Fastest of Three

Question

Three market makers, you and two rivals, respond to a request for quote. The three response times are independent draws from the same continuous distribution. What is the probability that your response is the fastest of the three?

Accepted Answer

Symmetry does everything. The three response times are independent and identically distributed, so the joint law is exchangeable: every ordering of the three times is equally likely, and ties have probability zero because the distribution is continuous. The fastest responder is therefore uniform over the three identities: P(textyou are fastest) = frac13. Integral confirmation, for the skeptical: if F is the common CDF with density f, then P(textyou fastest) = int (1 - F(t))^2 f(t), dt = int_0^1 (1 - v)^2, dv = frac13, substituting v = F(t). The answer is distribution-free: nothing about the shape of the latency distribution matters, only that the three draws are exchangeable. The trap is multiplying: P(textbeat A) times P(textbeat B) = frac12 times frac12 = frac14. Those two events are not independent, because both involve your own time. If your draw happens to be slow, you tend to lose both races at once; the positive dependence pushes the joint probability up from frac14 to frac13. The exchangeability reflex generalizes instantly: fastest of n is frac1n, the probability your time ranks k-th is frac1n for every k, and any question that depends only on relative order can be answere

Fastest of Three

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