A box contains two balls, one white and one black. Balls are…
A box contains two balls, one white and one black. Balls are drawn one at a time until the black ball appears; each time a white ball is drawn, it is returned to the box together with two more white balls. Find the probability that the black ball is not drawn in the first $50$ trials.
Explanation of Method
The composition of the box changes after every white draw, so the draws are dependent. We track the number of white balls before each trial: $1,3,5,\dots$ The probability that all first $50$ draws are white is a product of conditional probabilities (multiplication rule). The product of fractions $\frac{2k-1}{2k}$ can be written with a binomial coefficient and estimated with Stirling's formula.
Step 1: The box before each trial
At the start: $1$ white, $1$ black. After each white draw the white ball is returned and $2$ more are added, so the number of white balls grows by $2$. Given that trials $1,\dots,k-1$ were all white, before trial $k$ the box contains
\[ 2k-1 \text{ white balls and } 1 \text{ black ball}. \]
| Trial $k$ | White balls before the trial | $P(\text{white}\mid\text{earlier whites})$ |
|---|---|---|
| $1$ | $1$ | $\frac{1}{2}$ |
| $2$ | $3$ | $\frac{3}{4}$ |
| $3$ | $5$ | $\frac{5}{6}$ |
| $4$ | $7$ | $\frac{7}{8}$ |
| $k$ | $2k-1$ | $\frac{2k-1}{2k}$ |
Step 2: Multiplication rule
Let $W_k$ = "the $k$-th ball drawn is white". Then
\[ P(W_1W_2\cdots W_{50}) = \prod_{k=1}^{50}\frac{2k-1}{2k} = \frac{1\cdot3\cdot5\cdots99}{2\cdot4\cdot6\cdots100}. \]
Step 3: A closed form
Multiply numerator and denominator by $2\cdot4\cdots100=2^{50}\cdot50!$:
\[ \frac{1\cdot3\cdots99}{2\cdot4\cdots100} = \frac{100!}{(2^{50}\,50!)^2} = \frac{1}{2^{100}}\binom{100}{50}. \]
\[ \boxed{P = \frac{1}{2^{100}}\binom{100}{50}\approx0.0796.} \]
Step 4: Stirling's approximation
For large $n$, $\binom{2n}{n}\approx\dfrac{4^n}{\sqrt{\pi n}}$. With $n=50$:
\[ P\approx\frac1{\sqrt{50\pi}}\approx0.0798, \]
very close to the exact value.
Step 5: Interpretation
At first it seems that the black ball should appear quickly, because white balls "flood" the box. But the probability of surviving $n$ trials decreases only like $\frac1{\sqrt{\pi n}}$: after $50$ trials, the black ball has still not appeared in about $8\%$ of cases.
Result
| Answer | |
|---|---|
| Product | $\prod_{k=1}^{50}\frac{2k-1}{2k}$ |
| Probability | $\binom{100}{50}/2^{100}\approx0.0796$ |