Suppose there are 3 people A, B, C and a referee. A, B, C individually takes one number from [0,1] with the order A->B->C. B could see the choice of A, C could see the choice of A and B. After that, the referee randomly take a number $Y$ from U(0,1). People who chooses the number which is most closest to $Y$ wins. But people who did a later choice cannot take the same number as the previous one chooses.


  1. what's the strategy of A, B, and C to be the final winner, if it exists? if not, please state the reason?

  2. if A takes 0, what is the strategy of B? and is there a strategy to guarantee B is the winner? and the strategy of C?

For question 2), I think as long as B takes a positive number and makes it close to 0, B would be closer to the final number than A. But not sure how he could defeat C...is there anybody who can help me? Thanks!

  • $\begingroup$ I think you need to show some work yourself. I gave a hint below. $\endgroup$
    – Bob Jansen
    Apr 18, 2023 at 11:05
  • 2
    $\begingroup$ I am voting to close this question as it is not among the topics to be discussed in QF. $\endgroup$
    – Alper
    Apr 18, 2023 at 12:25
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    $\begingroup$ A cannot go into the interval because she would be squeezed between B and C, so A takes 0 (or 1), then B takes 2/3 (or 1- 2/3). This way, B guarantees her at least 1/3 chance of winning. C goes either immediately right or left of B giving her 1/3 chance. Equal chances for all and A has avoided a move were she would get 0 chance! $\endgroup$
    – Mats Lind
    Apr 18, 2023 at 12:26
  • $\begingroup$ Thanks for the reply @MatsLind, but why B takes 2/3 or 1-2/3? $\endgroup$
    – Xu Shan
    Apr 18, 2023 at 12:51
  • $\begingroup$ A takes 0: if B takes a point right of 2/3 it gets less chance than 1/3 and C takes a point left thereof and gets more than 1/3 chance. If B takes point left of 2/3 C takes a point immediately right thereof and gets more than 1/3 chance and B gets less than 1/3 $\endgroup$
    – Mats Lind
    Apr 18, 2023 at 13:35

1 Answer 1


First try to work out the case with 2 players, for that case figure out what Player B does given player A's choice. Given this you can determine the optimal choice for A, that is the choice that maximizes their chance of winning given they know B will choose optimally.

  • $\begingroup$ Consider situaiton of 2 people. For the first problem, if A chooses a, and suppose referee chooses Y~U(0,1). B would choose a number between a and Y. So B would guess the value of Y with the maximum prob. Since Y~U(0,1), Y would be any value in (0,1) with equal probability. So B could choose EY=0.5. Thus A would definitely minimize the length of [a,Y] or [Y,a]. So A would choose a=EY=0.5. Then what's B's strategy? since B cannot take the same number as A...does my logic make sense? $\endgroup$
    – Xu Shan
    Apr 18, 2023 at 11:32
  • $\begingroup$ In the case that A chooses $a = 0.5$, B should choose $0.5 \pm \varepsilon$ to get a chance of winning arbitrarily close to 50%. A is an epsilon more likely to win with that strategy. $\endgroup$
    – Bob Jansen
    Apr 18, 2023 at 13:07

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