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From: "Paul Beekhuizen" <news@pibb.demon.nl>
Newsgroups: alt.gambling
Subject: Re: Gambler's Ruin Problem
Date: Thu, 25 Nov 2004 21:51:49 +0100
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"Patrick D. Rockwell" <prockwell@thegrid.net> schreef in bericht
news:41A50D39.9050103@thegrid.net...
> Here is a statement of the Gambler's Ruin Problem.
>
> 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.
>
> If p=q=0.5, then Pr(of you going broke)=(B/(A+B))
>
> If p<>p and p<>0 and q<>0, then Pr(of you going
> broke)=(1-(q/p)^B)/(1-(q/p)^(A+B))
>
I don't think this formula is correct. Try A=B=1, then
(1-(q/p)^B)/(1-(q/p)^(A+B)) = [1-(1-p)/p] / [1-((1-p)/p)^2]
= [-1/p] / [ 1 - (1-2p+p^2)/p^2 ]
= [ -1/p ] / [ (p^2 -1 + 2p -p^2)/p^2 ]
= [ -1 / p ] / [ (-1 + 2p ) /p^2]
= -p / (1-2p ),
while obviously P(you go broke) = P( you lose the first and only game) =
1-p. Moreover, if 0<p<1/2, then -p / (1-2p )<0.
> If A=0 and B>0 then Pr(of you going broke)=1
>
> 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%.
>
> I'm thinking of writing a program for Gambler's Ruin, but I'm wondering,
> how would you
> define the value of Pr for
>
> p=1, A=0, B=0 or p=0, A=0, B=0, or p=0.5, A=0, B=0?
>
This is just a matter of definition. You'd best exclude these cases.
> I ask because in the above cases, A=0 which means you're broke,
> but B=0 so, the house is broke too. Would that mean that
> Pr=1 because your broke at the start regardless of what
> the house has?
Yes, but also P(you go broke) = P (house goes broke) =1. Normally we have
that P(you go broke) + P(house goes broke) =1, because P(you go broke and
house goes broke) =0. However, in this special case, it holds that P(you go
broke and house goes broke) =0, simply because you and the house ARE both
initially broke.
>
> How about this one.
>
> p=1, A=10, B=0?
>
>
P(you go broke)=0, P(house goes broke)=1, because the house IS already
broke.
Although you could also say that P(house goes broke) = 0, because the house
is broke and therefore it cannot GO broke anymore. It just depends on how
you interpret "going broke."
Kind regards,
Paul Beekhuizen
>
>
>
>
> --
> Patrick D. Rockwell
> prockwell@thegrid.net
> hnhc85a@prodigy.net
> patri48975@aol.com
>
>
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