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.

  • $\begingroup$ 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 Jul 9 '17 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 Jul 9 '17 at 15:12
  • 1
    $\begingroup$ "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 Jul 9 '17 at 16:44
  • 1
    $\begingroup$ 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 Jul 9 '17 at 16:48
  • $\begingroup$ Look it means exactly what i wrote. Money is created yes. $\endgroup$ – badmf Jul 9 '17 at 17:12

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

  • $\begingroup$ 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 Jul 9 '17 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 Jul 9 '17 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 Jul 9 '17 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 Jul 9 '17 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 Jul 9 '17 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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