Expected flips of a biased coin until the first tail

Question

A biased coin lands heads with probability 0.75 and tails with probability 0.25. You flip it until it lands tails for the first time. What is the expected number of flips?

Accepted Answer

Let E be the expected number of flips until the first tail. The "success" here is a tail, which has probability p = 0.25. Condition on the first flip: - With probability 0.25 it is a tail, and we are done after 1 flip. - With probability 0.75 it is a head: one flip used, and independence puts us back at the start, expecting E more. E = 0.25(1) + 0.75(1 + E) \implies E = 1 + 0.75E \implies 0.25E = 1, so E = 4. Equivalently, E = 1/p = 1/0.25 = 4. Takeaway. Identify which outcome is the "success" you are waiting for, then the expected wait is 1/p for that outcome; a rare event (p = 0.25) takes 4…

Expected flips of a biased coin until the first tail

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