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$A,B,C$ are playing a dice game. $A$ chooses a number from 1 to 30 first, then $B$ chooses another number from 1 to 30, finally $C$ chooses another number from 1 to 30. Therefore, the 3 numbers chosen are different. Then throw a 30 sided dice numbered from 1 to 30 with equal probability. The winner of the game is the one having the chosen number closest to the dice outcome. The winner will gain the number of dollars same as the dice outcome.

What is the optimal strategy for $A,B,C$ respectively? We assume that they all want to maximize their own expected payoff. I know the solution when we only have 2 people in the game. What about three people? Which player is in the most advantageous position?

How can we approach this problem?

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  • $\begingroup$ Two-player variation of this was asked at least $3$ times on the site: $[1]$, $[2]$, $[3]$. $\endgroup$ – Vepir Oct 13 '19 at 18:09
  • $\begingroup$ @Vepir I know two player. But this is three player $\endgroup$ – neverevernever Oct 13 '19 at 18:14
  • $\begingroup$ I just meant to collect relevant questions in one link for readers that aren't familiar with two-player variation. $\endgroup$ – Vepir Oct 13 '19 at 18:16
  • $\begingroup$ how you knows that the three chosen numbers are different? The players cannot choose the same number? $\endgroup$ – Masacroso Oct 13 '19 at 18:20
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    $\begingroup$ What happens if the outcome of the die roll is equally close to numbers chosen by different players? $\endgroup$ – Misha Lavrov Oct 13 '19 at 23:58
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The approach is to start by defining a function $C(a,b)$ which is the number $C$ picks given that the others have picked $a$ and $b$. It doesn't matter who picked which, so we only define it for $a \lt b$. $C(a,b)$ will either be $a-1, b+1$ or $b-1$ and you need to define the breakpoints where it shifts from one to another. Once you have that function, $B$ knows what $C$ will do based on what he does, so you can compute a function $B(a)$ which is the number $B$ can pick based on what $A$ did. Having done that, $A$ will know what the other two will do based on his play, so can assess the relative value of each play and take the best. Then $B$ takes $B(a)$ from that play, and $C$ takes $C(a,b)$ and we are done. It looks like a fair amount of work to me.

Alternately, a less sophisticated approach but maybe less work for the solver is just to write a program that computes the value to each player for each possible set of choices, then to use that to get $C(a,b), B(a),$ and $A$

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  • $\begingroup$ I think $b-1$ is preferable to $\lceil \frac{a+b}{2}\rceil$; either way, about half of the numbers in the range $(a,b)$ are closest to you, but when you pick $b-1$, those numbers are higher. $\endgroup$ – Misha Lavrov Oct 15 '19 at 3:46
  • $\begingroup$ @MishaLavrov: Good point. Updated. Thanks $\endgroup$ – Ross Millikan Oct 15 '19 at 4:00
  • $\begingroup$ Why isn't C choosing a+1 possible? $\endgroup$ – neverevernever Oct 15 '19 at 15:59
  • $\begingroup$ @neverevernever: it is possible but it will be dominated by choosing $b-1$ because you will win the same amount of the time but will win less when you do. This is just like Misha Lavrov's argument. $\endgroup$ – Ross Millikan Oct 15 '19 at 16:01

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