2020 CIME II Problems/Problem 6

Problem

An infinite number of buckets, labeled $1$ , $2$ , $3$ , $\ldots$ , lie in a line. A red ball, a green ball, and a blue ball are each tossed into a bucket, such that for each ball, the probability the ball lands in bucket $k$ is $2^{-k}$ . Given that all three balls land in the same bucket $B$ and that $B$ is even, then the expected value of $B$ can be expressed as $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

The probability that all three balls land in box $2n$ is $\frac{1}{64^n}$ . This means that the probability that the three balls land in the same even box is $\frac{1}{64} + \frac{1}{64^2} + \ldots = \frac{1}{63}$ . This means that the probability that all three balls land in box $2n$ $\emph{given that they land in the same even box}$ is simply $\frac{63}{64^n}$ . For events $a$ , $b$ , $c$ , $\ldots$ and probabilities $P_a$ , $P_b$ , $P_c$ , $\ldots$ , the expected value when conducting the experiment is $P_aa + P_bb + P_cc + \ldots$ . Thus, our expected value is just

$\frac{2 \cdot 63}{64} + \frac{4 \cdot 63}{64^2} + \frac{6 \cdot 63}{64^3} + \ldots$ $= 2(\frac{63}{64} + \frac{63}{64^2} + \ldots) + 2(\frac{63}{64^2} + \frac{63}{64^3}) + \ldots$ $= 2 \cdot 1 + 2 \cdot \frac{1}{64} + 2 \cdot \frac{1}{64^2} + \ldots$ $= 2 \cdot \frac{64}{63}$ $= \frac{128}{63}.$

Our answer is $128 + 63 = \boxed{191}$ .

~mathboy100

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2020 CIME II Problems/Problem 6

Problem

Solution