Seven right out of eight, exactly

Question

A brand-new strategy has taken 8 independent trades and won 7 of them. Under the null that it has no edge, each trade wins with probability $0.5$ .

Compute the exact one-sided p-value (probability of 7 or more wins by chance) and comment on how much to trust it.

Accepted Answer

Set up. Let X \sim 	ext{Binomial}(8, 0.5). The one-sided p-value is p = \mathbb{P}(X \geq 7) = \frac{\binom{8}{7} + \binom{8}{8}}{2^{8}}. Count and compute. Here \binom{8}{7} = 8 and \binom{8}{8} = 1, summing to 9, over 2^{8} = 256: p = \frac{9}{256} \approx 0.0352 \approx 0.035. The verdict, with a caveat. Formally, p \approx 0.035 is below 5%, so a no-edge strategy would win 7 or more of 8 only about 3.5% of the time. But eight trades is a razor-thin sample. A single win flipping to a loss (6 of 8) would push the exact p-value up to \mathbb{P}(X \geq 6) = (28 + 8 + 1)/256 = 37/256 \approx …

Seven right out of eight, exactly

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