pairwise vs mutual independence
When you have three or more events, 'independent' can mean two different strengths, and the gap between them traps the unwary. Pairwise independence means every PAIR of the events is independent: A and B multiply, A and C multiply, B and C multiply. Mutual (full) independence is stronger: not only every pair, but every sub-collection multiplies, including the triple — P(A and B and C) = P(A)P(B)P(C).
The surprising fact is that pairwise independence does NOT imply mutual independence. The famous counterexample: toss two fair coins, and define A = 'first is heads', B = 'second is heads', C = 'the two coins match' (both same). Each of A, B, C has probability 1/2, and any two of them are independent — for instance knowing the first coin is heads tells you nothing about whether they match. Yet the three together are not mutually independent: if you know A and B (both heads), then C is forced to be true, so P(C given A and B) = 1, not 1/2. The pairwise checks all pass while the joint relationship is rigidly determined.
This matters because many proofs and constructions quietly require FULL mutual independence — the multiplication rule P(A1 and ... and An) = product of P(Ai) only holds under mutual independence, not mere pairwise. When you assume 'the events are independent', be clear which you mean. Pairwise is enough for some second-moment arguments (variances of sums), but full independence is what you need for joint distributions to factor completely.
Two coins: A = first heads, B = second heads, C = match. Pairwise: P(A and C) = P(both heads) = 1/4 = (1/2)(1/2), and likewise for the other pairs — all independent. But P(A and B and C) = P(both heads) = 1/4, while P(A)P(B)P(C) = 1/8. Not equal, so not mutually independent.
All pairs multiply, yet the triple does not — pairwise independence is genuinely weaker.
Mutual independence of n events requires the product rule to hold for EVERY subset of size 2 up to n — there are 2^n minus n minus 1 such conditions, not just the pairwise ones.