2022 SSMO Speed Round Problems/Problem 1

Problem

Bobby is bored one day and flips a fair coin until it lands on tails. Bobby wins if the coin lands on heads a positive even number of times in the sequence of tosses. Then the probability that Bobby wins can be expressed in the form $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Consider the probability $P($ win $)$ as the sum of the probabilities of all sequences where Bobby wins: $P($ win $)=P(2$ heads and then 1 tails $)+P(4$ heads and then 1 tails $)+$ $P(6$ heads and then 1 tails $)+\ldots$

For any sequence with $2 k$ heads followed by a tail, the probability is: $\left(\frac{1}{2}\right)^{2 k} \times\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2 k+1}$

We sum this for $k=1,2,3, \ldots$ : $P(\text { win })=\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{2 k+1}$

Factor out the constant term $\frac{1}{2}$ : $P(\text { win })=\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^k$

This is a geometric series with the first term $a=\left(\frac{1}{4}\right)$ and common ratio $r=\left(\frac{1}{4}\right)$ $\sum_{k=0}^{\infty} r^k=\frac{a}{1-r}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Thus: $P(\text { win })=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$

The probability $P($ win) can be expressed as: $\frac{1}{6}$

In this case, $m=1$ and $n=6$ . Therefore, $m+n=1+6=7$ . Thus, the value of $m+n$ is: $\boxed{\text{7}}$

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2022 SSMO Speed Round Problems/Problem 1

Problem

Solution