Flip a fair coin 100 times, resulting in a sequence of heads (H) and tails (T). For each HH in the sequence, Alice gets a point; for each HT, Bob does, so e.g. for the subsequence THHHT Alice gets 2 points and Bob gets 1 point. Who is most likely to win?

An interesting conundrum in that the joint distribution of (A,B) need be considered for showing that Bob is more likely. Indeed, looking at the marginals does not help since the probability of the base events is the same. A solution on X validated (for a question posted when the Fiddler’s puzzle came out, Friday morn) demonstrates via a four state Markov chain representation the result (obvious from a quick simulation) that Alice wins 45% of the time while Bob wins 48%. The intuition is that, each time Alice wins at least a point, Bob gets an extra point at the end of the sequence (except possibly at the stopping time t=100), while in other cases Alice and Bob have the same probability to win one point.