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Dec 20, 2018 at 16:21 answer added Iosif Pinelis timeline score: 1
Dec 20, 2018 at 16:14 comment added Anthony Quas It’s $1/(4N+4)$: this is $\mathbb P(h_N^+<h_N^-).\mathbb P(c_N^-<e_N^+|h_N+<h_N^-)$. The first term is $1/2$ and the second is the probability of hitting $-(N+1)$ before $N+1$ given that you started at $N$. This is exactly $1/(2N+2)$: look up hitting times for gambler’s ruin.
Dec 20, 2018 at 15:20 history edited Vilhelm Agdur CC BY-SA 4.0
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Dec 20, 2018 at 15:13 history asked Vilhelm Agdur CC BY-SA 4.0