It’s always Alice and Bob. Why must it always be Alice and Bob? In any case, the two of them are tossing coins Until they hit a particular sequence: Alice until she hits a head then a tail, Bob until he hits two heads in a row.

Counter-intuitively, Alice will wait (on average) four tosses before she hits her target; Bob, by contrast, will take six.

This just seems wrong! Surely, since HT is just as likely as HH, the expected wait times should be the same?

### Why not (with algebra)?

Let’s let $h$ be the number of throws Alice expects to carry on throwing for, and $t$ be the number of throws after a tail.

After a head, Alice always throws at least once more; if she gets a tail, she finishes and if she gets a head, she’s back where she started - each of which happens half the time: $h = 1 + \frac{1}{2}h$, which means $h=2$.

Starting from a tail, she throws at least once more, getting a head or a tail with equal probability: $t = 1 + \frac{1}{2}h + \frac{1}{2}t$, so $t=4$.

Initially, she throws at least once, making $E(A) = 1 + \frac{1}{2}h +\frac{1}{2}t = 4$.

The only difference for Bob is in the first equation - the $h$ on the right becomes a $t$.

For Bob:

$h = 1 + \frac{1}{2}t$ $t = 1 + \frac{1}{2}h + \frac{1}{2}t$

Solving simultaneously gives $t=6$ and $h=4$ - and an expected wait time of six throws!

### Why not (with intuition)

Here’s a less formal explanation for why Bob can expect to throw for longer than Alice.

Think about the possible scenarios after any pair of throws: for Alice, after HT she has won; the other three are TH, HH and TT. In two of those three scenarios she has a chance of winning with the next throw.

As for Bob, he’s won after HH, but his remaining options are HT, TH and TT, only one of which allows him to win next time - so, generally speaking, Bob is less likely to be in a situation where he can win than Alice is.

* Many thanks to @cuttheknotmath for sharing the puzzle to begin with.