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This comes from Fivethirtyeight's riddler weekly challenge...

Toddler poker is played by two players. Each is dealt a “card,” which is actually a number randomly chosen uniformly from the interval [0,1]. (It could be 0.1, or 0.9234781, or 1/π, and so on.) The game starts with each player anteing a buck. Player A can then either “call,” in which case both numbers are shown and the player with the higher number wins the two bucks on the table, or “raise,” betting one more dollar. If A raises, B then has the option to either “call” by matching A’s second dollar, after which the higher number wins the $4 on the table, or “fold,” in which case A wins but B is out only his original buck. No other plays are made.

What is the optimal strategy for each player? Under those strategies, how much is a game of toddler poker worth to Player A?

My thinking is that you want to maximize some expected payoff. Since the 'cards' are uniform random numbers, the probability that player A's card is larger than player B's card is equivalent to the probability that $A-B>0$. The difference of two uniformly distributed random numbers is triangularly distributed, and so $Z := A-B$.

In the first turn, player A can either call or raise. If he calls, his expected payoff should be

$$E_A = \$1 P(Z) - \$1 P(1-Z)$$

A bucks if he has the larger number, and he loses his ante if not. If he decides to raise then his expected payout is

$$E_A = \$2 P(Z) - \$1 P(1-Z)$$

A similar argument holds for player B. Do you think I am on the right track? What is my stopping criteria? A hint would be lovely, or if you want you can throw down your solution.

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    $\begingroup$ I would say that you're on the right track, except that when Player B makes his decision, he has more information. He knows that Player A raised, which gives him some information about player A's card. I haven't worked out a solution to this one yet, but this is what I'm thinking about. $\endgroup$ – Duncan Feb 16 '17 at 1:10
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I think this problem is more related to game theory(math/econ). Anyway, you have to figure out the Bayesian Nash Equilibrium to see how this game plays out.

Assume that both players are risk-neutral, i.e. they only care about the expected win. Let $\sigma_A$ and $\sigma_B$ be the strategies of $A$ and $B$. From the structure of the game, the strategies are functions of each card values.

The Bayesian Nash Equilibrium $(\sigma_A^*,\sigma_B^*)$ is defined as $$\sigma_A^* = \arg\max_{\sigma_A} E_{X_B}[\pi_A(\sigma_A,\sigma_B^*)]$$ $$\sigma_B^* = \arg\max_{\sigma_B} E_{X_A}[\pi_B(\sigma_A^*,\sigma_B)]$$ where $\pi_A$ and $\pi_B$ are payoff function given opponents' type and strategy. In equilibrium, nobody has incentive to deviate from their strategies given that their opponents play the equilibrium strategy.

First, note that strategies can be characterized by a simple function that maps $[0,1]$ to $[0,1]$ where the range is the probability of choosing Raise($R$) for $A$ and the probability of choosing Call($C$) for $B$ given that $A$ raised. Note that $B$ has nothing to decide given $A$ called. So let's simply notate them by $P_A(x_A)$ and $P_B(x_B)$. (That's right. We are considering not just pure strategies, but also mixed strategies.)

Second, let's work out through $B$'s problem given $P_A(x_A)$. Given $x_B$ and that $A$ raised, $B$ maximizes $$E_{X_A}[\pi_B(P_A(x_A),P_B(x_B))]=P_B(x_B)(2P[x_A<x_B|A\ raised]-2P[x_A>x_B|A\ raised])-(1-P_B(x_B))$$with respect to $P_B(x_B)$. By definition of $P_A$, $$P[A\ raise|x_A<x_B]= \int_0^{x_B}P_A(t)dt$$ Thus, by Bayes rule, $$P[x_A<x_B|A\ raise] = \frac{\int_0^{x_B}P_A(t)dt}{\int_0^{1}P_A(t)dt}$$ Plugging in the above expression, we have $$E_{X_A}[\pi_B(P_A(x_A),P_B(x_B))] = P_B(x_B)[\frac{3\int_0^{x_B}P_A(t)dt-\int_{x_B}^1 P_A(t)dt}{\int_0^1 P_A(t)dt}]-1$$ and we conclude that the optimal $P_B(x_B)=1$ if $3\int_0^{x_B}P_A(t)dt-\int_{x_B}^1 P_A(t)dt>0$ and $0$ otherwise. Since the expression is monotone increasing in $x_B$, we know that the strategy must be characterized by a threshold $t_B$, where $B$ calls if $x_B>t_B$ and folds otherwise.

Given this, let's work on $A$'s problem which becomes a lot simpler with the threshold characterization of $B$'s strategy. Given $x_A$, $A$'s problem is to maximize
$$E_{X_B}[\pi_A(P_A(x_A),t_B)] = P_A(x_A)(2P[x_B<x_A,x_B>t_B]-2P[x_B>x_A,x_B>t_B]+P[x_B<t_B])+(1-P_A(x_A))(2x_A-1)$$ It is easy to see that $P[x_B<x_A,x_B>t_B]=\max(x_A-t_B,0)$ and $P[x_B>x_A,x_B>t_B] = 1-\max(x_A,t_B)$. So we have $$E_{X_B}[\pi_A(P_A(x_A),t_B)] = P_A(x_A)[4\max(x_A,t_B)-2t_B-2+t_B-2x_A+1]+(2x_A-1)$$ We have two cases to consider. First $x_A>t_B$. We have $P_A(x_A)=1$ if $x_A>\frac{t_B+1}{2}$. Second, $x_A<t_B$. We have $P_A(x_A)=1$ if $x_A<\frac{3t_B-1}{2}$. Note that the inequality $\frac{3t_B-1}{2}\le t_B \le \frac{t_B+1}{2}$ holds for $t_B\in [0,1]$. So the optimal strategy for $A$ is twofold as our intuition suggests. If he has small enough number, he would just raise to get lucky in case $B$ folds. If he has large enough number, he would also raise to get more out of his value. We conclude that $$P_A^*(x_B|t_B)=0 \quad \forall\frac{3t_B-1}{2}\le x_B \le \frac{t_B+1}{2}$$ and 1 otherwise.

Now let's get back to $B$'s problem of setting his $t_B$. We know that $$3\int_0^{t_B}P_A(t)dt=\int_{t_B}^1 P_A(t)dt$$ must hold. Plugging in the above result, we have $$3\frac{3t_B-1}{2}=(1-\frac{2t_B+1}{2})$$ and we finally get $t^*_B=\frac{2}{5}$.

So we conclude that in the Bayesian Nash equilibrium, $B$'s strategy is Call if $x_B>\frac{2}{5}$ and fold otherwise given that $A$ raised. And $A$'s strategy is Raise if $x_A < \frac{1}{10}$ or $x_A>\frac{7}{10}$, and Call otherwise. Expected values of each player can be easily calculated by drawing partitions on $X_A-X_B$ plane by each player's strategy. You can simply multiply the area and payoffs to get the expected values for each player. My calculation shows that the value is $\frac{1}{10}$ for player $A$ and $-\frac{1}{10}$ for $B$ as the game is zero-sum.

Partition of probability space Payoff at A = (-2,2), B = (-1,1), C = (1,-1), D = (2,-2). Area of A = $\frac{21}{200}$, B = $\frac{36}{100}$, C = $\frac{40}{100}$, D = $\frac{27}{200}$.

Intuitively, $A$ gets some advantage with the right to choose whether to raise or call. However, the game will be played much differently if $B$ is able to raise upon $A$'s call.

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