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Problem 607

AMC 10/12, early questions

Combinatorics Difficulty 3.7 Multiple choice CEMC Fermat · Canada · 2015

Amina and Bert alternate turns tossing a fair coin. Amina goes first and each player takes three turns. The first player to toss a tail wins. If neither Amina nor Bert tosses a tail, then neither wins. What is the probability that Amina wins?

Pick one

Official solution

If Amina wins, she can win on her first turn, on her second turn, or on her third turn.

If she wins on her first turn, then she went first and tossed tails.

This occurs with probability $\frac{1}{2}$ .

If she wins on her second turn, then she tossed heads, then Bert tossed heads, then Amina tossed tails. This gives the sequence HHT. The probability of this sequence of tosses occurring is $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{8}$ . (Note that there is only one possible sequence of Ts and Hs for which Amina wins on her second turn, and the probability of a specific toss on any turn is $\frac{1}{2}$ .)

Similarly, if Amina wins on her third turn, then the sequence of tosses that must have occurred is HHHHT, which has probability $\left(\frac{1}{2}\right)^5=\frac{1}{32}$ .

Therefore, the probability that Amina wins is $\frac{1}{2}+\frac{1}{8}+\frac{1}{32}=\frac{16+4+1}{32}=\frac{21}{32}$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.