Suppose three players play the following game: Player 1 picks a number in $[0,1]$. Then player $2$ picks a number in the same range but different from the number player $1$ picked. Player $3$ also picks a number in the same range but different from the previous two. We then pick a random number in $[0,1]$ uniformly randomly. Whoever has a number closer to the random number we picked wins the game. Assume all players play optimally with the goal of maximizing their probability of winning. If one of them has several optimal choices, they pick one of them at random.
1)If Player 1 chooses zero, what is the best choice for player 2?
2)What is the best choice for player 1?
I have some trouble seeing how this problem is well-defined. 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. Can someone help?
EDIT: I now understand better how the problem works, but I still have no idea how to approach this. Can someone give me some hints?
Best Answer
The first part. Let the three players choose $x,y,z$ in order. Suppose that $x=0$ and consider the third player's choice.
Option A: Choose $z>y$. Then the optimal choice is arbitrarily close to $y$ to make the winning probability arbitrarily close to $1-y$. In this case, the winning probability for the second player is arbitrarily close to (and a little larger than) $y/2$.
Option B: Choose $z<y$. The winning probability is always $y/2$, so the choice is a random element in $(0,y)$. In this case, the winning probability for the second player is $(1-y)+y/2=(1-y)/2$. [The probability that the random element is $\geq y$) $+$ (the probability that it is nearer than $z$ to $y$ conditioned on it being less than $y$)$\times$(probability of random number being less than $y$)].
The third player will choose A if $1-y>y/2$, i.e. if $y<2/3$ and will choose B if $y \geq 2/3$.
Whether the third player is made to choose option A or B, we can see that the optimal choice of $y$ is $2/3$.
The second part. This is an intuitive rather than analytic solution. After the first two players have made their choice, the maximum probability that the third player can get is among $x,(y-x)/2,1-y$. 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.