# 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?

• Hi. I have a similar problem, but the sequences are HTT and THH. My professor says it is a paradox as well, but I can't see how. I mean, each one has also a flat 50% chance of winning, unless we consider the value provided by the last flip. But he doesn't say if we should or not consider it. Do you have an idea of how this is a paradox? Thanks. – ddz Dec 13 '17 at 9:00
Failing to subtract the unknown entry fee from expected value $\frac{1}{2^n} \times 2^{n-1}$, before summing seems as obvious a blunder as blaming Isaac for newtonian determinism. The self-sustaining tunnel-teaching-fractal defies complexity at each iteration. Dave Arm.