Solving the Paradox of A & B's Money Game

[vinter]

I will write a few questions in succession such that the result of one contradicts that of the other and this renders the full situation fairly complicated. What I want you to do is to make the situation more comprehensible.

A & B are two close friends. They go to an ATM centre and withdraw arbitrary amounts of money, the range being from 0 to infinity. After they come out, what's the probability that A has more money than B? Assume that they had no money before goint to the ATM.
One way to approach would be to say that since A has no preference over B, that A has more money than B and that B has more than A, should be equally likely. They are mutually exclusive too. But they are not exhaustive since the case that A and B have equal amounts of money is not counted.
But whatever be the case, the two cases stated above ARE equally likely.

Now, consider this game :-
A & B take out their wallets and whoever has lesser amount of money gives all of it to the other person. That's it.
Consider this game from A's point of view. In any case, what he can lose is obviously less than what he can gain. So the game is advantageous for A. But the same argument holds for B. So the game is advantageous for B too! That's not possible!

 

[Zurtex]

Case 1: A and B are close friends thus their choices will not be independent of each other therefore it's impossible to tell. However if their choices are independent and A takes x money out first, then the probability of B taking out y money such that x < y is 1 by the trivial theorem of arithmetic.

 

[matt grime]

situations like this can usually be explained by showig you haev the wrong posteriors. in particular the second one is flawed because it assumes that all possible distributions of money in the wallets are equal. here is an explanation by keith devlin

http://www.maa.org/devlin/devlin_0708_04.html

 

[Zurtex]

situations like this can usually be explained by showig you haev the wrong posteriors. in particular the second one is flawed because it assumes that all possible distributions of money in the wallets are equal. here is an explanation by keith devlin

http://www.maa.org/devlin/devlin_0708_04.html

Very true, even for the first example I was thinking of a uniform probability distribution for taking their money out.

 

[vinter]

Yeah, thanks for the link. It clears things up.

 

[uart]

Now, consider this game :-
A & B take out their wallets and whoever has lesser amount of money gives all of it to the other person. That's it.
Consider this game from A's point of view. In any case, what he can lose is obviously less than what he can gain.

Whoo there, each player stands to lose all of their money, but since the one with the least money loses (and gives this lesser amount of money to the winner) then clearly the winner must win less than what he stands to lose! This in itself doesn't really spoil the "paradox", it's just you would have to put it the other way around - that is, that both A and B would find it disadvantageous to play.

Anyway a qualitative debunking of the paradox would be to point out that in this game the more money that a given player has then the more they stand to lose, but also the lower probability that they actually lose - true for any given probability distribution of the possible amounts of money.

 

[vinter]

Whoo there, each player stands to lose all of their money, but since the one with the least money loses (and gives this lesser amount of money to the winner) then clearly the winner must win less than what he stands to lose! This in itself doesn't really spoil the "paradox", it's just you would have to put it the other way around - that is, that both A and B would find it disadvantageous to play.

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One more Ooooooops! to make the msg longer than 10 characters such that it is accepted.

 

[toocool_sashi]

i do not really understand the paradox...whoever has more money after withdrawing from the ATM stands to gain...so at A time...there is only ONE person who is advantageous...where is the paradox in this!?

 

[matt grime]

because it is symmetric in the people involved. you hve no way of knowing which has more money. read the devlin article for a longer explanation of these kinds of paradoxes. essentially it states that if you are player A then it is in your interest to play, but the same could be said of player B since the situation is totally symmmetric thus they ought both to play since they'll both win, but then one of them must lose which is the contradction.

 

[gerben]

Your have X dollars the other has Y dollars.
There are two possible scenarios:
(1) X > Y you win Y
(2) X < Y you lose X
so the gain will be:
Gain = 0.5 (Y - X)

The gain will be positive or negative depending on whether X or Y is smaller, but whatever turns out to be the case you should expect to either win or lose half the difference between what the two of you have.

 

[toocool_sashi]

you hve no way of knowing which has more money. essentially it states that if you are player A then it is in your interest to play, but the same could be said of player B since the situation is totally symmmetric thus they ought both to play since they'll both win, but then one of them must lose which is the contradction.

how can they BOTH win...consider the starting of the 1st game...b4 even they withdraw money from the ATM. At THAT point...can we say that BOTH will win? i don't think so...we can only say that the person who has more money after the 1st game will win...which is not favouring both the ppl @ once...isnt it?? i still don't see any paradox in this!

 

[matt grime]

the ATM one wasn't a good example. go read the devlin article to get a clearer picture.

when i said they'll both win i was speaking loosely to attempt to explain the paradoxes of this type. it is relatively easy to set up a (zero sum) game for two players whereby if we *misapply* the rules of probability it appears that each players net gain is positive so that each player "wins" ie they both win.

in the ATM one you're seemingly favouring the person with most money but that is equally likely to be either person.

 

[toocool_sashi]

well in that case ill state a simpler *paradox*. Consider a game with 2 players A and B in which there cannot be any draw. now either A or B can win, equally likely and exhaustive too...so both have equal chances of winning...so its favouring both...but there's only 1 winner! so if the ATM thing is a paradox...so is this! haha..i know u will most probably kill me but this is just how i view the ATM question as!

 

[matt grime]

i did tell you that the ATM one wasn't well phrased indeed may not even succumb to this analysis but the explanation of these types of paradoxes still stands. go read the devlin articel to see what these paradoxes are attempting to describe. your example doesn't state that it is a zero sum game. it is prefectly possible for a game to exist where the expected win for both players is positive, but if its a zero sum game then it is a problem.

 

[matt grime]

anyway, the point i was trying to make (rather than explain someone else's example) was that misapplying prob. can lead to paradoxes. read thyat article by devlin.

here's another one, again from a devlin article.

well order the points of a dartboard (it is incountable) A nd B throw dasrts at it. A hits point X, there are only countably many points less than X hence B gets a bigger score almost certainly. hence B must win. but this is symmetric in A and B hence almost surely A must win.

 

[uart]

I think it's enlightening to look a concrete example of this "paradox" in action for a well defined probability density of the player initial money. Take for example the simple case of two players A and B each starting with an initial amount that is uniformly distributed between zero and the maximum that the automatic teller can dispense.

For convenience sake let's normalize this maximum value to unity. So assume player A has amount "a" and player B has amount "b", where "a" and "b" reals uniformly distributed in [0..1].

Now if we choose two people to play the game totally at random and have no idea of how much initial money that either has, then E(a) = E(b) = 0.5 and either player has a 50% chance (priori) of winning.

However if B is to win then amount "a" must be less than amount "b" and hence the posterior expectations will be :
E(a | a<b) = b/2
btw: this notation means "the expectation of a given that a<b"
Similarly E(b | b<a) = a/2

Thus B's expected wining is b/2 and A's expected wining is a/2. So A and B each only stand to gain one half of their current amounts if they win but if they loose then each stand to loose all of their money! So it seems that it is simultaneously dis-advantageous for both players to partake, which is a contradiction in a zero sum game.

As Matt said in his very first reply the apparent contradiction comes about from inconsistently mixing both priori and posterior expectations. Specifically A and B only have equal chance of winning if we know nothing about either of their current values (a and b). Once we know "a" then A's probability of winning is P(x<a) = a, for the uniform distribution. Similarly B's probability of winning is P(x<b) = b.

Use those probabilities and suddenly there is no contradiction.

B's expected win is b/2 with a probability of b, and B's expected loose is b with a probably of (1-b). So the net expected gain for B is : net = b*b/2 - (1-b)b = 3/2 b^2 - b.

Similarly for A the net expected gain is : net = 3/2 a^2 - a.

Looking a the function 3/2 x^2 - x it is easy to determine that either player stands to win on average only if they have more then 2/3 of the maximum amount, the maximum win occurs at x=1 in which case the expected gain is 0.5 . For less than 2/3 they always stand to loose, the maximum expected loose is 1/6 which occurs when x=1/3. So one third the maximum is the least desirable amount to have in this game and all of the maximum is the most desirable amount to have.

Note that if both players agree to always play the game whenever invited, irrespective of how much money they actually have, then the net expected win is integral( (3/2 x^2 - x), x=0..1) = 0, symmetrical for both players.