What's the probability of tossing two heads if the game started with a head?

Question

A player is tossing an unbiased coin until two heads or two tails occur in a row. What's the probability of heads winning the game if the game started with a head?

I looked at Two tails in a row - what's the probability that the game started with a head? and came up with this solution, but I'd like someone to confirm that it's correct or show some alternative/more elegant solution:

First toss was a head. Now there is an infinite number of ways how the game can end with heads winning:

• $H$ — probability is $0.5$
• $THH$ - probability is $0.5^3$
• $THTHH$ - probability is $0.5^5$
• $THTHTHH$ - probability is $0.5^7$
• $\ldots$

So, total probability of heads winning the game is: $$P(H) = \sum_{i=0}^{\infty}\frac{1}{2^{2i+1}} = \frac{0.5}{1 - 0.25} = \frac{2}{3}$$

I also checked that probability of tails winning is $\frac{1}{3}$ (calculated using pretty much the same method), so that's at least some sanity.

Answer 1

This is correct, but an easier way to see this is as follows: let $p$ be the probability of $H$ winning if an $H$ was just thrown. That means that $1 - p$ is the probability of $T$ winning when an $H$ was just thrown. (Implicit assumption: the probability that the game ends is 1.) By symmetry, the probability of $H$ winning when a $T$ was just thrown must also be $1-p$. Now if the first throw is an $H$, with probability $\frac12$ the next throw is also an $H$ (and $H$ wins); and with probability $\frac12$ the next throw is $T$, and then $H$ wins with probability $1-p$. Thus, $$ p = \frac12 + \frac12(1-p). $$ Now solving for $p$ gives $p = \frac23$.

Answer 2

If $p$ is the probability you're looking for, then $p= 0.5 + 0.5(1-p)$. Can you see why?