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Consider the following game.

You and your opponent is given a uniform random number in the interval $(0,1)$. Player 1 looks at his number and can either bet or fold. If he folds, he loses nothing and the round is over. If he bets and his opponent folds, he wins 1 dollar and the opponent loses nothing. If he bets and his opponent calls, the one with the highest number wins 2 dollars and the loser loses 1 dollar.

So this is an asymmetric game, where player 1 can bet or fold and player 2 can call(if given the chance) or fold. Both players looks at their number before action is taken.

Is this a game of psychology or is there a mathematical way to solve this game? How?

There is only 1 round in this game.

This game is like poker, I was wondering how to assess this situation mathematically? And in isolation.

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  • $\begingroup$ Barring communication between the players, it's a game of maths and has nothing to do with psychology as they have no information on each other. $\endgroup$
    – Shuri2060
    Commented Jul 9, 2017 at 15:09
  • $\begingroup$ @Shuri2060 Not sure I agree. Player $1$ can bluff. if $1$ draws $0$ but bets anyway, he can hope to trick $2$ into folding a winning hand. $\endgroup$
    – lulu
    Commented Jul 9, 2017 at 15:12
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    $\begingroup$ "If he bets and his opponent folds, he wins 1 dollar and the opponent loses nothing." Does this mean that they both win nothing? Back where they started? " If he bets and his opponent calls, the one with the highest number wins 2 dollars and the loser loses 1 dollar." Are you creating money out of thin air? $\endgroup$
    – DanielV
    Commented Jul 9, 2017 at 16:44
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    $\begingroup$ Assuming that you actually mean that money should be preserved, this game devolves into no one ever betting. Any strategy to bet on less than a 1 (such as betting on $1 - \epsilon$) would be beaten by a strategy only betting on $(1 - \epsilon / 2)$. There is a reason they require antes in poker, or on one would ever bet unless they got a royal flush. $\endgroup$
    – DanielV
    Commented Jul 9, 2017 at 16:48
  • $\begingroup$ Look it means exactly what i wrote. Money is created yes. $\endgroup$
    – badmf
    Commented Jul 9, 2017 at 17:12

1 Answer 1

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Player $1$s strategy is a function $p(x)$ which is the probability he bets given that his number is $x$. Similarly player $2$ has a strategy $q(y)$ which is the probability he calls if he can given that his number is $y$. Each player seeks to find a strategy which will maximize his outcome. Each has a floor of zero which is achieved by always folding. The analysis seems difficult particularly because of the continuous range of strategies for each player. I think I would start by doing the analysis for a game where each player gets one of a discrete set of numbers, starting with two, then three, and so on. I would hope to find a pattern in the probability distributions that I could extend to the continuous case. A challenge is that you need to have a rule about what happens with ties in the discrete case, but that should lose impact as the number of possible values increases and the chance of a tie decreases.

Added: Using a deterministic strategy where player $1$ bets when his number is above $x_0$ and player $2$ calls when his number is above $y_0$ with cutoffs forced to be a multiple of $0.05$ I find player $1$ should bet with a cutoff of $0.25$ and gets a minimum value of $0.375$ when player $2$ calls with a cutoff of $0.5$. With those values player $2$ gets $0.375$ as well. The figure below shows the region with player $1$s number on the horizontal axis and player $2$s number on the vertical axis. The numbers in the square are the payoff to player $1$ in each region. There is a change in the calculation when $y_0 \lt x_0$ because the trapezoid region changes over. I just computed the average player $1$ payoff for each combination of $(x_0,y_0)$. For each $x_0$ I took the minimum over all $y_0$s, so that is the minimum player $2$ can let player $1$ have by choosing $y_0$. I then took the maximum over $x_0$ to get the result. I did not prove that this maximizes player $2$s expectation.enter image description here

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  • $\begingroup$ Is ther eany reason the strategy shouldnt be deterministic? Can you prove thats an advantage? I actually computed some for the discrete case, I computed the optimal strategy based on the strategy for the opponent, then i computed the other guys strategy based on this new strategy etc, what I found is no convergence of strategy but oscillating regularly, how to interpret this? $\endgroup$
    – badmf
    Commented Jul 9, 2017 at 15:18
  • $\begingroup$ You could certainly have a $p(x)$ that is $0$ for $x$ less than some threshold and $1$ above that. I suspect that is not optimal, but it might be. If you require both players to play deterministically the analysis for each to find the threshold should not be hard. $\endgroup$
    – Ross Millikan
    Commented Jul 9, 2017 at 15:22
  • $\begingroup$ Can you explain abit how you computed this? What if player 2 knows player 1 s cutoff, can he achieve better by changing his own? $\endgroup$
    – badmf
    Commented Jul 9, 2017 at 16:09
  • $\begingroup$ This seems to be a Nash equilibrium, I verified your result, why was their expectation values equal is this a coincidence? $\endgroup$
    – badmf
    Commented Jul 9, 2017 at 18:21
  • $\begingroup$ I don't see any particular reason for the payoffs to be the same. In my diagram the payoff to player $2$ is $0$ where the payoff to player $1$ is $1$ and the $2$ and $-1$ are swapped. $\endgroup$
    – Ross Millikan
    Commented Jul 9, 2017 at 18:32

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