Solving Gambler's Ruin with Martingale and Electric Networks

[@LetsSolveMathProblems]

Correction: For the conditional expectation at 4:35, to use the correct definition of a martingale, we should condition on all the preceding terms (so (q/p)^(S_0), ..., (q/p)^(S_(n-1))), not just the immediately preceding term. Despite this, the same argument goes through since X_n is independent of all such previous terms.

[@sajidrizvi4665]

I'm sure I'm gonna enjoy this one!

[@circe3976]

It was very enjoyable to watch, thank you.

[@ccolombe]

Love your videos man! Glad to see another in the feed

[@replicaacliper]

I think at 7:00 its also crucial that the probability of ruin/winning is nonvanishing at T increases, i.e. you can put a lower bound on the probability that is independent of T. If the probability of escaping decreased quick enough with T then I'm sure you can have a nonzero probability of having an infinite sequence

[@LetsSolveMathProblems]

I'm guessing you meant T as in just the generic time step (not stopping time), and in that case, you are right that it is crucial that the probability of ruin/winning is bounded away from 0 at any time step (which is true here since the entire "game board" is finite).

[@replicaacliper]

@@LetsSolveMathProblems indeed

[@joeblogg9937]

There is a limit as r->1. 2/5 ?

[@LetsSolveMathProblems]

Yes, that's correct! In the electric network argument, we do not even have to take the limit (and try to justify the limit we get is actual probability at r = 1): Just plug in r = 1 into expressions for x and y (see 21:55).