# A seeming paradox in a coin-flipping game

This is related to my other question on a similar topic.

Suppose we play the following game: we flip a coin repeatedly and record the outcomes. For example we might get HHTTTHTTHHTTT.... Now Alice and Bob each choose distinct patterns of the same length, called $A$ and $B$, respectively. Which ever player's pattern appears first wins the game.

Now suppose Alice chooses $A=$HHHH and Bob chooses $B=$HHHT. First, let's notice that the expected number of flips to obtain $A$ is 30, but for $B$ it's only 16. This would seem to imply that Bob is very likely to win this game most of the time.

However, thinking about the game in another way, neither Alice nor Bob can win until HHH occurs. And after this, the game ends on the next flip with each player winning equiprobably.

This seems counter-intuitive to me: we have two events, one expected to occur much sooner than the other, but relative to each other the ordering is 50-50. What am I missing?

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