Question

A and b toss a coin, the sequence ht is required such that whoever throws a tail in the last will win. What is the probability of a to win, if a start the toss?

Answer 1

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The game begins with player A, who either tosses a head or a tail. By the total probability theorem:

P(A)=12P(A∣H)+12P(A∣T)

If A first tosses a tail, then the game repeats with player B starting first. By symmetry then:

P(A∣T)=1−P(A)

If A first tosses a head, then the next player to toss a tail wins. If B does not do so on the next toss, then at that point B has the same probability of winning as A did before, thus: P(A∣HH)=1−P(A∣H) and hence:

P(A∣H)=12(1−P(A∣H))

#answerwithquality #bal

Answer 2

By applying the total probability theorem:

P(A)=12P(A∣H)+12P(A∣T)

If A first tosses a tail, then the game repeats with player B starting first. By symmetry then:

P(A∣T)=1−P(A)

If A first tosses a head, then the next player to toss a tail wins. If B does not do so on the next toss, then at that point B has the same probability of winning as A did before, thus: P(A∣HH)=1−P(A∣H) and hence:

P(A∣H)=12(1−P(A∣H))