Probability question

Question

Two players, A and B, alternately and independently flip a coin and the first player to obtain a head wins. Assume player A flips first

(a) if the coin is fair, what is the probability that A wins?

(b) Suppose that P(head) = p₁not necessarily 1/2, what is the probability that A wins?

(c) Show that for all p₁₀<p<1, P(A wins)>1/2 (hint: try to write P(A wins) in terms of the evens E_1,E_2,....., where E₁ = {head first appears on ith toss}）

Answer 1

P(A wins) = P(H) + P(TTH) + P(TTTTH) + P(TTTTTTH) + ... =

p + (1-p)²p + (1-p)⁴p + ...

if p = 1, of course, P(A) = 1

If p = 0, the game never stops.

Otherwise, this is a geometric sum with first term p and ratio (1-p)²

A geometric sum has value a/(1-r), where a is the first term and r is the ratio.

Thus, the sum is p/(1 - (1-p)²) = p/(1 - (1 - 2p + p²)) =

p/(2p - p²) = p/(p(2 - p)) = 1/(2 - p)

a) 1/(2 - p) = 1/(2 - 1/2) = 1/(3/2) = 2/3

b) P(A) = 1/(2 - p) for 0 < p < 1, 1 for p = 1, and does not exist for p = 0

c) for 0 < p < 1, 1 < 2 - p < 2, and 1/2 < 1/(2 - p) < 1

Thus, 1/2 < P(A) < 1