by Richard G. Burke

Fred likes to fish, so he and his wife rented a Reno resort hotel room for a week. She enjoys  the pool while he—armed with flies, rods, reels, waders, and a sack lunch—fishes in the nearby mountains’ catch-and-release streams. After supper they head to the casino, she to play bingo or the slots and he to the nightly low-entry-fee, no-limit hold’em tournament in the poker room.

 You need to understand that Fred doesn’t wail about his poker results; win or lose, he has a ready smile and a cheery attitude. Just the same, he came as close to wailing as I ever heard when he called me that night.

 His tournament started at 7 p.m. with five full tables of ten, paying five places. Each player started with $1,500 in tournament chips. The blinds started at $25-50 and went up every 20 minutes. After two-and-a-half hours of solid play, catching a few hands, and stealing blinds upon occasion in the right spot, Fred moved to the final table.

by Richard G. Burke

In our last column, responding to Larry Duplessis’s question, we found that ace-ten led or tied at a ten-handed table 62.7 percent of the time prior to the flop. In this column we find ace-ten improving as the number of players decrease from ten to two.

 Each deck has twenty Broadway cards: four each aces, kings, queens, jacks, and tens. Given that you hold ace-ten, we have 153 doubletons possible from 18 remaining Broadway cards, C(18,2). If the dealer happened to pitch one or more of those danger doubletons to an opponent then, as the table shows, 60 of them dominate ace-ten, and ace-ten leads or ties the other 93. (If you have three or fewer outs if you both whiffed the flop, then we call your hand “dominated” by having a win probability of 30 percent or less.) However, your ace-ten leads over the 84 doubletons of K-Q, K-J, K-T, Q-J, Q-T, and J-T; your ace-ten ties with the other 9 possible ace-tens.

 You can bet on players to win the WSOP’s main event at a sports book in Las Vegas, or on-line in other countries. Can we find betting opportunities in those lines? Let’s see.

by Richard Burke

Jesse Lancos reported in an e-mail that he flopped straight flushes back-to-back in his local card room and asked about the odds of that event.

We have three ways to figure the odds of back-to-back hands: given that he just flopped a straight flush, the probability of flopping another one on the very next hand equals 40/2,598,960, for odds against of 64,973-to-1; before he looked at his cards, the probability of flopping straight flushes backto- back equals 1600/6,754,593,081,600, for odds against of 4,221,620,675-to-1, about 4.2 billion-to-1; or, we could compute the odds of back-to-back straight flushes during a poker session of N hands, which seems much the more interesting.

Every night after supper, Fred plays no-limit hold’em for low-stakes. Because he read somewhere that Doyle Brunson’s favorite hand was 5-3 suited, he e-mailed us beaucoup questions about playing suited wheel cards, the 24 double tons of ranks five, four, three, and two, shown here:

You need to know the chance of making a five-card flush by the river when starting with a suited hand equals 5.8 percent, about 16-to-1 against. You should know by now that even in a ten-handed game your five-card flush prevails 76 percent of the time. Of course with four or more trumps among the community cards your baby flush has little or no chance.

Your chances for a straight depend on which double ton you hold. The 5-4 double ton can make a straight four ways;the 5-3 and 4-3 doubletons three ways; and the others two ways. The chance of a straight by the river equals 13.1 percent,9.8 percent, and 6.5 percent, respectively, for those doubletons.

Linda Mae braced me just outside our local poker room and said, “You made a big mistake in your column titled ‘Trumped-Up Flops’ (PokerPlayer Newspaper, July 18, 2011). You wrote that when the dealer flops all trumps, then the chance of your having a flush equals 4 percent. And you also wrote that when you hold a suited hand, the chance of your flopping a flush is less than 1 percent! Both statements can’t be right, so which is it?”
 
Both statements correctly describe your chances. It just depends on what you know and when you know it. Given that you have a suited hand, a probability of 4/17, the dealer flops you a flush less than one percent of the time. Because you have two trumps, only eleven trumps remain out of the other fifty cards in the deck, so the probability of flopping a flush equals C(11,3)/C(50,3) , which equals 0.0084, or about once in 110 times.
 

In their books, Harrington on Cash Games, Harrington and Robertie wrote that you have a bluffing opportunity heads-up when the dealer flops all the same suit. Let’s see why.
 
If you hold a suited hold’em hand, then the probability that the dealer flops three of your suit equals 0.008418, about 118-to-1 against, as given by C(11,3)/C(50,3). The probability that the dealer pitched you two suited cards equals 4/17, or 0.23529. So, before you look at your hand the probability of your flopping a flush equals their product, 0.00198, about 504-to-1 against. (You could offer side bet odds of 250-to-1 and make a tidy profit.)
 
If you have a suited hand then the probability that the dealer flops three of a different suit equals 0.04378, from 3*C(13,3)/C(50,3), or about 22-to-1 against.
 

Fred wrote to tell me his latest bad-beat story. He played in a large, free-roll tournament held at his local casino’s ballroom and squeaked through the early levels by stealing blinds and catching an occasional hand.

Linda Mae upbraided me outside our local poker room. “You assured us that suited connectors have a 50 percent chance of hitting the flop, and that you would explain it later, but you never did. What’s up with that!?”

The Station Casinos with poker rooms in metropolitan Las Vegas offer a bonus of $250 for any hold’em player who makes a royal flush with both hole cards playing. We thought you might like to see the details, so we didn’t hurry through any steps in the following computations.

We have only four royal flushes possible; three of the five royal cards must lie among the five community cards; the other two come from the 47 remaining cards. The probability obtains from 4*C(5,3)*C(47,2)/C(52,5) which after factoring common terms reduces to

4 * 5*4/2  *  47*46/2  *  5*4*3*2*1/52*51*50*49*48

which equals 0.016637 rounded to six places, about 59-to-1 against. So, about one deal in sixty, the dealer places three royal cards on the table.