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 .545 .556
1 .444 .452 .461 .47 .479 .488 .498 .508
2 .398 .405 .413 .421 .43 .439 .448 .457
3 - .356 .363 .37 .378 .386 .394 .403
4 - - .31 .316 .323 .33 .338 .345
5 - - - .259 .265 .271 .277 .284
6 - - - - .204 .209 .214 .219
7 - - - - - .143 .146 .15
8 - - - - - - .075 .077


  10 11 12 13 14 15 16
0 .567 .578 .59 .603 .616 .629 .643
1 .519 .53 .541 .553 .566 .578 .592
2 .467 .477 .488 .499 .511 .523 .536
3 .412 .421 .431 .442 .453 .464 .476
4 .353 .362 .37 .38 .389 .4 .41
5 .291 .298 .305 .313 .322 .33 .34
6 .224 .23 .236 .242 .249 .256 .263
7 .154 .158 .162 .166 .171 .176 .181
8 .079 .081 .083 .086 .088 .091 .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 .404 .413 .422 .432
1 .334 .341 .348 .356 .364 .372 .381 .39
2 .296 .302 .309 .316 .323 .331 .339 .347
3 .257 .262 .268 .274 .281 .287 .294 .302
4 .217 .221 .226 .232 .237 .243 .249 .257
5 - .179 .184 .188 .192 .197 .202 .207
6 - - .139 .143 .146 .15 .154 .158
7 - - - .096 .099 .101 .104 .107
8 - - - - .05 .051 .053 .054


  11 12 13 14 15 16 17 18
0 .443 .454 .465 .477 .49 .503 .517 .532
1 .4 .41 .421 .432 .444 .456 .469 .483
2 .356 .365 .374 .384 .395 .406 .418 .431
3 .31 .318 .326 .335 .348 .355 .366 .377
4 .262 .269 .276 .284 .292 .301 .31 .32
5 .213 .218 .225 .231 .238 .245 .253 .261
6 .162 .166 .171 .176 .181 .187 .193 .2
7 .11 .113 .116 .119 .123 .127 .131 .135
8 .056 .057 .059 .061 .063 .065 .067 .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

\begin{displaymath}\frac{9-m}{46-n}.\end{displaymath}


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