Chance a losing decision still shows more wins than losses, by number of plays: exact binomial, two ways
Our own calculation, 8 November 2026 (UTC), for the article "A Winning Move Can Still Be a Bad Decision". Each play pays +1 or -1; a decision that wins with probability p has expected value 2p - 1. The chance it shows more wins than losses after n plays (n odd, so no ties) is the binomial tail P(wins > n/2). Way 1: the exact sum of binomial terms in exact fractions. Way 2: a step-by-step count of every win total after each play (dynamic programming, exact fractions) up to 251 plays, and a normal approximation with continuity correction for the thresholds. Way 1 and way 2 agree exactly on every cell checked.
What we found
- A decision that wins 20% of the time (EV -0.6) shows a net win after one play 20.0% of the time, after five plays 5.8% (about one time in 17) and after eleven plays 1.2% (one time in 86).
- At 45% (EV -0.1) the chance of a net win after eleven plays is 36.7%; the first odd number of plays at which it falls under 5% is 269 (4.991%; at 267 plays it is 5.054%); the normal approximation gives 4.964% at 269.
- At 48% (EV -0.04) the first odd number of plays under 5% is 1,691 (4.991%; at 1,689 plays it is 5.001%); the normal approximation gives 4.986%.
- At 40% (EV -0.2) after 251 plays the chance is 0.068%, shown in the table as under 0.1%.
| Plays | Win 20% (EV -0.6) | Win 40% (EV -0.2) | Win 45% (EV -0.1) | Win 48% (EV -0.04) |
|---|---|---|---|---|
| 1 | 20.0% | 40.0% | 45.0% | 48.0% |
| 5 | 5.8% | 31.7% | 40.7% | 46.3% |
| 11 | 1.2% | 24.7% | 36.7% | 44.6% |
| 25 | under 0.1% | 15.4% | 30.6% | 42.0% |
| 51 | under 0.1% | 7.4% | 23.6% | 38.7% |
| 101 | under 0.1% | 2.1% | 15.6% | 34.3% |
| 251 | under 0.1% | under 0.1% | 5.6% | 26.3% |
| 501 | under 0.1% | under 0.1% | 1.2% | 18.5% |
What this cannot tell you: real games, where payoffs are not +1 or -1 and the chance of winning is not known exactly.