17
$\begingroup$

Three prisoners are seated at a table. Each of them has a mobile phone on their lap, and they are not allowed to look at anyone else’s phone (and obviously no other form of communication is allowed).

Each phone displays a number from 0 to 10 inclusive. They know no two prisoners have the same number. Assume that every number is equally likely (i.e. uniform distribution for the math nerds among you). Each prisoner must make a bet between 1 and 100 chips that they have the highest number.

Wins and losses are tallied and the prisoners are freed if and only if their net winnings are positive (Bets are submitted via mobile phone so no information about someone else’s bet can be used for one’s own strategy).

Example: A,B,C have numbers 3,5,8 respectively. They bet 30, 42, 53 respectively. C wins 53 but A and B lose a total of 72 and the prisoners are not freed.

What is the Lap Theory Optimal strategy for the three prisoners? And what are the chances they win freedom? Can you prove your answer is indeed optimal?

Assume the prisoners cooperate and there is no “envy” towards whoever wins their individual bet.

NOTE: the puzzle title is based on the concept of Game Theory Optimal (GTO) – there is a single best decision for every conceivable betting scenario in any form of Poker (whether it involves Holdem, Stud, Razz or removing items of clothing every time you fold a winning hand). The actual question is inspired by a cheating scandal involving Mike Postle and Stones’ Gambling Hall, which I only found out about very recently.

NOTE: I'm not sure if hat-guessing is an appropriate tag but I can't think of anything better.

$\endgroup$
  • $\begingroup$ Initially I thought " if their net winnings are positive" meant A is freed if A's net winnings are positive etc., and the puzzle did not make sense at all. Maybe you could clarify that part. $\endgroup$ – Retudin Sep 27 at 10:58
  • $\begingroup$ how many times do they play the game? is there any limit or they can play it forever until they all have positive wins? $\endgroup$ – Oray Sep 27 at 11:50
  • $\begingroup$ Will money be doubled if have the highest number? $\endgroup$ – Helena Sep 27 at 16:39
  • $\begingroup$ Brilliant puzzle! Using smartphones to display the numbers seems counterproductive for this puzzle: the setup is more intuitive if there's a shuffled deck of eleven (distinct) cards, and each contestant draws (and keeps) one. $\endgroup$ – Bass Sep 28 at 5:54
9
$\begingroup$

Their best strategy is

(number drawn $\rightarrow$ amount they should bet) $0\rightarrow 0,1\rightarrow 0,2\rightarrow 1,3\rightarrow 2,4\rightarrow 4,5\rightarrow 7$ $6\rightarrow 12,7\rightarrow 20,8\rightarrow 33,9\rightarrow 54,10\rightarrow 88$
if placing a $0$ bet is allowed.

Otherwise we have to replace the series with something like
$0\rightarrow 1$
$1\rightarrow 1$
$2\rightarrow 1$
$3\rightarrow 3$
$4\rightarrow 5$
$5\rightarrow 9$
$6\rightarrow 15$
$7\rightarrow 25$
$8\rightarrow 41$
$9\rightarrow 67$
$10\rightarrow 100$.
This is not unique. The only requirement is that of any three distinct numbers drawn the largest will bet more than the two lower ones combined or if this cannot be achieved then to have as few exceptions as possible. In scenario 2 we have two exceptions: $(0,1,2)$ and $(8,9,10)$.

Their chances with this strategy are

$100\%$ in the first scenario and $1 - \frac 2 {\left(\begin{matrix}11 \\ 3\end{matrix}\right )}\approx 98.8\%$ in the other.

Optimality

For this we need to show that there is no strategy that makes us lose in fewer than two outcomes. The critical case is one bad outcome. We would be able to eliminate that single case by rather knee-jerkly removing one of its numbers drawn from the pool of admissible numbers leaving $10$ drawable numbers and a $100\%$ success rate. But $10$ still cannot be separated even with the tightest packing: $1,1,3,5,9,15,25,41,67,109$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I think that 0 is not allowed as "Each prisoner must make a bet between 1 and 100 chips" but otherwise this seems good $\endgroup$ – hexomino Sep 27 at 10:39
  • $\begingroup$ @hexomino Yeah, also it seems more interesting, since we have to decide where to optimally compromise. $\endgroup$ – Albert.Lang Sep 27 at 10:42
  • $\begingroup$ This does not seem the full solution. After all, the solution is not unique, and the prisoners are not allowed to talk to each other. So it is not guaranteed that the the fellow prisoners choose the same 'solution'. Somehow solutions have to be used (and maybe non solutions?) with certain chance in GTO style to maximize the chance of success. $\endgroup$ – Retudin Sep 27 at 11:22
  • 2
    $\begingroup$ @Retudin I think the usual idea of cooperation is that the players can conspire to form a strategy before the game starts, otherwise it's very difficult to cooperate. $\endgroup$ – hexomino Sep 27 at 11:39
  • $\begingroup$ @hexomino Ok, that is true. Otherwise, with a common goal, mentioning cooperation makes no sense. $\endgroup$ – Retudin Sep 27 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.