The Wallet Paradox

Bill Gates meets Warren Buffett at a dinner party and the host tells them to play a game. Each person will place his wallet on the table. The person with less money in his wallet wins all the money.

Is anyone favored to win this game?

What happens when the game is repeated?

The answers are after the break.

A matter of perspective

At first glance, the game seems favorable for Bill Gates.

He might think as follows. He could either have more or less money, so he has a 50/50 chance of winning the game. If he loses, then he would lose only the money in his wallet. If he wins, however, then he would win more money than what he has in his wallet.

That is, Bill Gates can wager the money in his wallet (say x dollars) and have a 50 percent chance of winning the money in Warren Buffett’s wallet, which is more than x dollars. This is obviously a winning gamble, and thus Bill Gates will want to play as he seems favored.

On the other hand, Warren Buffett thinks about the game. By symmetry, he can reason exactly the same thing. He feels that he has an even chance of winning more money than he has in his wallet, and thus he is favored by the game.

Both billionaires find the game advantageous. And yet this cannot be. The game is a zero-sum game (the profits of one are exactly the losses of another) and thus it is impossible for both to expect a profit at the same time.

How can we resolve the reasoning and resolve the apparent paradox?

Careful accounting

The setup is similar to a wager I previously wrote about called the necktie paradox. The wallet paradox can be resolved in a couple similar ways.

Resolution 1: fix the reasoning

Bill Gates and Warren Buffett are unlikely to reason so wrongly. They would see the error in the above logic.

The problem with the reasoning is that each is considering his wallet as being both the one with more money and less money at the same time. In reality, a wallet can either be the one with more money or it can be the one with less money.

The correct logic is:

–If I have the wallet with more money, and I play the game, then I will lose my more valuable contents

–If I have the wallet with less money, and I play the game, then I will win the more valuable contents of the other wallet

There is an equal chance of being in either situation. Therefore, the game does even out as either player can win or lose the wallet with more money. The game could end with Bill Gates winning half the time, and Warren Buffett winning half the time. Thus neither is favored and the game is fair.

Resolution 2: make a proper expectation

Another way to see the game is fair is to write out the math carefully. Suppose the more valuable wallet has x dollars.

To write the expecation, it is necessary to know two things.

First, what is the chance a player wins the bet? This is the chance the person’s wallet has more money. Assuming the game was not rigged, either Bill Gates or Warren Buffett could have more money than the other. It would be reasonable to assign a 50 percent chance on either being the winner of the game.

Second, what are the payoffs in the game? One case is when a player wins the game, in which case he wins x dollars from the other player’s wallet. The other case is when a player loses, so he loses the more valuable contents of his wallet of x dollars.

Putting these two facts together, we get:

The math demonstrates the game has a zero expected payoff to either player, and thus the game is fair.

Adding a twist with repeated play

Afterwards, it turns out both Bill Gates and Warren Buffett loved the game.

They add a twist as follows. They will play the game every day for a month. To manage the stakes, they agree to carry an average (mean) of $100. Is this game fair? What is the best possible strategy, given you know what your opponent is doing?

Winning strategy in repeated play

The game gets more interesting when repeated. Even when players agree to carry the same average money, the game is not necessarily fair. There will be times that one person will be favored to win.

Consider the example of Bill Gates and Warren Buffett agreeing to carry $100 on average. Suppose they limit their strategies to the following three:

Strategy A: carry $100 every time
Strategy B: carry $75 half the time, and $125 half the time
Strategy C: carry $75 two-thirds of the time, and $150 one-third of the time

We can evaluate how one strategy performs against another.

It turns out there is a rock-paper-scissors type of situation: strategy A is profitable against B, which is profitable against C. But then strategy C is profitable against A!

This cyclical nature gives a hint about the nature of the game. More on that in a bit. First, here is the math that confirms it.

Strategy A vs. B:

Strategy A means always carry $100. The way the game plays out depends on the realization of strategy B.

The wallets can either have amounts (100, 75) or (100, 125), each with probability 0.5.

In the (100, 75) case, the second wallet has less money, so employing strategy A means a loss of $100.

In the (100, 125) case, the first wallet has less money, so employing strategy A means a profit of $125.

The expectation of employing strategy A against B is thus: 0.5 ( -100 + 125) = 12.5 > 0. Thus strategy A is profitable against B.

Strategy B vs. C:

Calculating this expectation is just a bit more involved, as both strategies are probabilistic.

There are four possible outcomes, each with different probabilities of occurrence. Here is how strategy B fares:

(75, 75) –> happens 1/2 * 2/3 = 2/6, and is a draw
(75, 150) –> happens 1/2 * 1/3 = 1/6, and strategy B profits 150
(125, 75) –> happens 1/2 * 2/3 = 2/6, and strategy B loses 125
(125, 150) –> happens 1/2 * 1/3 = 1/6, and strategy B profits 150

Putting this together, the expectation is (2/6 * 0 + 1/6 * 150 + 2/6 * -125 + 1/6 * 150) = 8.33… > 0.

Thus we can see that strategy B is profitable when played against strategy C.

Strategy A vs. C:

There are two possible outcomes, each with different probabilities of occurrence. Here is how strategy A fares:

(100, 75) –> happens 2/3, and strategy A loses 100
(100, 150) –> happens 1/3, and strategy A profits 150

Putting this together, the expectation is (2/3 * -100 + 1/3 * 150) = -16.66… < 0

Thus it is seen that strategy A is a losing decision against C!

There is no one best strategy

The fact that strategy A is preferred to B, B is preferred to C, and C is preferred to A is a clue that the game might not have a single winning strategy. This hunch turns out to be true.

For any given strategy* with a fixed mean, one can find another strategy with the same mean that is profitable and is better. (*there are some technical restrictions that the strategy is well-defined)

The details and rigorous math can be found in this paper: The Wallet Paradox Revisited.

As a side point, this means the game has no Nash equilibrium. These types of games can be fun an interesting. Here is another game I wrote about with a similar flavor: the dollar auction game.

And I will close with something obvious but noteworthy. If someone asks you to play the game, you should definitely be skeptical. They probably have little or no money in their wallet and are hoping to make a quick buck.

It is only sensible to play the game in a fun party setting when both people are taken by surprise. It’s even better after a night of drinking where both people are less likely to remember exactly how much is in their wallet.



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  • Jin

    I think, in the real scenario, Bill Gates will win the wager. He will have lots of credit cards rather than paper money since he is tech savvy. And as far as i know Warren Buffett always likes to keep his liquidity high, hence he will have lots of cash with him (or bond papers) :)

  • anomdebus

    I am not much of a mathematician, so I will not be able to rise to your level of deduction. However, in the second scenario, there is no rule that they must continue to follow the simple rule they started out with.
    When you bid low early, you are amassing a debt that must be paid later. However, the number of wins can have more of an impact than the worst loss. If ‘A’ starts out betting $100, but then repeats their opponent’s last bet, the opponent can bet $99 to start and decrease a dollar each day ‘A’ decreases until the last day when the opponent bets $535 and wins a net of $1959 (assuming 30 day month). Note a $2/day decrease still nets $1147, a $3/day decrease nets $335 and a $4/day decrease is a net loser.
    Is there a strategy that ‘A’ can deploy to take advantage of the fact that the opponent owes $1 to the pot? It may hinge on the switch from net win to net loss between $3 and $4.

    Also, in the final paragraph, it only makes sense for a person to play when they are surprised and they can be absolutely certain the other person is surprised (note: this is perhaps impossible).

  • Joey

    I thought it was very interesting that the extension of the game ended up being no more than a rock-paper-scissors kind of game. Just as a tangent, I looked on the web and I was surprised that there was actually a strategy for a RPS game.

  • http://www.sedimentblog.com The Sediment Blog

    But given the status of the guests, which wine would you take to the dinner party?

    http://bit.ly/h5Jeq1

  • Champ Bailey

    Why would each players’ wining/losing probability be 50/50. What if they have equal amount of cash?

  • rageguy

    Realistically, I just see something banter like this happening at the party:

    http://goo.gl/ww0f3

  • Travis

    I think the reasoning in Resolution #1 is wrong.

    The idea of winning or losing “the most valuable wallet” is meaningless. Suppose Bill Gates plays the game and has $1000 in his wallet. If he loses, he loses $1000. If he wins, he wins more than $1000. Those are facts and he knows them. The incorrect assumption is not that he is wagering $1000. The incorrect assumption is that the chances are 50-50. The more money Bill Gates has in his wallet the higher the probability of losing, and the less money he has the higher his probability of winning. This negative feedback is what drives the game to zero-sum.

  • john

    I sort of agree with Travis.

    Resolution #1 isn’t necessarily wrong, but it certainly isn’t a correction of the previous reasoning. You start out with Bill Gates’ perspective, then give Warren Buffett’s. Assuming they each know what is in their own wallet but not what is in the others, they will have to use some probability distribution for the dollar amount in the other person’s wallet to estimate a probability of winning and the payoff.

    To illustrate, let’s define some dollar values. a and b such that a<b, and x is the value in Warren Buffett's wallet. Now, let's define p1 as the probability that x<a; p2 as the probability that a<x

    If Bill Gates has b dollars in his pocket, he can conclude that his probability of winning is p3. The expectation for his payoff would then be e3p3-b(p2+p1+p(a)) (where p(a) is the probability that Mr. Buffett has exactly a in his wallet). If on the other hand Bill gates had a dollars in his wallet, his probability of winning would be p3+p2+p(b), and the expectation for his payoff would be e3p3+e2p2+p(b)b-ap1. This is clearly a more favorable situation for Mr.Gates as the expectation for his payoff is greater by p2(e2+b)+(p(a)+p(b))b+p1(b-a).

    So, the error in the original reasoning was that the probability of winning was considered independent of the value in the wallet. Resolutions #1 doesn't correct the original reasoning, it instead gives a 3rd perspective: that of a third party. And resolution #2 is really the same as #1, just writing it out more formally. The reasoning given in the 2 resolutions is correct if you assume the third person knows absolutely nothing about the cash carrying habits of the 2 playing the game, or if he/she knows they on average carry the same amount, but knows nothing more. Otherwise, your arguments for the repeated game apply here too. For instance, if one typically strays very little from the average they both carry, and the other carries about 75% 2/3 of the time and 150% the other third, then the game isn't fair. As you've pointed out, the second person has an expectation for his payoff equal to one sixth of the average.

    So far I disagree on 2 points: the resolutions aren't a correction of the previous reasonings, and the game is in general not fair. My third disagreement is with the statement that the repeated game isn't fair. In the one time game, different cash carrying habits can give an edge to one of the players, and if each knows what he is carrying, neither is likely to come up with an expectation of 0 for the payoff. In the repeated game, any strategy used by one person is a strategy the other could have chosen. You seem to imply that the game isn't fair because there is no winning strategy. I think that , if anything, makes it more fair. If there were a winning strategy and only one player knew it, that could be considered unfair. If there is no winning strategy, neither can use such a strategy.

    However, not all strategies are equal. While there may not be a winning strategy, there definitely is a losing one. This strategy is likely to end in a loss, but could end in a tie. A crude assessment (crude because it is difficult to assign a probability to all possible strategies) of the expectation value for the loss would be the average (or $100 in your example). The strategy would be to carry all the money once and nothing the rest of the time (in your example of a $100 average over 30 days, that would be $3000 once and nothing the rest of the time).

    Whatever the other person is carrying in their wallet the day you have the total is the amount you will lose overall (unless the other person is using the same strategy and carries the total on the same day). A tie will result any time the other person carries all or nothing the day you carry all. Also, if you pick your day randomly and assume there is no chance that the other person is using the same strategy, then the expectation from the last paragraph becomes exact.

    I have one final disagreement, although you may actually already agree with me here. The apparent disagreement could be due to the first disagreement.

    The fact that it is a zero sum game does not mean that proper reasoning will necessarily mean that if Bill Gates thinks the game is in his favor, Warren Buffett will necessarily agree. Following proper reasoning, it is not unlikely that each will believe he himself is favored, or that each will believe the other has the advantage.

    You incorrectly assumed that the monetary value in each wallet didn't matter. Because of this, the reasoning you give for each is based on essentially the same information. When both have the same information relating to a zero sum game, They do indeed have to agree. It seems a lot of people (not necessarily you) have trouble understanding that probability is based on available information. 2 people who have access to different information about an event and who each reason through the information correctly will generally come up with different probabilities for the outcome of the event.

    If both Bill Gates and Warren Buffett are carrying very little in their wallets that day, each should think he himself is favored. Just as 2 poker players, one with a royal flush, and one with a full house, aces over kings.

    If Bill Gates thinks he has a 90% chance of winning and the expectation for his win is $50, then a zero sum game just means that Bill Gates thinks Warren Buffett has a 10% chance of winning (less the chance of a tie), and an expectation to lose $50. But it doesn't prevent Warren Buffett from thinking he is the one with the 90% chance to win. And he might think his expectation to win is $40. Then he would consider Bill Gates' chance to win to be 10% (less the chance of a tie), and the zero sum game would mean that he would give Bill Gates a $40 expectation for a loss.

    P.S. About the repeated play version, you ask "What is the best possible strategy, given you know what your opponent is doing?", but I don't see a specific answer. You do indicate that if you know the other person's strategy, you can find a strategy which will give you an expectation of a profit. If you know exactly what the other is playing each day, your best strategy is to carry a penny (or a dollar if pennies aren't counted) less on everyday but one, and make sure that one is the day the other is carrying the least. This will net you the total minus twice the other's minimum (which will be at most the average) less a penny (or dollar) per day plus a penny (or dollar). So for $100 average for 30 days, the minimum (when the other uses your strategy A) would be $2771.

  • john

    Sorry to have to add a second post like this, but the 2nd half of the 2nd paragraph in the previous post has vanished. So here is a repeat of that whole second paragraph.

    To illustrate, let’s define some dollar values. a and b such that a<b, and x is the value in Warren Buffett's wallet. Now, let's define p1 as the probability that x<a; p2 as the probability that a<xb. Finally, let’s define e1 as the expectation value for x given p1=1; e2 as the expectation value for x given p2=1; and e3 as the expectation value for x given p3=1. By necessity, this results in e1<a<e2<b<e3.

  • john

    Once again, part of the text I submitted has vanished. It was fine in the editor, but something happened when I clicked the “submit Comment” button.

    In the middle of the paragraph just before “Finally”, there is “a<xb." The following text should be added between the x and the b: "”

    Could the less than symbol at the beginning and the greater than symbol at the end be read as brackets? And could that be causing the problem?

  • john

    Ok, the “brackets” do seem to be the problem, so I’ll rewrite it whitout them. I could probably switch the order and use a less than at the end, but I don’t want to risk having to rewrite this yet again. So the “a(less than symbol)xb” needs to be replaced by:

    a less than x less than b; and p3 as the probability that x greater than b.

    Hope this does it. If my original comment could be edited to include the full second paragraph as I wrote it, and if my subsequent comments could be removed, that would be appreciated (I assume by more than just me – but maybe no one else is reading these comments any more).

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Thanks for your comments John–sorry the comments keep getting messed up in the formatting. I will take a look at it and hopefully can fix it!

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