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Gambler's ruin: biased dice

Problem

Say you start with a capital of 25. Then you fight your rival. If you win you add 1 to what you have, if you lose you subtract 1 from what you have. If you reach 0 you’re extinct. If your probability of winning is 49.5% calculate ~ the median number of fights you will last before going extinct. This will teaches you something about survivorship, background extinction and the fate of empires.

Notation

P(increase by 1) = P(+1) = .495
P(decrease by 1) = P(-1) = .505
C(n) = capital at step n. C(0) = 25.
\(C(n) → 0)\ is an event where at step n, capital becomes 0 but not before.

Expected number of turns to ruin

E[capital after 1 turn] = 25 -.01
E[capital after n turns] = 25 - .01n
For n=2500, ruin in expected.

Median

Simulation

R code here.

Theoretical bound attempt

Statement: Find n such that Pr(capital C(n)=0) = 1/2.
For n<25, Pr(C(n)=0) = 0
For n=25, \(Pr(C(n)=0) = P(-1)^{25}\)
For n=25+k where k is odd, \(Pr(C(n) → 0\) is impossible.

Proof

Beyond 25 losses, it is impossible to have exactly k/2 wins as it won’t be a whole number.

At n=25+2k, C(n) becomes 0 only if there are k wins and 25 + k losses, and if for any t<n, C(t) was not 0. An upper bound on the number of such events is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). We consider this n=25+2k case below.
\(P(C(n) → 0) \leq \binom{n}{k}P(-1)^{25+k}P(+1)^{k}\)
\(P(C(n) → 0) \geq P(-1)^{25+k}P(+1)^{k}\)
- Find m for which \(\sum_{m}^{k=1} P(-1)^{25+k}P(+1)^{k} = 0.5\).
TODO: incomplete