# Flushing Dreams: Part II

Brian Alspach

Poker Digest Vol. 3, No. 13, June 16 - 29, 2000

In Part I of this series we determined the probabilities of a player holding three suited cards amongst her first three or four cards in seven-card stud making a flush. We now move to the cases of the player having four suited cards in her hand. The next two tables give the probabilities of a player holding four suited cards for her first four cards finishing with a flush.

If she has seen n other cards of which m are in her suit (again not counting her cards), then there are 48-n unseen cards of which 9-m are in her suit. The total number of completions of her hand is C(48-n,3). The only way she can avoid ending up with a flush is if all three cards are not in her suit. Since the number of cards not in her suit is 48-n-(9-m)=39-n+m, there are C(39-n+m,3) hand completions which do not give her a flush. We then get the probability in the obvious way.

The columns are headed by n and the rows correspond to m.

 2 3 4 5 6 7 8 9 0 .488 .497 .506 .515 .525 .535 0.545 0.556 1 .444 .452 .461 .47 .479 .488 0.498 0.508 2 .398 .405 .413 .421 .43 .439 0.448 0.457 3 - .356 .363 .37 .378 .386 0.394 0.403 4 - - .31 .316 .323 .33 0.338 0.345 5 - - - .259 .265 .271 0.277 0.284 6 - - - - .204 .209 0.214 0.219 7 - - - - - .143 0.146 0.15 8 - - - - - - 0.075 0.077

 10 11 12 13 14 15 16 0 0.567 0.578 0.59 0.603 0.616 0.629 0.643 1 0.519 0.53 0.541 0.553 0.566 0.578 0.592 2 0.467 0.477 0.488 0.499 0.511 0.523 0.536 3 0.412 0.421 0.431 0.442 0.453 0.464 0.476 4 0.353 0.362 0.37 0.38 0.389 0.4 0.41 5 0.291 0.298 0.305 0.313 0.322 0.33 0.34 6 0.224 0.23 0.236 0.242 0.249 0.256 0.263 7 0.154 0.158 0.162 0.166 0.171 0.176 0.181 8 0.079 0.081 0.083 0.086 0.088 0.091 0.094

Now suppose the player has four suited cards amongst her first five cards. If she has seen n cards from the other players and m of those are in her suit, then there are C(47-n,2) completions of her hand and C(38-n+m,2) of the completions do not give her a flush. So we get the probability of the player ending up with a flush by subtracting C(38-n+m,2) from C(47-n,2) and dividing the result by C(47-n,2). The following tables give these probabilities for some values of m and n. As before, the columns are headed by values of n and the rows by values of m.

 3 4 5 6 7 8 9 10 0 .371 .379 .387 .395 0.404 0.413 0.422 0.432 1 .334 .341 .348 .356 0.364 0.372 0.381 0.39 2 .296 .302 .309 .316 0.323 0.331 0.339 0.347 3 .257 .262 .268 .274 0.281 0.287 0.294 0.302 4 .217 .221 .226 .232 0.237 0.243 0.249 0.257 5 - .179 .184 .188 0.192 0.197 0.202 0.207 6 - - .139 .143 0.146 0.15 0.154 0.158 7 - - - .096 0.099 0.101 0.104 0.107 8 - - - - 0.05 0.051 0.053 0.054

 11 12 13 14 15 16 17 18 0 0.443 0.454 0.465 0.477 0.49 0.503 0.517 0.532 1 0.4 0.41 0.421 0.432 0.444 0.456 0.469 0.483 2 0.356 0.365 0.374 0.384 0.395 0.406 0.418 0.431 3 0.31 0.318 0.326 0.335 0.348 0.355 0.366 0.377 4 0.262 0.269 0.276 0.284 0.292 0.301 0.31 0.32 5 0.213 0.218 0.225 0.231 0.238 0.245 0.253 0.261 6 0.162 0.166 0.171 0.176 0.181 0.187 0.193 0.2 7 0.11 0.113 0.116 0.119 0.123 0.127 0.131 0.135 8 0.056 0.057 0.059 0.061 0.063 0.065 0.067 0.069

Finally, if a player has four suited cards amongst her first six cards, then the probability of the player making a flush is easy to calculate. In this case we shall not give a table because the formula is so easily evaluated. Suppose the player has seen n cards of which m are in her suit, where we do not include her own cards in counting n and m. Then there are 46-n unseen cards of which 9-m are in her suit. This implies the probability of her making the flush with the last card is

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