"AngleWyrm" <no_spam_anglewyrm@hotmail.com> wrote in message
news:8Zppd.150351$R05.111548@attbi_s53...
> "Patrick D. Rockwell" <prockwell@thegrid.net> wrote in message
> news:41A50D39.9050103@thegrid.net...
>> Here is a statement of the Gambler's Ruin Problem.
> [clip]
>> 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.
> [clip]
>> If p<>p and p<>0 and q<>0, then Pr(of you going
>> broke)=(1-(q/p)^B)/(1-(q/p)^(A+B))
> [clip]
>> 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))
>
Your're right. I checked my old finite math book. The formula should be
(1-(p/q)^B)/(1-(p/q)^(A+B)).
which means that the formula DOESN'T fail at p=1 or p=0.
> 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.
Thanks. Hmm, If I had to guess, I'd have thought that inf/inf = 1 because
the limit as x goes to infinity of (x./x)=1.
--
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Patrick D. Rockwell
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