MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!topic/ , which address similar problems, though neither seems to address this simple game with such a surprising result.

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.

share|cite|improve this question

closed as off-topic by Steven Landsburg, Andy Putman, Chris Godsil, Andrey Rekalo, David White Jun 27 '13 at 18:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Steven Landsburg, Andy Putman, Chris Godsil, Andrey Rekalo, David White
If this question can be reworded to fit the rules in the help center, please edit the question.

If the result is correct then maybe it is "highly counterintuitive" because of the counterintuitive folding rule "If either player folds, each player receives 0." – tergi Aug 20 '12 at 13:40
I find the terminology counterintuitive. Folding is usually defined so that if one player folds and the other player does not fold, then the player that does not fold gets a larger reward. – tergi Aug 20 '12 at 14:35
For the argument to work, you need more than rationality, you need common knowlede of rationality. And that this assumption might be violated and be a bad predictor for "real life behavior" is not new and very well known. See for example Basu's traveller's dilemma at – Michael Greinecker Aug 20 '12 at 19:10
It's not clear what game you mean when you say the pot is $\$100$ and you would have to pay $\$1$ to play. The normal interpretation is that if you choose to play and your opponent refuses then you win the pot, but I think you mean something strange to happen instead. You should explicitly state the payoffs in all cases. – Douglas Zare Aug 20 '12 at 19:27
This is not a math question. – Steven Landsburg Jun 26 '13 at 23:25

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 $n$.

share|cite|improve this answer
I don't think this answers the question when alpha is > 0. We don't both need to have above average values for us both to bet. For instance, if we both agree that there is a 1/3 probability that I'll win, we should both bet if alpha > 1. – Robert James Aug 20 '12 at 18:52
In other words, your analysis works for alpha = 0, where my result is quite intuitive. But, for alpha > 0, your analysis doesn't help. In fact, it highlights what's counterintuive about the result. – Robert James Aug 20 '12 at 18:54
No, the gain from winning can be arbitrarily large, my argument still works. – Michael Greinecker Aug 20 '12 at 19:02
Can you clarify it? You write "not all players can receive above average values". For alpha > 0, even if my value if less than average, it still could make sense to bet. Thus, to both play, we don't need to agree to disagree (ala Aumann) - we may both maintain that you win with probability 2/3, and yet it's still optimal for each of us to play. – Robert James Aug 20 '12 at 19:31
The average refers to the average of the drawn values, not their expectation. And it is indeed impossible that $x > (x+y)/2$ and $y > (x+y)/2$ both hold. – Michael Greinecker Aug 21 '12 at 12:11

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 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.

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.

share|cite|improve this answer
Right, when I first started with alpha = 0, I got the above result, but wasn't surprised. What shocked me was that it holds with any positive alpha. I accept your argument that poker doesn't work this way. But please see my updated question that, with or without resembling poker, large values of alpha lead to a large discrepancy between human behavior (even rational) and Nash equilibriums. – Robert James Aug 20 '12 at 18:47
I wonder how that 11.8% is calculated - do you know anything about it? – domotorp Nov 16 '14 at 12:58
@domotorp: Yes, it's not something you can calculate directly. It depends on the particular game being played, Texas Hold'em, since that affects the equity when your 90th percentile hand runs into a 95th percentile hand, for example, and the blocking effect that holding a hand like AT means there are fewer combinations of AA, TT, and AK hands. An iterative method is used that is quite fast when you have the cached results of each pair of hands. – Douglas Zare Nov 17 '14 at 2:56
One complication is that the hands are not linearly ordered. The 90th percentile hand for calling against a wide range is not necessarily the same as the 90th percentile hand for calling against a tight range, and the pushing and calling hands are different. However, a common shortcut for these sorts of Nash equilibrium calculators is to assume a particular ordering of the hands, in which case there is just one parameter needed to specify a range. This isn't necessary for heads-up calculations but it is normal for multiway calculations, some of which are computationally hard. – Douglas Zare Nov 17 '14 at 2:59
@Douglas: I've posted a question about this:… – domotorp Nov 17 '14 at 12:50

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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
That makes no sense. In a NE, every player plays a strategy that is optimal given the strategies chosen by the other players. – Michael Greinecker Sep 5 '12 at 0:37
In common usage, "optimal" is often used to mean "exploitatively optimal," or optimal against whatever you know your opponent is doing, which is not assumed to be rational. It's not deep, but not nonsense. – Douglas Zare Sep 5 '12 at 3:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.