# Factoring risk of ruin into binomal calculation

Say I'm flipping coins and betting £1 on each flip. After 100 flips with infinite money, I have an 18.4% chance of being up by £10 or more (binomial with p = 0.5, n = 100, x = 55).

If I start with a finite amount of money, say £5, and cannot continue if I go broke, presumably my probability of finishing up £10 or more is less.

How would a statistician factor this into the model, to get a better estimate of probability of winning at least £X?

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What is the "x = 55" in the first paragraph? – gung Oct 9 '12 at 1:12
@gung number of successes; we require P(X >= x) – Luigi Plinge Oct 9 '12 at 1:33
If you continue until you reach £15 (up £10) or £0 (down £5), and then stop, then the probability of reaching £15 first is $\frac{5}{5+10} = \frac13$, provable by a simple induction. – Henry Oct 9 '12 at 7:12
@gung: $55 = 10 + \frac{100-10}{2}$ – Henry Oct 9 '12 at 7:26

Personally I would calculate the recursion, with $$P(X_{n+1}=x)= \frac12 P(X_{n}=x-1) + \frac12P(X_n=x+1)$$ but with $P(X_{n+1}=0)= P(X_{n}=0) + \frac12P(X_n=1)$ and $P(X_{n+1}=1)= \frac12 P(X_{n}=2)$, and starting at $P(X_{0}=5)=1$.
Then as far as I can see, $$P(X_{100} \ge 15)= \sum_{x=15}^{105} P(X_{100} =x)\approx 0.1556568.$$
You also have $P(X_{100} = 0) \approx 0.6172994.$