Lap Theory Optimal Betting Game

Question

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.

Answer 1

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: $1,1,3,5,9,15,25,41,67,109$.