 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2003/calculus-probability-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/5/b/ii/solution

Let

I

be the number of heads in the initial allocation tosses. Then

I

n

and

1/2

. Given

I = i

, the winning probability is

i / n

, including the already finished games

i = 0, n

P (W) = i = 0 \sum n P (I = i) \frac{i}{n} = \frac{E I}{n} = \frac{n /2}{n} .

Thus, for

n \geq 1

P (H finishes with all coins) = \frac{1}{2} .

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