I'm trying to work out the solution to a variant of the gambler's ruin. Here's my version:

There are two very unlucky but friendly gamblers A and B who decide to pool their money together to form a common budget with starting amount $b$, a positive integer. They roll a weighted die to decide who will play the next game. Therefore, gambler A will play a round with probability $p_A$. Likewise gambler B plays with probability $p_B=1-p_A$. Now, A and B are bad at gambling and either break even or lose money whenever they play—say \$1. So their pool of money can only decrease. However, they are not equally unlucky. Gambler A breaks even (does not lose or make money) $q_A$ of the time and loses otherwise. And gambler B breaks even $q_B$ of the time and loses otherwise.

When they've totally exhausted their funds, I want to know how much money each gambler is individually responsible for losing.

For example, say they started off with \$1000 and gambler A plays 1/3 of the time, and breaks even 1/3 of the times he plays. Gambler B plays (therefore) 2/3 of the time and breaks even 1/2 of the time she plays. (By simulation) gambler A is likely responsible for about \$400 lost and gambler B is responsible the remaining \$600.

I'd appreciate any hints.

  • 3
    $\begingroup$ The amount of money person A loses is proportional to the fraction of games he played by A, the average loss per game (likelihood of him losing, if the loss rate is constant), and the number of games played by both players. Note the last consideration is irrelevant if you just want it as a fraction of B's losses. $\endgroup$ – Meow Feb 20 '13 at 15:40
  • $\begingroup$ "Now, A and B are bad at gambling and either break even or lose money whenever they play..." That violates probability. The notion of "unlucky" is very sloppy probability. The probability $p_A$ is the probability of winning, and if $p_A>0$ the only way to always lose is to keep playing until you've lost. $\endgroup$ – Thomas Andrews Feb 20 '13 at 15:53
  • $\begingroup$ You can pretend that they are bad gamblers in the sense that they engage in a game that costs nothing to play, but the payoff is either zero or $1 lost. You could also think of them as very hesitant left-ward walkers that sometimes stay put when it's their turn to move, or walk down the number line one step at a time. Saying they are unlucky was just meant as a narrative flourish. Sorry. $\endgroup$ – jreyes Feb 21 '13 at 18:36

Say each round a player either loses \$1 or breaks even. The amount player $A$ loses each round is a Bernoulli trial with $p=p_A(1-q_A)$. So the amount he loses in $N$ rounds is a sum of $N$ Bernoulli trials, or in other words a Binomial with parameters $N$ and $p_A(1-q_A)$.

The expected loss for player $A$ after $N$ rounds is therefore $Np_A(1-q_A)$. And the expected loss for player $B$ is $Np_B(1-q_B)$. The proportion of the total loss attributable to player $A$ at any given step is therefore $$\frac{Np_A(1-q_A)}{Np_A(1-q_A)+Np_B(1-q_B)}$$ $$=\frac{p_1(1-q_A)}{p_A(1-q_A)+p_B(1-q_B)}$$ This gives exactly 40% for your example so agrees with your simulation.


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.