"Patrick D. Rockwell" <prockwell@thegrid.net> wrote in message
news:41A50D39.9050103@thegrid.net...
> Here is a statement of the Gambler's Ruin Problem.
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> Let A be your amount of money, and B be how much money the house has.
> Let p be your probability of winning 1 game, and q=1-p.
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> If p<>p and p<>0 and q<>0, then Pr(of you going
> broke)=(1-(q/p)^B)/(1-(q/p)^(A+B))
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> BUT...
>
> if A>0 and B>0 and p=1, doesn't the formula fail? If p=1, then
> Pr=(1-(q/p)^B)/(1-(q/p)^(A+B))=1
> which implies that if your chances of winning 1 game are 100%, then the
> formula predicts that Pr(of you going broke)=100%.
Did you get the p's and q's backwords, or did I?
The formula above might make a bit more sense if we express it like so:
lose = 1-win;
ProbYouLose =
(1-(win/lose)^houseMoney)/(1 - (win/lose)^(yourMoney+houseMoney))
If win=1 then lose = 0 by definition. This means that the fraction
"win/lose" evaluates as 1/0. Debates rage on about what that expression
means; I tend to think of it as positive infinity, but I could be wrong. If
we dare to go there:
(1-(+inf)^houseMoney)/(1-(+inf)^(yourMoney+houseMoney) )
If we accept that infinity raised to a real power is still infinity, then
we can remove the power expressions to get:
(1 - (+inf)) / (1 - (+inf) )
Now without any training in this department, I can only suggest my hobbyist
opinion at this point that subtracting positive infinity from a real number
results in negative infinity. If this seems reasonable, (real - (+inf) ) =
(-inf), then we can reduce:
(1 - (+inf)) / (1 - (+inf) ) = -inf / -inf
And going into the even more bizarre waters, my guess is that:
(-inf / -inf) = +inf
For a final result of positive infinity.
As to writing a program to solve it, the terminal conditions that result in
infinity involve lose=0. So the valid range of the math part is if win<>1,
otherwise a div/0 error will result. You could trim any random input by
clamping the input range for win to [0,1], and then return 0 if win=1.
-:|:-
AngleWyrm
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