More coin flipping

After their last game, Alice and Bob decide to have a bit more fun with their fair coin. Bob proposes the following game: considering a flip can come out as heads or tails, they will both predict one sequence of three consecutive flips each (both sequences must be distinct). Alice picks her sequence first. Bob will then start repeatedly flipping the coin, until the last three flips match one of the predicted sequences, and the predictor will be declared the winner.

Alice doesn't have enough time to calculate all of the probabilities, so you probably shouldn't do that in the answer either. She just wants you to convince her of which moves are good enough. What sequences can she pick so she has at least a 50% chance of winning?

• To be clear, Alice chooses a sequence of 3 first, then Bob chooses his and then the flipping commences? Apr 8 '16 at 21:56
• @Reti43 Exactly.
– ffao
Apr 8 '16 at 21:58

She can't have more than 50% chance of winning. If she chooses a sequence, Bob can easily take the opposite sequence (change heads to tails and tails to heads) which has the same probability.

Moreover Bob can always find a better sequence as he can choose after her...

Let's see in details how Bob always wins ( I explore the cases with 0 or 1 Tail, by symmetry the other cases are the same )
• If Alice chooses HHH, Bob can choose THH. As soon as there is a T, the sequence THH will always appear before HHH. If HHH is not the first sequence, Alice looses.
• If Alice chooses HHT, Bob can choose THH (same reasoning as HHH)
• If Alice chooses HTH, Bob can choose HHT. They both need a sequence that starts with a H. If there is an other H (50%), Bob wins, but if there is a T, Alice does not always win (HTTHHT for example) so Alice has less than 50% chance of winning.
• If Alice chooses THH, Bob can choose TTH. They both need a sequence that starts with a T. If there is an other T, Bob wins but if there is an H, Alice does not always win (THTTH) so Alice has less than 50% chance of winning.

None: this is Penney's game, and the second picker always has an advantage because the win condition is nontransitive: that is, there are several instances of "Option A beats option B, which beats option C, which beats option A".

• +1, but the Wikipedia page only lists out the actual probabilities, which are hard to compute. I was looking for a simpler proof...
– ffao
Apr 8 '16 at 22:20