Nash Equilibrium of simple betting game [closed]

Question

I'm trying to find the Nash Equilibrium of a simple betting game, and have come up with a very surprising result which I'd like to solicit comment on.

The game is simple: Two players each receive a secret real number selected randomly (uniform distribution) from $(0,1)$. Each player secretly chooses to bet or fold. If either player folds, each player receives $0$. If both players bet, the player with the higher number receives $1 + \alpha$, and the player with the lower number recieves $-1$ (i.e. looses 1).

(Note: To simplify discussion, I am ignoring ties and other probability 0 events.)

If there is no difference in expected values, both players would slightly prefer to fold than to bet.

My result: No matter what $\alpha$ is, the only Nash Equilibrium is for both players to always fold!

Proof sketch:

1. Assume there exists a value of $\tau$ that Player 1 will bet iff he believes his prob. of winning is above $\tau$. (Generally, $\tau * (1 + \alpha) > (1 - \tau)$.)
2. Before the game-theoretic regress, Player 1 computes that he will have prob. $> \tau$ if his secret number is $> k$.
3. By symmetry, Player 2 will only bet if his value if also $> \tau$, which means that his secret number is $>k$.
4. Which of courses raises Player 1's value of $k$ in an infinite regress. The formula is $k_{n+1} = (1-k_n)*\tau + k_n$, which converges at $1$.

Is this correct? It means that even if $\alpha$ is very large, both players will always fold, which is highly counter-intuitive.

NOTE: In researching this, I came across this excellent post Finding the Nash Equilibrium of $0-1$ poker with one betting round and https://groups.google.com/forum/?fromgroups#!topic/rec.gambling.poker/wFZ_aJqVY_A%5B1-25%5D , which address similar problems, though neither seems to address this simple game with such a surprising result.

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UPDATE: What's unique and still unexplained?

The responses have confirmed my result. However, I believe the question still stands. I believe the result is unique in that it shows a huge discrepancy between game theory and real life rational behavior, which none of the responses addresed or even explored. Let me illustrate:

You and an opponent are each dealt a hand and shown a pot of $100. After receiving your hand, you can choose to play, which costs one dollar, or walk away. Afterwards, your opponent is given the same choice. If both bet, winner takes all. If either one walks away, the pot goes back to the house.

You see that you have a very good hand (4 of a kind, let's say). I would certainly play. I do not know a single person that I think would refuse to play. I would play even if my opponent was a Fields medal laureate. If someone told me they'd refuse, I would think it foolish. Yet, the only Nash equilibrium behavior (that is, the only rational behavior given common knowledge of both players rationality) is to walk away. This is true even if the pot was $100,000.

This gross discrepancy between what even seemingly highly rational people would do in real life and what game theory tells us is extraordinary.

This is not true for the prisoners dilemma (cited by some responses), where many people would indeed confess in real life. And it has nothing to do with the fact that the game doesn't model real life poker when folding (as other responses mentioned). It's a case where game theory tells us one thing, and even the most rational intelligent players do something else.

To answer this question, I'd like to see this discrepancy explained or even explored. Or even similar examples which are discussed elsewhere cited.

Answer 1

Outside Lake Wobegon, not all players can receive above average values. Intuitively, what a player does when betting can only matter when the other player bets too. But if the player knows that the other player is betting, the player knows that the other player is confidet not to lose- which should dampen your confidence. UNless you are really confidemt, but then the other player knows you only bet because you are really confident, which should dampen her confidence...

The phenomenon is well known an related to the literature on the impossibility of common knowledge agreeing to disagree under a common prior, as pioneered by Robert Aumann.

You can simplify the analysis by assuming that there is a natural number $n$ and each players number is uniformly and independently drawn from $\{1,2,\ldots,n\}$. For simplicity, assume that both players get $0$ under a draw. Take any Nash equilibrium. Let $m$ be the smallest number such that any player who has drawn $m$ would bet. Suppose $m< n$. If there is a player willing to bet at $m$, it is the unique best response of the other player to play when getting $n$. But this means that the player betting with $m$ can never win but potentially lose with positive probability, which gives a worse value than folding. Since everyone is supposed to play a best response to the equilibrium strategy profile, we get a contradiction, players never bet with values smaller than $n$.

Answer 2

In poker, the result for $\alpha=0$ is stated, "There needs to be something in the pot for players to get involved." It is unusual that you are considering positive values of $\alpha$ and find that this is still true. Usually in poker the players are risk-averse, or there is some rake (paying the house), so $\alpha \lt 0$.

If you want to get a nondegenerate game, put something in the pot which is won when one player bets and the other folds. In poker, this is the ante or the blinds or what was bet on previous rounds. When there is something in the pot, folding all of the time loses to someone who bets all of the time, and betting all of the time is still suboptimal. Poker handles the case where no one bets differently, but I think I recall model games where the pot is discarded when no one bets.

A useful tool for a model of poker in which one can only bet or fold, but the players act in order, is the Nash equilibrium calculator at HoldemResources.net which produces good numerical approximations. For example, with antes $1/100$ of the stacks, the player acting first should bet with the top $11.8\%$ of hands, and the player acting second should bet with the top $5.9\%$. You can check that players should bet with a tighter range of hands as the stacks get larger in relation to the antes.

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Edit: To understand the difference between untrained human behavior and the Nash equilibrium, see the well-studied $2/3$ game in which people try to guess $2/3$ of the average value of the guesses.

Answer 3

As for "similar examples which are discussed elsewhere" - see the No-trade theorem.

Answer 4

To make this example more realistic (less counter-intuitive) we can assume that eg.: a) with small probability your opponent will make the decision whether to bet or fold in a random way, or will make a mistake, in other words sometimes he won't be so hyperrational; of course, the higher the alfa the smaller this probability can be for you to bet, b) you get some positive utility from the fact that alfa amount is transferred from the current owner to your opponent, c) players get some pleasure (utility) from the thrill of betting itself, d) the game can be repeated in the future (iterated game), e) there's some possibility for a cooperation between two opponents (side-payments).

The counter-intuitive nature of the conclusion is a direct consequence of the fact that the assumptions of the original model are unrealistic. In practice, you can always expect some of the above assumptions to hold.

Answer 5

Your result does not seem counter-intuitive to me. A Nash equilibrium does not imply you have an optimal strategy, it just means it's impossible for your opponent to exploit your strategy, the best they can do is use the same strategy.