Flop runout count
After four known hole cards and three board cards, 45 cards remain and C(45,2)=990 unordered runouts.
Learn how combinations, conditional probability, best-five comparison, split-pot equity, and seeded simulation fit together. This poker probability guide follows the same transparent definitions used by the browser calculator rather than treating an odds result as a black box.
How the result is built
Concrete unordered card combinations and abstract starting-hand classes answer different counting questions. They must not share a denominator.
A flop with two missing board cards needs unordered turn-river sets for showdown equity. Counting both A-then-B and B-then-A as separate runouts would double the same final board.
Hole cards and board cards leave the sample space before enumeration or sampling. Range calculations also remove blocked concrete combinations before selecting hands.
A sole winner receives one unit of an outcome. Two tied winners receive one half each; three receive one third. Across players, equity sums to 100% within floating-point tolerance.
Monte Carlo samples legal runouts without replacement. More iterations reduce sampling error, while the seed makes the selected pseudo-random sequence reproducible.
After four known hole cards and three board cards, 45 cards remain and C(45,2)=990 unordered runouts.
When the board is the best hand for all three players, each receives 33.333…% equity.
It is biased, hard to reproduce, and unnecessary; partial Fisher–Yates sampling is used.
No. It means every legal modeled outcome was counted.
It reproduces the same finite simulation for testing and comparison.
Split shares and finite sample counts often do not terminate in base ten.