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Bertrand’s box paradox
You will learn: Compute the posterior under random drawing and deliberate showing protocols.
Start with: Conditioning and independence
Three boxes contain two gold coins, one gold and one silver coin, and two silver coins. Choose a box uniformly, choose one of its coins uniformly, and observe gold. What is the probability that its other coin is gold?
The tempting answer is one half: the silver-only box is impossible, leaving two boxes. But those boxes were not equally likely to produce the observation. The gold-only box offered two chances to see gold; the mixed box offered one. Under this experiment, the answer is two thirds.
Look at the observation weights
Evidence probability under the original sampling scheme: 0.5000000.
Filled left bars: original probability. Open right bars: probability after the stated observation. The same underlying outcomes remain in the table, including excluded ones.
| Outcome | Prior | P(observation | outcome) | Joint mass | Posterior |
|---|---|---|---|---|
| GG | 0.3333333 | 1.000000 | 0.3333333 | 0.6666667 |
| GS | 0.3333333 | 0.5000000 | 0.1666667 | 0.3333333 |
| SS | 0.3333333 | 0.000000 | 0.000000 | 0.000000 |
Other coin gold: 0.6666667. G means gold and S silver. Original boxes are chosen uniformly from the counted boxes. The forced protocol first excludes SS and then samples a remaining box uniformly; its reported evidence mass is the fraction of boxes eligible for that selection.
Filled bars give box-type probabilities before observing gold; open bars give probabilities afterward. With one box of each type, GG has prior 1/3 and observation likelihood 1. GS has prior 1/3 and likelihood 1/2. SS has likelihood zero. Gold has total probability 1/3 + 1/6 = 1/2, and GG contributes 1/3. Dividing gives (1/3)/(1/2) = 2/3.
Alternatively, label the individual coins. Among the six equally likely physical draws, three show gold. Two of those gold coins have a gold partner. Labeling the coins does not change the experiment; it makes its equally likely outcomes explicit.
Showing gold can mean a different experiment
Switch to Select a gold-containing box; show gold. Exclude SS boxes first, choose uniformly among eligible boxes, then deliberately show a gold coin. With one GG and one GS box, the answer is 1/2. Every eligible box can now produce the display, so GG has no likelihood advantage.
The table’s evidence probability for this protocol is the fraction of original boxes that are eligible. After conditioning on eligibility, deliberately displaying gold adds no further information about box type. This differs from choosing among all boxes and drawing blindly.
The observation can look identical to an onlooker. What changes is its probability under each possible hidden box. The selection instructions therefore belong in the probability question.
More boxes, same reasoning
Let a, b, and c count GG, GS, and SS boxes. For uniform box and coin selection, the probability of a gold partner after seeing gold is 2a/(2a + b), provided the denominator is positive. The c boxes affect how often gold is observed but contribute nothing to the selected gold sample.
Set GG to two, GS to three, and SS to five. Four observable gold coins come from GG and three from GS, giving 4/7. Under deliberate showing, two of five eligible boxes are GG, giving 2/5. Increasing SS alone makes random gold observations rarer while leaving either conditional answer unchanged.
Make a prediction
Do ten additional silver-only boxes make seeing gold weaker evidence for GG?
Explore the answer
No. Those boxes cannot generate gold in either protocol. They reduce the random-coin observation probability, but the relative selected masses of GG and GS stay the same. Evidence rarity and discrimination between the remaining hypotheses are different quantities.
Boundaries and assumptions
With no boxes there is no experiment; the display reports an invalid setup. With only SS boxes, gold is impossible and the posterior is undefined. With GG but no GS, observing gold identifies GG with probability one. With GS but no GG, the other coin must be silver.
Unequal box-selection probabilities or a preference for a coin position would change the likelihoods. The exact table assumes the uniform choices described above. No finite sample frequency is substituted for those probabilities.
Reference and connection
Grinstead and Snell, Introduction to Probability, chapter 4 includes the box paradox among its exercises. Compare Monty Hall: a host’s selection policy similarly determines what a reveal tells you about a hidden choice.