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This is really a poker tournament/payout/bankroll question but I think I can frame it in a dice game for easier understanding. If a player starts with $X$ units, and plays a game where on each throw the cost/payouts are:

1,2 or 3 = -1

4: = 0

5: = +1

6: = +3

If the players wealth is never allowed to rise above $X$, ie, hits a 6 with a current wealth of $(X-1)$ the new wealth is only $X$, the extra two units are ditched. Is there an algorithm/formula to predict/plot the chance of ruin over many throws?

There is a Risk of Ruin formula for when the wealth is unrestricted and this, I think, is

$RoR(X) = e^{-aX}$, where $a$ is a constant related to the relative likelihood and expectations but is there a simple formula for when the max wealth is capped?

Or any clues to how to find a reasonable solution?

In the real results I would like to solve there are more outcomes - perhaps 20 or so +ve outcomes out of 100.

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If the player keeps playing, she will be ruined with probability $1$. – André Nicolas May 15 '12 at 20:46
True, ruin is certain with enough throws, I should have asked for the probability of survival after n throws. – Pete M May 15 '12 at 21:17
A more or less explicit procedure for determining this has been supplied by Ross Millikan. From this you can determine a fairly simple dominant term for the probability that you ask for. – André Nicolas May 15 '12 at 23:14
up vote 2 down vote accepted

One approach is to make a stochastic matrix where $P_{i,j}$ is the probability of transitioning from $i$ to $j$. For $i$ and $j$ far enough from $0$ and $X$, you will have $P_{i,i-1}=1/2,P_{i,i}=1/6,P_{1,i+1}=1/6,P_{i,i+3}=1/6$ there will be modifications at the ends. Your starting vector has $1$ in the last position and $0$ everywhere else. Then the probability distribution after $n$ plays is $P^n$ times your starting vector. The calculation can be eased if you diagonalize the matrix so $P=SDS^{-1}$ where $D$ is diagonal. Then $P^n=SD^nS^{-1}$

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Thanks Ross. I'll look into this. – Pete M May 16 '12 at 10:19

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