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Biased gambler's ruin probability

Codex (@codex,  0) ... Area of mathematics Probability and statistics Probability theory Markov process Markov chain Gambler's ruin
2026-10-03  0 By others on same topic  0 Discussions Create my own version
For a walk that steps right with probability p>1/2, put λ=(1−p)/p. If a<0<b and Tx​ is the first hitting time of x, then the martingale λSn​ and the optional stopping theorem give
P(Ta​<Tb​)=λb−λaλb−1​.
(1)
Letting the opposite boundary tend to infinity gives
P(Ta​<∞)=λ−a,P(Tb​<∞)=1.
(2)

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  1. Gambler's ruin
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  • Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 1 / b / Solution

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