**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

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