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2008 Mock ARML 1 Problems/Problem 4

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Problem

There are 4 black balls and 1 white ball in a hat. A turn consists of picking a ball from the hat and replacing it with one of the opposite color. Compute the probability that, after a sequence of turns, there are 5 black balls in the hat before there are 5 white balls.

Solution

Let f(n) denote the probability of reaching 5 black balls before 5 white balls from a position of n black balls and 5-n white balls. The probability that we have 5 black balls after a turn is \frac 15 (note that f(5) = 1), and that we have 3 black balls is \frac 45. Similarly, if we start with 3 black balls, after a turn there is a \frac 25 probability of ending with 4 black balls and a \frac 35 probability of ending with 2 balls. Thus, we have the recursions \begin{align}f(4) &= \frac 15 \cdot [1] + \frac 45 \cdot f(3) \\f(3) &= \frac 25\cdot f(4) + \frac 35 \cdot f(2) \end... By symmetry, f(2) = 1-f(3); substitution into (2) yields 8f(3) = 2f(4) + 3. Now, (1) and (2) form a two equation linear system which can be solved to find that f(4) = \boxed{\frac{5}{8}}.

See also

2008 Mock ARML 1 (Problems, Source)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8
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