Alternating coin toss – redux

I want to revisit a probability puzzle from a previous post: see Alternating coin toss game.

You play a game where you alternate tossing a coin with a friend, and the first person to toss heads wins. But let’s make it a little more interesting by making the coin weighted, or biased, so that it lands heads only 25% of the time. If you were to start, what would be your chances of winning?

Let’s call the probability of tossing heads $p$ and the probability of tossing tails $q$.

On the starter’s first toss, his chances of tossing heads are $p$. The chances he’ll toss heads on his second toss are $q^2\times p$. On his 4th toss it’s $q^4\times p$ and so on.

So his chances of winning the game is the sum $s$ of the geometric progression $p\times q^0\;+\;p\times q^2\;+\;p\times q^4\;+$…

Let’s use the shifting approach to solve this. The idea is to subtract out the first element, and then shift the rest of the elements to the left. So first subtract:
$s-p\times q^0=p\times q^2\;+\;p\times q^4\;+$…

Now to shift the terms to the left, we need to divide by $q^2$
$\frac{s-p\times q^0}{q^2}=p\times q^0\;+\;p\times q^2\;+$…

So…
$\frac{s-p\times q^0}{q^2}=s$
$s=\frac{p}{1-q^2}$

This gives us a general solution to the alternating coin toss problem. If we plug in the unweighted coin probabilities, we get $s=\frac{2}{3}$ which is what we’d worked out before. For the case where the probability of tossing a head is 1/4, then we get $s=\frac{4}{7}$, which is the probability of winning for the starter.