Last White Kitten with Three Black Remaining
A box contains $15$ kittens: $8$ black and $7$ white. You draw kittens one at a time uniformly at random (without replacement) until the box is empty.
What is the probability that when the last white kitten is drawn, exactly $3$ black kittens remain in the box?
Hints
- Think of the draw sequence as a random permutation of $8$ B's and $7$ W's. What pattern must the last four positions follow?
- The sequence must end in $WBBB$: position $12$ is the last white kitten drawn, and positions $13$-$15$ are the remaining black kittens.
- Count the favorable arrangements of the first $11$ positions ($6$ W's among $11$ slots) and divide by the total number of arrangements $\binom{15}{7}$.
Worked Solution
How to Think About It: Think about the full sequence of $15$ draws as a random permutation of $8$ B's and $7$ W's. The condition "the last white kitten is drawn with exactly $3$ black remaining" means: the last $3$ positions in the sequence are all B, position $12$ is W, and the first $11$ positions contain the remaining $6$ W's and $5$ B's. So this is a straightforward counting problem.
Quick Estimate: The last white kitten is drawn at position $12$ (out of $15$). Roughly, the probability that position $12$ is the last W is moderate -- there are $15$ possible positions for the last W, and it needs to land at exactly position $12$. Quick ballpark: something on the order of $1/15$ to $1/10$, so maybe $5\%$-$10\%$.
Approach: Count favorable permutations over total permutations using combinations.
Formal Solution:
The total number of distinguishable sequences of $8$ B's and $7$ W's is:
$$\binom{15}{7} = 6435$$
For the favorable outcome, we need the sequence to end in $W B B B$ (positions $12, 13, 14, 15$), where position $12$ is the last W and positions $13$-$15$ are the last $3$ B's.
The first $11$ positions must contain exactly $6$ W's (the remaining whites) and $5$ B's (the remaining blacks). The number of ways to arrange these is:
$$\binom{11}{6} = 462$$
(Choose $6$ of the $11$ positions for W; the rest are B.)
The probability is:
$$P = \frac{\binom{11}{6}}{\binom{15}{7}} = \frac{462}{6435} = \frac{14}{195} \approx 0.0718$$
Verification: $462 / 6435 = 462 / 6435$. Simplify: $\gcd(462, 6435) = 33$, so $462/33 = 14$ and $6435/33 = 195$. Thus $P = 14/195$.
Answer: $\dfrac{14}{195} \approx 7.2\%$.
Intuition
The key insight is to reframe the drawing process as a random permutation. Instead of thinking about a sequential process, think about all $\binom{15}{7}$ equally likely arrangements of colors. The event "last white kitten drawn with 3 black remaining" translates to a specific structural constraint on the permutation: it must end $WBBB$. Once you see this, the counting is straightforward -- you just need to count how many ways to fill the unconstrained positions. This "permutation reframing" trick is extremely useful for problems about order statistics in sampling without replacement. Whenever a problem asks about the position of the last (or first, or $k$-th) occurrence of something in a random sequence, think about the full permutation rather than the sequential draws.