If they play in the same subgame, it must be the larger one, of size $1-x_1$, since Player $3$ wouldn't play in the smaller game if Player $2$ has already played there. Either the other two players play in the same subgame, or in different subgames. In the general case, by playing at $x_1$ Player $1$ effectively creates two new games on either side of $x_1$, one scaled down by $x_1$ and one scaled down by $1-x_1$, and these subgames work like the game in part $1$), since they're both delimited by a play by Player $1$ on one end and a normal boundary on the other.Īgain, without loss of generality assume $x_1\le\frac12$. We can use the first case to solve the general case recursively.Īs the existing answers have established, the answer to part $1$) is $2/3$, and the payoffs in this case are $(\frac16,\frac12,\frac13$). I thought about this some more and realized why the question is posed the way it is, in two steps. Equating the three of them, we get $x=1/4,y=3/4$, so both $1/4$ and $3/4$ should be optimal for the first player. After the first two players have made their choice, the maximum probability that the third player can get is among $x,(y-x)/2,1-y$. This is an intuitive rather than analytic solution. Whether the third player is made to choose option A or B, we can see that the optimal choice of $y$ is $2/3$. if $y<2/3$ and will choose B if $y \geq 2/3$. In this case, the winning probability for the second player is arbitrarily close to (and a little larger than) $y/2$. Then the optimal choice is arbitrarily close to $y$ to make the winning probability arbitrarily close to $1-y$. Suppose that $x=0$ and consider the third player's choice. Let the three players choose $x,y,z$ in order. Can someone help?ĮDIT: I now understand better how the problem works, but I still have no idea how to approach this. For instance, if Player $1$ picks $0$ and Player $2$ picks 1, then I cannot see what the optimal choice would be for the last player since he has to pick different numbers. I have some trouble seeing how this problem is well-defined. If one of them has several optimal choices, they pick one of them at random.ġ)If Player 1 chooses zero, what is the best choice for player 2? Assume all players play optimally with the goal of maximizing their probability of winning. Whoever has a number closer to the random number we picked wins the game. We then pick a random number in $$ uniformly randomly. Player $3$ also picks a number in the same range but different from the previous two. Then player $2$ picks a number in the same range but different from the number player $1$ picked. Suppose three players play the following game: Player 1 picks a number in $$.
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