Taking turns until head then tail
A and B take turns tossing one fair coin: A, then B, then A, and so on. The game stops the first time a tail comes right after a head in the sequence of tosses, and the player who tossed that tail wins.
Find the probability that A wins.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Once a head has appeared, the game ends at the very next tail. Who tosses it?
Two situations matter: the last toss was a head, or it was not (which includes the start). In each, find the probability that the player about to toss wins.
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Only one thing about the past matters: whether the last toss was a head. Write one equation for each of the two situations, from the point of view of the player about to toss.
After a head. Let be the probability that the player about to toss wins, given that the previous toss was a head. They win at once with a tail. With a head, the other player is now about to toss after a head, and wins with probability . So In particular, the player who has just tossed a head wins with probability .
Before any head. Let be the probability that the player about to toss wins when the previous toss was not a head (or at the start). With a head, they become the player who has just tossed a head: probability . With a tail nothing has changed, except that the other player is now about to toss, so they win with probability . So which gives , so .
Answer. A tosses first with no head yet, so A wins with probability and B with . Going first is a disadvantage: a head is needed before a tail can win, and whoever tosses the first head hands the winning toss to the opponent.