# Prove that the chance of ever getting more heads than tails with an unfair coin is 1/(p-1) where the chance of getting a head with each toss is 1/p

Imagine a game scenario in which you toss a coin indefinitely until the cumulative number of heads exceeds the cumulative number of tails, upon which you stop. Given the general case that there is a $\frac{1}{p}$ chance of getting heads on each toss, what is the probability that the game will EVER stop?

It is hypothesized that there is a $\frac{1}{p-1}$ chance of ever achieving more heads than tails, when $p\geq 2$(doesn't need to be an integer). This is on the basis of spreadsheet experiments.

Can anybody prove this? Thanks.

• Condition on the first result: it is heads with probability $1/p$ and then you win with probability $1$, and it is tails with probability $1-1/p$ and then you have to win twice. Thus the probability $w$ to win solves $$w=1\cdot1/p+w^2\cdot(1-1/p),$$ that is, $(w-1)((p-1)w-1)=0$. If $p>2$, there is positive probability to never win hence $w\ne1$, which implies that indeed $w=1/(p-1)$. If $1\leqslant p\leqslant2$, $w=1$ (only root between $0$ and $1$). This avoids summing infinite series or requiring extrinsic knowledge. – Did Jun 26 '15 at 13:12

## 1 Answer

We just have to count the number of strings over the alphabet $\{H,T\}$ with the property that the number of heads is exactly one more than the number of tails, and in every prefix the number of heads is $\leq$ the number of tails. Let $E$ be such a set. This is clearly related with the ballot theorem and Catalan numbers.

In particular, there are $\frac{1}{n+1}\binom{2n}{n}$ elements of $E$ with length $2n+1$, every one of them occurring with probability $\frac{1}{p^{n+1}}\left(1-\frac{1}{p}\right)^n$. So, assuming $p\geq 2$, we have that the probability that the game stops, soon or later, is given by:

$$\frac{1}{p}\sum_{n\geq 0}\frac{1}{n+1}\binom{2n}{n}\left(\frac{1}{p}-\frac{1}{p^2}\right)^n =\frac{1}{p-1}$$ as conjectured.

• Thanks for this proof. I understood most of it, but could you elaborate on the last part, where I haven't been able to see how LHS simplifies to RHS. – Thomas Delaney Jun 27 '15 at 1:06
• @ThomasDelaney: oh, sure. That happens because: $$\sum_{n\geq 0}\frac{1}{n+1}\binom{2n}{n}x^n = \frac{1-\sqrt{1-4x}}{2}.$$ Plugging in $x=\frac{1}{p}-\frac{1}{p^2}$ and simplifying you have my claim. You can find many proof of the last line here on MSE. – Jack D'Aurizio Jun 27 '15 at 9:06