Two Boxes, a Die and a Transferred Ball
Two boxes stand on a table in a school mathematics club demonstration. Box A contains 2 red balls and 1 blue ball; Box B contains 1 red ball and 3 blue balls. A fair six-sided die is rolled once. If the result is 1, 2 or 3, a ball is chosen uniformly at random from Box A and moved to Box B; if the result is 4, 5 or 6, a ball is chosen uniformly at random from Box B and moved to Box A. After the transfer, one ball is drawn uniformly at random from Box B and its colour is recorded. The club scores 10 points if that final ball is red and loses 4 points if it is blue. One member claims that because the die is fair and the procedure looks symmetric, the final colour is essentially decided by Box B's original composition and the die roll can be ignored. Another member wants to know, given that a red ball was drawn at the end, how likely it is that the transfer went from Box A to Box B.
1. Find the probability that the final ball drawn from Box B is red.
Read the worked answer
Split on the transfer, then on the colour transferred.
Transfer A → B (probability 1/2). Box A has 2 red, 1 blue.
- Red moved (2/3): B becomes 2 red, 3 blue, so P(red) = 2/5.
- Blue moved (1/3): B becomes 1 red, 4 blue, so P(red) = 1/5. Contribution: (1/2)[(2/3)(2/5) + (1/3)(1/5)] = (1/2)(4/15 + 1/15) = (1/2)(1/3) = 1/6.
Transfer B → A (probability 1/2). Box B has 1 red, 3 blue.
- Red moved (1/4): B becomes 0 red, 4 blue, so P(red) = 0.
- Blue moved (3/4): B becomes 1 red, 3 blue, so P(red) = 1/4. Contribution: (1/2)[(1/4)(0) + (3/4)(1/4)] = (1/2)(3/16) = 3/32.
Total: 1/6 + 3/32 = 16/96 + 9/96 = 25/96 ≈ 0.260.
The die cannot be ignored: the two branches give quite different red probabilities, and the member's appeal to symmetry confuses a symmetric procedure with equally likely outcomes.
2. Given that the final ball was red, find the probability that the transfer was from Box A to Box B.
Read the worked answer
Use the conditional-probability definition with the pieces already computed.
P(red and transfer A → B) = (1/2)(1/3) = 1/6 = 16/96. P(red) = 25/96.
Therefore
P(A → B | red) = (16/96)/(25/96) = 16/25 = 0.64.
Interpretation: although the two transfer directions are equally likely a priori, a red final ball is evidence for the A → B branch, because Box A is the red-richer box and moving a red ball from it raises Box B's red share. The remaining probability 9/25 belongs to the B → A branch, which produces red only when a blue ball is moved out of B.
3. Find the expected score for one round and say whether the club should play.
Read the worked answer
The score is 10 for red and −4 for blue, with P(red) = 25/96 and P(blue) = 1 − 25/96 = 71/96.
E = 10·(25/96) + (−4)·(71/96) = (250 − 284)/96 = −34/96 = −17/48 ≈ −0.354.
Since the expectation is negative, the game loses points on average, so a rational club declines to play unless the point values are renegotiated. The break-even condition would be 10p − 4(1 − p) = 0, i.e. p = 4/14 = 2/7 ≈ 0.286; the actual red probability 25/96 ≈ 0.260 falls short, which is exactly why the expectation is negative. A common error is to average 10 and −4 as +3; that ignores the unequal probabilities.