Linda Mae frowned as she studied the 'random hand' promotion that our local casino's marketing department dreamed up for the poker room. They added $50 to the jackpot for every day without a winner. With the jackpot at $2,750, Linda Mae wondered what the odds were that of the five cards shown, including their suits, she would have two in her hand and three among the community cards in a ten-player hold 'em game. "What were the odds of that?" she asked, "and would I please explain it in English." (Linda Mae, you will recall, is mathematically challenged.)
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The requirement that two of the cards be in your hold 'em hand means that three of them must be on the tableau. There are ten combinations possible when dealing three out of five, C(5,3). Two of the 47 other cards must be on the tableau, C(47,2). That works out to 10,810 boards which qualify for the promotion. There are 2,598,960 possible ways to deal 5 cards from a standard deck, so the probability that any 3 of those 5 exact cards will be on the table is their quotient, 0.0041594, about 240-to-1 against.
Linda Mae allowed that she would take my word for it. "We're not done yet," I told her, "now we have to figure the probability that the other 2 cards are among the 20 cards dealt to the ten players." The probability of that is given by the formula, C(45,18)/C(47,20), 0.17576, about 5-to-1 against.
"So, it's about 1,200-to-1 against," she concluded proudly, "5-to-1 times 240-to-1 equals 1,200-to-1."
"We're still not done yet," I told her, "now we have to figure the probability that the other 2 cards are in one hand. Here's how that works. The first of 20 cards can go to anyone; 19 remain, so pick one. The first of the remaining 18 cards can go to anyone else; 17 remain, so pick one, and so on: 19*17*15*13*11*9*7*5*3*1 obtains the number of ways to deal 2 cards each to ten players, or 654,729,075. We want the probability that one player at the table has both key cards. If one player has them both, then the first of the remaining 18 can go to anyone else, and 17 remain, and so on. 17*15*13*11*9*7*5*3*1 obtains the number of ways to deal 2 cards each to the nine other players at the table, or 34,459,425. Their quotient is the probability, 1/19, so it's 18-to-1 against.
"Now we multiply the three to arrive at the probability that anyone at the table could have the jackpot hand, namely, 0.000038477, or 25,989-to-1 against. In other words, about once in 26 thousand deals, those exact cards will appear among the seven cards available to anyone at the table, and also meet the requirement of having two in someone's hand."
"So, it's about 26,000-to-1 against my having the jackpot hand, not 1,200-to-1 against," she concluded. "Is that harder or easier than a royal flush," she asked.
"It's four times harder than making a royal when both hole cards must play," I responded, "and furthermore, those are the odds against anyone at the table in a ten-handed game. The odds against you in particular are ten times worse, about 260,000-to-1, on any one deal."
"That's about as unlikely as my winning a bad-beat jackpot," she wailed, "only one chance in a quarter-million?! At least for a bad-beat jackpot, there's a table-share! For this stupid promotion only one player at the table wins anything, and not all that much either!"
"Well," I said, "bad-beat jackpots are somewhat less likely than this 'random hand' promotion, and besides the poker players themselves pay for their bad-beat jackpot. For this promotion, casino marketing supplies the prize money, so it's a nice extra."
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
