**Brian Alspach**

**4 May 2000**

We compute the probabilities for a variety of beginning hands in
7-card stud finishing with a flush.

It is easy, though tedious, to determine the probability of a 7-card stud hand ending up with a flush. We do not make any assumptions about the number of players in the game. Instead, we consider the number of cards the player has seen.

Let's first consider a player who finds her first three cards
are suited. That means there are 10 other cards in her suit.
If the other players in the game have *m* cards from the suit
showing amongst their *n* upcards, then there are 10-*m*unseen cards from her suit. She needs to get 2 of them within
the next 4 cards dealt to her. Altogether there are 49-*n*unseen cards since she has *n* opponents. There are
possible sets of 4 cards to complete her
hand. She will not get a flush if she receives either 0 or 1
card from her suit. There are 39-*n*+*m* unseen cards which are
not in her suit. Thus, there are
ways
she can be dealt 4 cards none of which are in her suit. The
number of ways she can be dealt exactly 1 card from her suit is
.
We sum the latter two numbers
and subtract the sum from
to get the
number of ways she can be dealt 4 cards which give her a flush.
We then get the probability in the obvious way.

The following table gives the probabilities for a player
starting with 3 suited cards in 7-card stud to finish with a
flush given that the player sees *n* opponent cards of which
*m* are in the player's suit (*m* and *n* are not counting the
player's own cards). The columns correspond to values of *n*and the rows correspond to values of *m*.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |

0 | .187 | .194 | .201 | .209 | .218 | .226 | .236 | .245 |

1 | .155 | .16 | .167 | .173 | .18 | .188 | .196 | .204 |

2 | - | .129 | .134 | .14 | .145 | .151 | .158 | .165 |

3 | - | - | .104 | .108 | .113 | .118 | .123 | .129 |

4 | - | - | - | .08 | .083 | .087 | .091 | .095 |

5 | - | - | - | - | .057 | .06 | .063 | .066 |

6 | - | - | - | - | - | .037 | .039 | .041 |

7 | - | - | - | - | - | - | .02 | .021 |

8 | - | - | - | - | - | - | - | .007 |

The next table gives the probability of a player holding 3 suited
cards amongst her first 4 cards finishing with a flush. We again
let *n* be the total number of cards she has seen in the other player's
hands of which *m* are in her suit. Neither *m* nor *n* are counting
the player's own cards. Thus, the number of unseen cards is 48-*n*and the number of unseen cards from her suit is 10-*m*. The total
number of ways her hand can be completed is
.
There are
ways her hand can be completed with
all 3 cards in her suit, and there are
completions which give her precisely 2 cards in her suit. Adding
the two ways of completing her hand to a flush yields the number of
ways she can achieve a flush. We then obtain the probability in the
obvious way.

The table is broken into two parts where the columns are headed by
values of *n* and the rows by values of *m*.

2 | 3 | 4 | 5 | 6 | 7 | 8 | |

0 | .115 | .119 | .125 | .13 | .136 | .142 | .149 |

1 | .093 | .097 | .101 | .106 | .111 | .116 | .121 |

2 | .074 | .077 | .08 | .084 | .088 | .092 | .096 |

3 | - | .059 | .061 | .064 | .067 | .07 | .074 |

4 | - | - | .045 | .047 | .049 | .051 | .054 |

5 | - | - | - | .032 | .033 | .035 | .036 |

6 | - | - | - | - | .02 | .021 | .022 |

7 | - | - | - | - | - | .011 | .011 |

8 | - | - | - | - | - | - | .004 |

9 | 10 | 11 | 12 | 13 | 14 | 15 | |

0 | .156 | .164 | .172 | .181 | .19 | .2 | .212 |

1 | .127 | .134 | .141 | .148 | .156 | .164 | .174 |

2 | .101 | .106 | .112 | .118 | .124 | .131 | .139 |

3 | .077 | .081 | .086 | .09 | .095 | .1 | .106 |

4 | .056 | .059 | .062 | .066 | .07 | .074 | .078 |

5 | .038 | .04 | .042 | .045 | .047 | .05 | .053 |

6 | .023 | .025 | .026 | .027 | .029 | .031 | .033 |

7 | .012 | .013 | .013 | .014 | .015 | .016 | .017 |

8 | .004 | ..4 | .005 | .005 | .005 | .005 | .006 |

We now move to the cases of the player having 4 suited cards in
her hand. The next two tables give the probabilities of a player
holding 4 suited cards for her first 4 cards finishing with a flush.
These calculations are simpler than those for 3 suited cards above.
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
.
The only way she can avoid ending
up with a flush is if all 3 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
hand completions which do not give her a
flush. We then get the probability in the obvious way.

As in the examples above, 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 we move to the situation in which the player has 4 suited cards
amongst her first 5 cards. If she has seen *n* cards from the
other players and *m* of those are in her suit, then there are
completions of her hand and
of the completions do not give her a flush. So we get the
probability of the player ending up with a flush by subtracting
from
and dividing the result
by
.
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 4 suited cards amongst her first 6 cards,
then the probability of the player making a flush is easy to
calculate. In this case we shall not give a table because it is
easy to write down a single formula capturing all the information.
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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