Days until one of two limit orders fills

Question

You and a competitor each rest a limit order in the same stock. On any given day, your order fills with probability $0.4$ and the competitor's fills with probability $0.5$ , independently of each other and across days. You watch until at least one of the two orders fills.

On average, how many days until the first fill?

Accepted Answer

First find the daily chance that at least one order fills, then treat days as independent trials. The chance neither fills on a day is (1 - 0.4)(1 - 0.5) = 0.6 	imes 0.5 = 0.3, so the chance at least one fills is p = 1 - 0.3 = 0.7. The expected wait is 1/p: E = \frac{1}{0.7} = \frac{10}{7} \approx 1.43 	ext{ days.} Takeaway. For the first of independent daily events, use p = 1 - \prod(1 - pi) = 0.7, so the expected wait is 10/7 \approx 1.43 days.

Days until one of two limit orders fills

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