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A year or so back, on the verge of falling asleep, I thought up this question:

You have come to me ready to gamble. I have two envelopes on the table, one containing the amount of my bet, and one containing the probability I will win the bet (possibly in the form of some sports game result to check, etc. Something whose result is unknown to both of us, but whose odds have been estimated more or less exactly). There are no ties (the unknown probability is from a Bernoulli distribution). Clearly, I know what is inside both of these envelopes, but the only one I let you open is the one showing how much money I have bet, not the one containing the probability I win.

What is the optimal bet for you, given that you do not intend to refuse the bet?

If we play $n$ times, is there a winning strategy? (How about in the limit?)

I haven't really come up with a satisfying answer yet (if there is one), and I don't think I'll get around to it, so I'm sharing.


  • The winner of the bet receives the full amount of the other person's bet (for that round, if the game is repeated)
  • Both players have a finite amount of money, that has a minimum increment.
  • In the infinite game version, players may choose to repeat the bet with new parameters (odds selected by me, your bet amount selected by you) until one player is bankrupted.
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Go back to sleep. –  Did Mar 1 '13 at 9:08
I feel like it's similar to the cut-choose problem, but I have no real justification for that. –  Trevor Alexander Mar 1 '13 at 9:12
What's the payoff? –  joriki Mar 1 '13 at 9:51
If I win your bet regardless of my bet, why isn't it simply optimal to bet as little as I can? What happens to my bet -- is that what you receive if you win? –  joriki Mar 1 '13 at 10:37
It seems everyone agrees that the game is entirely trivial if played only once; so perhaps you should elaborate on why you think this would change in the iterated version -- I dont' see how it could. –  joriki Mar 1 '13 at 11:18
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1 Answer

up vote 2 down vote accepted

An equivalent description of your game can be given using the following steps:

  1. Before play begins, you announce your bet, a non-negative amount of money $C$. This step is really not part of the game, and only affects it insofar as it changes the payoffs (which, under optimal play, it turns out not to do).

  2. We both write down our moves and seal them. Your move is a number $0 \le p \le 1$ giving the probability that you win, while my move is a non-negative bet $B$.

  3. The moves are then unsealed; with probability $p$, I pay you the amount $B$, otherwise you pay me the amount $C$.

This is a single-round zero-sum game with simultaneous moves, with a convex and compact strategy space and a linear expected payoff function. Thus, it has a unique minimax solution, which is also its unique Nash equilibrium.

Indeed, the solution is very simple and obvious: regardless of my move, your optimal move is to always choose $p = 1$, while my optimal move is to choose $B = 0$. Thus, you will always win the bet, but the sum you win will be zero.

If we constrain the strategy space to require that $B \ge \epsilon$ for some $\epsilon > 0$, or that $p \le \delta$ for some $\delta < 1$, then the optimal strategy is still for me to choose $B = \epsilon$ and for you to choose $p = \delta$. However, in this case, you clearly should also choose your initial public bet $C$ to be as small as possible. In any case, in the game as stated, there's never any incentive for you to select a larger $C$ or a smaller $p$ than you absolutely have to, since you know that, even if you did so, my optimal strategy would still be to choose as small a $B$ as I can.

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Dang. I guess I need to work on this problem a little bit to make it more interesting. Thanks though. –  Trevor Alexander Mar 2 '13 at 5:44
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