Parity Lock

Question

Two fair dice are rolled and you are told that the sum is even. What is the probability that the two dice show the same parity (both odd or both even)?

Accepted Answer

The sum of two integers is even if and only if the two integers have the same parity: odd + odd = even, even + even = even, and a mixed pair is always odd. So the event you are asked about is identical to the event you conditioned on: P(textsame parity mid textsum even) = 1. To see it in raw counts: of the 36 ordered outcomes, 18 have an even sum, 9 with both dice odd and 9 with both dice even, and every single one of those 18 is a same-parity pair. The trap The reflex is to start a conditional-probability calculation, get frac1836 somewhere in the numerator, and report frac12. The question is testing whether you inspect the events before grinding: if A supseteq B then P(A mid B) = 1, and here A = B exactly. Why interviewers ask it Trading interviews reward people who notice structure before computing. A surprising number of candidates burn a minute on Bayes' rule for a question whose answer is forced by definition, the same failure mode as repricing an instrument from scratch when an arbitrage identity pins the answer.

Parity Lock

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