This problem is much easier to deal with if the presentation is
simplified. Let $.25 be one unit. Then wagers can be 1 unit or 2
units. A deposit of 20 units or 40 units is made initially, 80%
depositing 20 units and 20 % depositing 40 units. A player stops when
his account reaches 0 units (bust) or 100 units (cash out). If we let
p(n) denote the probability of cashing out when you have n units in
your account, then the answer to the problem is .8*p(20) + .2*P(40).
The problem is that we can't simply solve for p(20) and p(40), because
the different p(n)'s are so deviously interrelated. We can only solve
for one of the p(n) if we solve for all of them.
Again, for simplicity of presentation, let u = .97 and v = .03, and
probability of a win on each wager (if I understand the problem
correctly).
We get a recursion formula
p(n) = (u/2)*p(n+2) + (u/2)*p(n+1) + (v/2)*p(n-1) + (v/2)*p(n-2)
We know p(100) = p(101) = 1, and the account will never get more than 101.
Also p(0) = p(-1) = 0, because the game ends when you bust.
From the recursion formula we get p(99) = u + (v/2)*p(98) + (v/2)*p(97).
Also p(98) = u/2 + (u/2)*p(99) + (v/2)*p(97) + (v/2)*p(96). Substitute
the known formula for p(99) into this last equation, and we get a
formula of the form
p(98) = c + a*p(97) + b*p(96), where a, b, and c are constants in terms of u and v.
In general, we can get a second recursion formula of the form
p(n) = c + a*p(n-1) + b*p(n-2) for n = 99, 98, 97, ..., 1. If you
continue this process for several steps, then you may be able to
conjecture a pattern (and prove it). Then you can write down the
formula for p(1) without having to write down all 99 of the second
recursion formulas.
Once we have the second recursion formula for p(1), we can find its
value because p(0) and p(-1) are known (= 0). Then p(1) and p(0) are
known, so we can again use our second recursion formula to get the
value of p(2). Continue this process, and you can calculate p(n) for
any n up to 99. To answer your specific question, you can stop at
p(40).
I'll work out an answer next weekend if nobody else does it before then. |
