Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 11F / d

Solution

Let

E_{j}

be the expected additional number of tosses when the current terminal run contains

j

consecutive heads. Then

E_{k} = 0

, and for

0 \leq j < k

E_{j} = 1 + p E_{j + 1} + (1 - p) E_{0} .

For

0 < p < 1

, iterating this recurrence from

j = k - 1

down to

0

gives

E_{0} = (1 + (1 - p) E_{0}) \frac{1 - p ^{k}}{1 - p} .

Solving,

E_{0} = \frac{1 - p ^{k}}{( 1 - p ) p ^{k}} .

When

p = 1

, the waiting time is deterministically

k

, which is also the continuous limit of this formula as

p \to 1

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