. Should you switch? The answer is clearly that it doesn't matter, because the other cup either contains $X or $2X! And even if you were to look inside cup A and see it contained $1, this should not change the logic one bit.

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wlcm

i think the idea in economics is that you have to decide whether the first cup contained sufficient money to your satifaction.

lets say you looked in a cup and say $10. if you think $10 is a good number, there is no reason to risk losing half of it.

If you think that $10 is a small number, then go ahead and switch for a chance at a higher amount.


sorta like saying. would you rather have $10 now or $20 later?
or would you rather have $1k now or $2k later? but, in this case replace later with "a .5 chance of getting double, or end up with half"

orange monkeyee

I'd grab both cups and run like hell.

KnifeMissile

Think about what you're saying here, very carefully. You pick cup A and you think you haven't looked inside but you really have! I mean, you obviously haven't but you might as well have when you continued with your line of logic. Why? Because you've already assigned the cup a value and that value is X! (exclamation, not factorial)

Now, you totally misunderstood what I was trying to say. I would have tried to clarify it to you but I'm having a hard time trying to figure out why you didn't understand me in the first place. In particular, whenever I read your response, I can't figure out what it is you were thinking when you said it. It barely even looked like a response to my post. Perhaps it would help if you were to tell me what you think I was trying to say?

telekinetic2

I agree with the original logic completely. You are looking at it wrong...the problem assumes the values based on a known. You are getting it wrong: There isn't X and 2X: Cup A contains X, and then in Cup B there is either 2X or .5X. Statistically, it is a good idea to risk .5X for an even chance of gaining 1X.

The paradoxical nature of this problem is the unrepeateable way in which it is set up, and repeatability is the only thing that statistics can help with.

You are looking at it as if someone is holding a nickel in one hand and a dime in the other, and you are offered a choice between the two. That is not how the problem is set up...both quantities are unknown.

To repeat the problem over and over, you would have to set it up that you are repeatedly given a dollar. Each time you are given a dollar, you are offered the chance to trade it for either two dollars or fifty cents, based on a coin flip, or keep it. If you were offered this choice 100 times, you would either have 100 dollars if you kept it, or (50*2+50*.5) 125 dollars if you swapped each time.

Make sense? It isn't a paradox at all, your brain is just having a hard time resetting the problem properly.

SVT01Cobra

Yeah but when you flip a coin it doesnt always come out to 50/50.

You may have 20*2 and 80*.5, leaving you with $80.

It's pretty much just an error in mathematics, there are plenty.

lordjeebus

One problem is that, to figure out the long term payoff you want to consider the repeated case where you pick over and over.

But each event needs to be independent, so the amount of $$ you see doesn't give you a hint as to how likely the other hand is to have more money.

To do this, the amount of money in each trial needs to be completely random, with no upper limit because if there was a limit high amounts of money would give you an idea of how much is in the other hand.

Hence infinite amounts of money would have to be possible, and any amount of money from 1 cent to infinity would have to be equally likely -- this sort of setup is impossible.

Hence you cannot think in terms of repeating the trial over and over.

I think.

rsl12

Knifemissile: Sorry if I wasn't clear. Let me start by explaining the most important point I was trying to make: There are two games to consider:

Game 1: The original game, as stated in the 1st post. You are allowed to look under one of the cups.

Game 2: You are given a certain amount of money, and told that under a box, there is either 1/2 or double that amount (50/50 chance), and you can keep the amount you have already, or switch for what's in the box.

You have stated that the two games are identical, but I believe that they are completely different, because in Game 1, it makes no difference if you switch after looking, but in Game 2, you should always switch. I will prove it via a thought experiment: let's suppose there are two dummies, dummy X and dummy Y, who play Game 1 100 times. The amounts under the cups are always the same (cup A = $1 and cup B = $2, respectively, but because X and Y are dummies, they don't know any better). Dummy X plays game 1 always switching (but randomly looking at cup A or B) and dummy Y plays game 1 never switching (but randomly picking cup A or B). Who gets more money? Near as I can tell, they should both end up with about $150. So switching makes no difference in this game.

Now let's say that dummies X and Y play game 2. They are both shown $1 dollar, and the box contains either $0.50 or $2. Dummy X always switches to the money in the box and Dummy Y always keeps the $1. If both dummies play the game 100 times, dummy X should end up with about $125, while dummy Y ends up with $100. So switching makes a difference in this game.

That's why the two games are not the same, and why the logic in my original post is not correct!

telekinetic: the experiment is repeatable, as shown here.

lordbejeesus: i thought that was the answer originally too--that it has to do with the fact that the amount can't go up to infinity, but after thinking about it more, I figured the logic I used above is more satisfying an answer...

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gloveshot3

pick your cup, accept your choice, get on with life.

maxero

personally i wouldve poured one of the cups into the other and taken the full one.

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gilada

I was told there would be no math...

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