In article <3f4301\$rho@babyblue.cs.yale.edu> in rec.puzzles, David Moews <dmoews@fastmail.fm> wrote:

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I assume that you have to bet your bankroll one dollar at a time, so that it takes m gambles to dispose of a \$m bankroll, and that each chip you receive is worth one dollar.

ObPuzzles:

1. (easy) In the long run, do you expect to gain or lose?

Betting a dollar gets you an expected return of one dollar. Therefore, given your bankroll at time t, your expected bankroll at time t+1 is the same. In the long run, you neither gain nor lose.

2. (easy) There's nothing special about the number 100; it can WLOG tend to infinity without affecting much.
That's true. In that case, the return from a \$1 bet is a Poisson-distributed random variable with mean 1, so the return from m \$1 bets is the sum of m of these variables, which is Poisson with mean m.

3. (hard) Approximately what are the chances that you are bankrupt after t time units? [I assume that this is in the 100 -> infinity limit.]
From the above, if you start with \$m, the chance of having \$n at the next time unit is e^(-m) m^n/n! . From this, it's an easy induction that if you start with \$m, your chance of going bankrupt in t time steps is e^(-m alpha[t]), where

alpha[1]=1, alpha[t+1]=1-e^(-alpha[t]).

As t becomes large we have

alpha[t] = 2/t + O(log t/t^2),

so your chance of going bankrupt after t time units is

1 - 2/t + O(log t/t^2),

for big t.