# Probability of drawing a red ball before a blue ball, after already drawing the first blue ball

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?

• It's not stated if you mean the chance, per the overall process, or, the chance, after we see the first blue. The two are totally different. Commented Apr 12, 2023 at 19:29

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

$$\frac{E\{R\}}{3+E\{R\}} = \frac{16}{31} \neq \frac{1}{2}$$

• Thanks for the response. I believe that each ball has equal but NOT independent probability of being in each red slot. But the thing is that linearity of expectation has nothing to do with independence so the expected number of red balls in the first slot is $1/5*4 = 4/5$ and the expected number of red balls afterwards is $4 - 4/5 = 3\frac{1}{5}$, and so I disagree with your $E[R]$. But your argument of $E[\frac{R}{3+R}] != \frac{E[R]}{3+E[R]}$ does make sense and that's probably where my logic is wrong. Commented Apr 12, 2023 at 14:32
• Yes, I had a calculation error. I do get $E[R] = 3 \frac{1}{5}$. Commented Apr 13, 2023 at 0:19
• Why don't you count the white balls? Commented Apr 13, 2023 at 13:27
• How do you know if didn't at least one red ball extract before the first blue ball? Commented Apr 13, 2023 at 13:30
• @VadimChernetsov The white balls have no effect on the answer. Only the order of the red and blue balls relative to each other matters. Discarding the white balls when they're drawn works out the same as removing them before we start drawing any. I don't understand your second question. "At least one red ball extract before the first blue ball" is the event $R>0$. Commented Apr 13, 2023 at 21:55

The probability that the ball immediately following the first blue ball is red is $$1/2$$.

We will disregard the white balls since they have no effect on the sequence of the blue and red balls. Number the red balls $$R_1, R_2, R_3, R_4$$, and similarly for the blue balls. We would like to find the probability that the ball following the first blue ball is $$R_1$$. There are $$8!$$ possible sequences of the blue and red balls, all of which are equally likely. We want to count the sequences in which the ball after the first blue ball is $$R_1$$. To do so, imagine removing $$R_1$$ from the bucket. There are $$7!$$ possible sequences of the remaing balls, and in each case there is only one place to insert $$R_1$$ so it immediately follow the first blue ball. So the probability that $$R_1$$ immediately follows the first blue ball is $$\frac{7!}{8!} = \frac{1}{8}$$

The same is true for $$R_2, R_3$$ and $$R_4$$; in each case, the probability that $$R_i$$ immediately follows the first blue ball is $$1/8$$. So the probability that a red ball follows the first blue ball is $$4 \times \frac{1}{8} = \frac{1}{2}$$

• Thanks. This is a good way to think about it. I was just confused about why my own logic did not work out and it seems that the previous answer shed some light on that. Commented Apr 12, 2023 at 14:34
• @wwyws Going from probabilities to expected value is straightforward, at least in theory, but reasoning from expected values to probabilities is tricky and prone to mistakes. Commented Apr 12, 2023 at 14:48