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House of Cards
by Ashley Adams
 

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Straight Skinny: Regal Findings

Linda Mae swung into our local poker room proudly wearing a nice-looking jacket from another casino. We greeted her derisively—and heartily suggested that if she wore the jacket, then she ought to patronize that casino’s poker room.
 
She dismissed us with a sniff and explained that she and a girlfriend went to Laughlin for the weekend to splash in the Colorado, sun by its side, and play some hold’em. She won the jacket when she made a royal flush holding the club queen, she explained.
 
Kc-10c-4h-Ac-Jc
 
We responded to that announcement with hoots and whistles, because we all knew that no one awarded a jacket unless both hole cards played. We argued that the both-hole-cards-must-play rule made it harder to win a jacket or any other prize.
 
Linda Mae declared, “Au contraire, the casino told me the odds against making a royal with one card in hand exactly equal those when both hold cards play.”
 
Let’s see if the casino erred. First we calculate the probability of making a royal when both hole cards play. The probability of two cards to a royal in your hand equals 4*C(5,2)/C(52,2), which equals 0.03017. The probability of the necessary three cards on the table after dealing out all the cards equals C(47,2)/C(50,5), which equals 0.00051. We multiply them to find the probability of making a royal when both cards must play, namely 1.5391*10-5, or odds against of 64,973-to-1.
 
Now we calculate the probability of having exactly one honor card in your hand and the four other royal cards on the table. The probability of one honor card equals C(20,1)*C(32,1)/C(52,2), which equals 0.48265. The probability of the necessary four cards on the table equals C(46,1)/C(50,5), which equals 2.1711*10-5. Again we multiply to find the probability, 1.047979* 10-5, about 95,000-to-1 against.
 
You can also make a royal when you hold unsuited honor cards, including pairs. You can hold two honor cards in 190 ways, as given by C(20,2). Forty of those ways have suited honors, so 150 ways remain to hold two unsuited honor cards in your hand: that probability equals 150/C(52,2), or 0.11312. The probability of the necessary four cards on the table equals 2*C(46,1)/C(50,5), which equals 4.3422*10-5. Multiplying, we find the probability equals 4.9120*10-6, about 200,000-to-1 against.
 
Too many numbers for you? We have only one step left. We add the two probabilities of making a royal with four royal cards on the table to find the probability of making a royal that way: 1.5391*10-5, exactly the same as we calculated for making a royal when both hole cards must play!
 
Before you look at your cards, you have just as good a chance of making a royal using one of your cards as you do with both hole cards. The casino got it right.
 
We got it right too. The both-hole-cards-must-play rule does make it harder to win a prize, in this case of making a royal, exactly twice as hard.
 
We also discovered that you should have one honor card in your hand 48 percent of the time. You can expect two honor cards, including pocket pairs, 14 percent of the time.
 
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 burkecaltech@cox.net

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Wendeen H. Eolis

World Series of Poker


September 4, 2014 - 10:31am
August 28, 2014 - 9:45am
August 25, 2014 - 9:44am
August 22, 2014 - 9:45am
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