2008 Mock ARML 1 Problems/Problem 4

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{align} 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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