7-Card Stud Flush Completion

Brian Alspach

4 May 2000

Abstract:

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-munseen cards from her suit. She needs to get 2 of them within the next 4 cards dealt to her. Altogether there are 49-nunseen cards since she has n opponents. There are ${{49-n}\choose{4}}$ 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 ${{39-n+m}\choose{4}}$ 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 $(10-m){{39-n+m}\choose{3}}$. We sum the latter two numbers and subtract the sum from ${{49-n}\choose{4}}$ 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 nand 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-nand the number of unseen cards from her suit is 10-m. The total number of ways her hand can be completed is ${{48-n}\choose{3}}$. There are ${{10-m}\choose{3}}$ ways her hand can be completed with all 3 cards in her suit, and there are $(38-n+m){{10-m}\choose{2}}$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 ${{48-n}\choose{3}}$. 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 ${{39-n+m}\choose{3}}$ 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 ${{47-n}\choose{2}}$ completions of her hand and ${{38-n+m}\choose
{2}}$ of the completions do not give her a flush. So we get the probability of the player ending up with a flush by subtracting ${{38-n+m}\choose
{2}}$ from ${{47-n}\choose{2}}$ and dividing the result by ${{47-n}\choose{2}}$. 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 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

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


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