Problem 7.1 Easy Free

Conditioning on Past Coin Flips

A fair coin is flipped repeatedly, and $H_n$ denotes the number of heads among the first $n$ flips. Compute $\mathbb{P}[H_9 = 6 \mid H_6 = 5]$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Write $H_9 = H_6 + (H_9 - H_6)$ , where $H_9 - H_6$ counts the heads among flips $7$ , $8$ , $9$ . These three flips are independent of the first six, so $H_9 - H_6 \sim \mathrm{Bin}(3,\tfrac12)$ independently of $H_6$ . Hence

\mathbb{P}[H_9 = 6 \mid H_6 = 5] = \mathbb{P}[H_9 - H_6 = 1 \mid H_6 = 5] = \mathbb{P}[H_9 - H_6 = 1] = \binom{3}{1}\frac{1}{2^3} = \frac{3}{8}.

\boxed{\tfrac{3}{8}}

Key idea. Increments over disjoint blocks of flips are independent, so conditioning on $H_6$ leaves a fresh $\mathrm{Bin}(3,\tfrac12)$ count for the remaining flips.

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