Given we have $20$ balls in a bucket: $4$ blue, $4$ red, rest are all white. We draw randomly from the bucket without replacement until we see the first blue ball. Now we continue drawing. Which is more likely to happen first – seeing a red ball, or seeing another blue ball?
Now, I did some simulations and the answer seems to be that both are equally likely to happen. But I'm having trouble seeing this mathematically. My thought process is as follows:
After the first blue ball is drawn, the probability of drawing a red or another blue ball first is determined by how many red and blue balls are left in the bucket. Now, obviously there are only $3$ blue balls left. But the expected number of red balls left are $3\frac{1}{5}$. This is because, as shown below,
, we can think of the order of us drawing balls as a permutation of the balls, and so each of the red ball have equal likelihood of being placed in one of the red slots, and so expected number of red ball in each red slot is $4/5$, and thus the expected number of red balls after the first "B" is $4/5*4 = 3\frac{1}{5}$. So we are expecting more red balls than blue balls remaining in the bucket. Then shouldn't that mean we have higher probability of draw red ball first before anothehr blue ball?
Best Answer
For other ways to reason with this problem, see this very similar question about the card after the first queen in a deck of cards.
A straightforward approach: If $R$ is the number of red balls left after the first blue ball is drawn, we find
$$ \begin{align*} P(R=4) &= \frac{4}{8} \\ P(R=3) &= \frac{4}{8} \cdot \frac{4}{7} \\ P(R=2) &= \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{4}{6} \\ P(R=1) &= \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{4}{5} \\ P(R=0) &= \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5} \end{align*} $$
So the probability the next ball is red is
$$ P(N_R) = \frac{4 \cdot 4}{8 \cdot 7} + \frac{4 \cdot 4 \cdot 3}{8 \cdot 7 \cdot 6} + \frac{4 \cdot 3 \cdot 4 \cdot 2}{8 \cdot 7 \cdot 6 \cdot 5} + \frac{4 \cdot 3 \cdot 2 \cdot 4 \cdot 1}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4} = \frac{1}{2} $$
The "expected number of red balls left" is not useful for this question. Given a value of $R$, the probability the next draw is red is $P(N_R \mid R = r) = \frac{r}{3+r}$, so the overall probability is $P(N_R) = E\left\{\frac{R}{3+R}\right\}$. But since this is not a linear function of $R$, that's not generally the same as $\frac{E\{R\}}{3+E\{R\}}$
$$ E\{R\} = \frac{4 \cdot 4}{8} + \frac{4 \cdot 4 \cdot 3}{8 \cdot 7} + \frac{4 \cdot 4 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6} + \frac{4 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{8 \cdot 7 \cdot 6 \cdot 5} = \frac{16}{5} $$
which agrees with your reasoning. But
$$ \frac{E\{R\}}{3+E\{R\}} = \frac{16}{31} \neq \frac{1}{2} $$