# 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 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 nand the rows correspond to values of m.

 1 2 3 4 5 6 7 8 0 .187 .194 .201 .209 .218 .226 .236 0.245 1 .155 .16 .167 .173 .18 .188 .196 0.204 2 - .129 .134 .14 .145 .151 .158 0.165 3 - - .104 .108 .113 .118 .123 0.129 4 - - - .08 .083 .087 .091 0.095 5 - - - - .057 .06 .063 0.066 6 - - - - - .037 .039 0.041 7 - - - - - - .02 0.021 8 - - - - - - - 0.007

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 . 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 0.149 1 .093 .097 .101 .106 .111 .116 0.121 2 .074 .077 .08 .084 .088 .092 0.096 3 - .059 .061 .064 .067 .07 0.074 4 - - .045 .047 .049 .051 0.054 5 - - - .032 .033 .035 0.036 6 - - - - .02 .021 0.022 7 - - - - - .011 0.011 8 - - - - - - 0.004

 9 10 11 12 13 14 15 0 0.156 .164 0.172 0.181 0.19 0.2 0.212 1 0.127 .134 0.141 0.148 0.156 0.164 0.174 2 0.101 .106 0.112 0.118 0.124 0.131 0.139 3 0.077 .081 0.086 0.09 0.095 0.1 0.106 4 0.056 .059 0.062 0.066 0.07 0.074 0.078 5 0.038 .04 0.042 0.045 0.047 0.05 0.053 6 0.023 .025 0.026 0.027 0.029 0.031 0.033 7 0.012 .013 0.013 0.014 0.015 0.016 0.017 8 0.004 ..4 0.005 0.005 0.005 0.005 0.006

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 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 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 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 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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