Marble Transfer: Conditional Probability
You have two bins. Bin 1 contains $r_1 = 4$ red and $b_1 = 3$ blue marbles. Bin 2 contains $r_2 = 5$ red and $b_2 = 3$ blue marbles.
You draw one marble uniformly at random from Bin 1 and drop it into Bin 2. Then you draw one marble uniformly at random from the now-enlarged Bin 2.
What is the probability that the marble you draw from Bin 2 is blue?
Hints
- The outcome from Bin 2 depends on what color marble was moved there -- split the problem into two cases based on the color of the transfer.
- Use the Law of Total Probability: $P(B_2) = P(B_2 \mid B_1)P(B_1) + P(B_2 \mid R_1)P(R_1)$.
- After the transfer, Bin 2 always has 9 marbles total. Compute how many are blue in each case, then weight by the probability of each transfer color.
Worked Solution
How to Think About It: The draw from Bin 2 depends on what you transferred -- if you moved a blue marble over, you boosted the blue count; if you moved a red, you did not. So the first thing to do is split on the color of the transferred marble. This is the Law of Total Probability in its most natural form: condition on the thing that creates the uncertainty, compute each branch, and weight by the branch probabilities.
A quick gut check before calculating: Bin 1 is mostly red (4 red, 3 blue), so most of the time you are adding a red marble to Bin 2, which makes the blue draw less likely. Bin 2 starts at $3/8 = 0.375$ blue. The transfer nudges that up or down depending on what moves. Since red is more likely to transfer, the answer should be slightly below 0.375. Let us see.
Quick Estimate: Probability of transferring blue from Bin 1: $3/7 \approx 0.43$. Probability of transferring red: $4/7 \approx 0.57$.
- If blue transfers: Bin 2 has 4 blue, 5 red (9 total). $P(\text{blue draw}) = 4/9 \approx 0.44$.
- If red transfers: Bin 2 has 3 blue, 6 red (9 total). $P(\text{blue draw}) = 3/9 = 0.33$.
Weighted average: $(0.43)(0.44) + (0.57)(0.33) \approx 0.19 + 0.19 = 0.38$. So the answer is around $0.38$, which is close to the baseline $3/8 = 0.375$ -- makes sense since the transfer is a wash in expectation (it does not systematically bias toward blue or red in Bin 2).
Approach: Apply the Law of Total Probability, conditioning on the color of the marble transferred from Bin 1.
Formal Solution:
Let $B_1$ be the event that the marble drawn from Bin 1 is blue, and $R_1$ the event it is red. Let $B_2$ be the event that the marble drawn from Bin 2 is blue.
From Bin 1 (7 marbles total): $$P(B_1) = \frac{b_1}{r_1 + b_1} = \frac{3}{7}, \quad P(R_1) = \frac{r_1}{r_1 + b_1} = \frac{4}{7}$$
After the transfer, Bin 2 always has $r_2 + b_2 + 1 = 9$ marbles.
- If a blue was transferred: Bin 2 has $b_2 + 1 = 4$ blue marbles. $$P(B_2 \mid B_1) = \frac{b_2 + 1}{r_2 + b_2 + 1} = \frac{4}{9}$$
- If a red was transferred: Bin 2 has $b_2 = 3$ blue marbles. $$P(B_2 \mid R_1) = \frac{b_2}{r_2 + b_2 + 1} = \frac{3}{9} = \frac{1}{3}$$
By the Law of Total Probability: $$P(B_2) = P(B_2 \mid B_1)P(B_1) + P(B_2 \mid R_1)P(R_1)$$ $$P(B_2) = \frac{4}{9} \cdot \frac{3}{7} + \frac{3}{9} \cdot \frac{4}{7} = \frac{12}{63} + \frac{12}{63} = \frac{24}{63} = \frac{8}{21}$$
The general formula is: $$P(B_2) = \frac{b_1(b_2 + 1) + r_1 b_2}{(b_1 + r_1)(b_2 + r_2 + 1)}$$
Answer: $P(B_2) = \dfrac{8}{21} \approx 0.381$
Intuition
Notice that in the numerical answer, both branches contribute exactly $12/63$ -- the blue-transfer and red-transfer terms are equal. This is not a coincidence. Transferring a marble from Bin 1 to Bin 2 before drawing is equivalent, in expectation, to just drawing the second marble from a combined pool. More precisely, the expected probability of a blue draw from Bin 2 equals the blue fraction after accounting for the stochastic transfer, and the two branches balance because you are equally likely to add a blue (boosting the draw probability) or a red (hurting it), weighted exactly right.
This type of conditioning -- split on the unknown, compute conditional probabilities, weight and sum -- is the workhorse of discrete probability and shows up constantly in quant problems: market making under adverse selection (condition on whether the counterparty has information), sequential sampling (condition on the first draw to get a recursion), and Bayesian updating (condition on observed data). Whenever you see a two-stage process where the first stage affects the second, reach for the Law of Total Probability first.