Why the order of wins carries no information

Question

You record three trades as wins or losses, each independently a win with probability $p$ . You are told only that exactly two of the three were wins.

Given that there were two wins, what is the probability the pattern was win-win-loss, and what does the answer reveal about why the win count is a sufficient statistic for $p$ ?

Accepted Answer

List the equally likely patterns. With exactly two wins among three trades, the possible orderings are win-win-loss, win-loss-win, and loss-win-win. Each has the same probability of occurring, p \cdot p \cdot (1 - p) = p^2(1 - p), because multiplication does not care about order. Condition on the count. Given that the total is two wins, the probability of any one specific pattern is that pattern's probability divided by the total over all two-win patterns: P(	ext{win-win-loss} \mid 	ext{2 wins}) = \frac{p^2(1 - p)}{3\,p^2(1 - p)} = \frac{1}{3}. The punchline: p cancels. The answer is 1/3 no…

Why the order of wins carries no information

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