Posted by Richard Reid on June 03, 1997 at 17:40:55:The General Risk of Ruin formula in Allan Wilson's "The Casino Gambler's Guide" determines the probability of losing ones entire bankroll before doubling ones bankroll. I've been thinking that perhaps a Goal Oriented Risk of Ruin (ROR) formula that allows one to determine the ROR for a user specified bankroll percentage might come in handy. In other words, there may be times when one wishes to determine the probability of losing z% of their bankroll before winning y%. Equation [19] will allow a person to calculate a Goal Oriented ROR for these kinds of questions.
We start the proof presuming as Wilson does in "The Casino Gambler's Guide", that the solution to the difference equation can be shown in the following simplified form:
[1] r (x) = A + B (1/S)^x
where:
p = probability of winning on a single play
q = probability of losing on a single play
S = p/q
A and B are arbitrary constants that depend upon:
1) the initial capital of the player,x = the amount of capital the player has at any given point in time
2) the amount the player is willing to lose and
3) the amount the player wishes to win
Let's set up a game where a player starts with 'a' units of capital and his opponent starts with 'b' units of capital.
What we are looking for is the equation that tells us the probability of the player losing 'c' units of his own capital before winning the opponents 'b' units.
For any a >= b and a>=c, if the player starts with 'a' units and loses 'c' of those units, then he can be said to have immediately lost the game. In other words, if his capital gets reduced by 'c' units, then the probability of loss is 1.
If the player starts with 'a' units and wins his opponents 'b' units, then the game is immediately won. In other words, if the player's capital increases by 'b' units, then the probability of loss is 0.
We can state things mathematically as follows:
[2] r(a-c) = A + B (1/S)^(a-c) = 1
[3] r(a+b) = A + B(1/S)^(a+b) = 0
If we let b = y*a, where y is the percentage of 'a' that we wish the player to win and if we let c = z*a, where z is the percentage of 'a' that gives us the portion of bankroll we can lose. Then, we can rewrite the equations as follows:
[4] r(a-za) = A + B (1/S)^(a-za) = 1
[5] r(a+ya) = A + B(1/S)^(a+ya) = 0
Now we have to solve for A and for B.
From equation [4] we get:
[6] B = (1 - A) / (1/S)^(a-za)
Substituting [6] into [5] we get:
[8] A + [(1 - A) / (1/S)^(a-za)] * (1/S)^(a+ya) = 0
Multiplying both sides of [8] by (1/S)^(a-za) gives us
[9] A * (1/S)^(a-za) + (1 - A) * (1/S)^(a+ya) = 0
[10] A * (1/S)^(a-za) + (1/S)^(a+ya) - A * (1/S)^(a+ya) = 0
[11] A * [(1/S)^(a-za) - (1/S)^(a+ya)] = -(1/S)^(a+ya)
[12] A = -(1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^(a+ya)]
Now, substituting A from [12] into [4] gives us:
[13] -(1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^(a+ya)] + B (1/S)^(a-za) = 1
Multiplying both sides of [13] by [(1/S)^(a-za) - (1/S)^(a+ya)] gives us
[14] -(1/S)^(a+ya) + B (1/S)^(a-za) * [(1/S)^(a-za) - (1/S)^(a+ya)] = [(1/S)^(a-za) - (1/S)^(a+ya)]
[15] B = [(1/S)^(a-za) - (1/S)^(a+ya) + (1/S)^(a+ya)] / [ (1/S)^(a-za) * [(1/S)^(a-za) - (1/S)^(a+ya)] ]
[16] B = (1/S)^(a-za) / [(1/S)^(a-za + a-za) - (1/S)^(a-za + a + ya)]
[17] B = (1/S)^(a-za) / [(1/S)^(2a-2za) - (1/S)^(2a + ya - za)]
Now if we substitute A from [12] and B from [17] into [1] and with x set to a, we get:
[18] r (a) = -(1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^(a+ya)] +
[ (1/S)^(a-za) / [(1/S)^(2a-2za) - (1/S)^(2a + ya - za)] ] * (1/S)^aRearranging [18] a bit gives us:
[19] ] r (a) = [ (1/S)^(2a-za) / [(1/S)^(2a-2za) - (1/S)^(2a + ya - za)] ] - (1/S)^(a+ya) / [(1/S)^(a-za) - (1/S)^(a+ya)]
[QED]