Desiree, a poker dealer in a southern California card room, wrote to ask two hold 'em bad-beat questions. The first question was tougher than this one and we'll wait for a later column to answer it. She asked what the odds were against a steel wheel losing to a royal flush.
We think the math is fun and logical so we'll break it into small chunks for easy digestion. You should skip to the end if you're not at all interested in combinatorial mathematics. Here goes.
There must be an ace on the table in any one of four suits. There must be two wheel cards out of four on the table and there are six ways to do that, C(4,2). There must be two royal cards out of four on the table and there are also six ways to do that. There are 144 boards similar to the one shown that qualify.
There are 2,598,960 ways to deal five cards from a standard deck, C(52,5), so the probability that a board will qualify is 5.54068*10-5.
Forty-seven cards remain. The other four cards that complete the straight flushes must be dealt among the nine players. There are 78,378,960,360 ways to do that, C(43,14), out of 4,568,648,125,690 ways that one could deal 18 cards from a stock of 47 cards, C(47,18).
Now we come to a tricky bit. In general, we don't care who gets the first card from the 18 hole cards dealt among nine players. 17 cards remain, so there are 17 choices to go along with the first card. We don't care who gets the next card from the remaining 16, and there are 15 choices for that player's second hole card. Continuing in a like manner, we find there are 17*15*13*11*9*7*5*3*1 ways that any 18 cards could be dealt among nine players. (That odd product becomes 17!! In math shorthand.)
For this bad beat, the four key cards must be in two players' hands and there are 3!! ways to do that. The remaining 14 cards can go any which way to the seven others, and we know there are 13!! ways to distribute them, since we care not who gets which of those cards.
The four key cards could have been dealt to any two players these three ways: Kh-5h and Qh-2h; Qh-5h and Kh-2h; or, 5h-2h and Kh-Qh. Only the last way would make the hand a bad beat.
That's all we need to answer Desiree's question. We just need now to find the probability and then convert it to odds. The probability is given by this expression, 144/C(52,5)*C(43,14 )/C(47,18)*3!!*13!!/17!!/3. To find the answer to this expression, we turn to our handy spreadsheet and let our computer do the arithmetic. You could do it with pencil and paper-there are many terms that cancel each other-we didn't because it would be too easy to make a mistake.
If you want to follow along, then key these three expressions into separate cells of your Excel® spreadsheet:
= 4*combin(4,2)*combin(4,2)/combin(52,5)
= combin(43,14)/combin(47,18)
= factdouble(3)*factdouble(13)/factdouble(17)/3
Multiply the three cells and you should find the same answer we did for the probability, 3.72764*10-9.
We find the odds by inverting the probability to arrive at this number, 268,265,914.6, then rounding it and subtracting one. On any hold 'em deal with nine active players, the odds against a steel wheel losing to a royal flush are 268,265,914-to-1.
Mr. Burke is the author of Flop: The Art of Winning at Low-Limit Hold 'Em, on sale at amazon.com & kokopellipress.com. E-mail your Hold 'Em questions to richardburke@comcast.net
