Practical Probability — Part XII

In my last column, I left you with some probability problems that involve calculating the chance that either (or both) of two events will result in a certain outcome. I recommended using a sequential method for this type of problem. First you calculate the possibilities for the first event. Then, do the calculations for the second event, given that the key outcome did not already occur in the first event. This sequential method often works well for poker problems.

Here are the problems that you were asked to solve (this is technically called calculating the union of two events): Calculate only the possibility that your opponent hits a flush in problem No. 1 or quads in problem No. 2. Don’t worry about other ways that these hands might be won or lost. The easiest approach is: (a) Figure out the percentage of the time that an out hits on the turn. Then, figure out how often it doesn’t, and for those times that it doesn’t, (b) figure out how often it hits on the river, then add (a) and (b) together.

1. You are ahead after the flop in a hold’em hand. Your opponent has a flush draw. He can win by hitting a heart on the turn or a heart on the river to make a flush. Assuming that you don’t have any hearts, what is the chance that he will make a flush?

2. A girl at the bar claims that she flopped a set of aces against an opponent who flopped a set of sixes. Her opponent can win by hitting a 6 on the turn or on the river. What are the chances of this happening?

3. You win seats in supersatellites 20 percent of the time. There will be two more supersatellites before the main event. What is the chance that you will win a seat? (If you win a seat in the first one, you will skip the second one in order to get some rest.)

In the first problem, there are 45 cards left in the deck. Nine of them are hearts and 36 of them are not hearts. Therefore, your opponent will make his flush on the turn 9/45ths of the time, or exactly 20 percent of the time. The remaining 80 percent of the time, there will still be nine hearts left, but only 44 cards remaining. Therefore, he will hit his flush on the river 9/44ths of the time, or slightly more than 20 percent. Don’t make the mistake of adding these together, and thinking he will make the flush 40 percent of the time. He will make it on the river only 20 percent of the 80 percent (when he missed on the turn). He will make a flush 20 percent plus 16 percent, or about 36 percent of the time.

It is common to hear players say, “I had nine outs twice.” This type of thinking overestimates the chance of making the hand. Remember, there is no bonus for making your flush twice — on both the turn and the river — yet some players do their calculations as if there were.

The second problem is almost identical to the first, except that the player with bottom set has only one out. In this case, the chance of hitting a 6 on the turn is 1/45, or about 2.2 percent. The chance of hitting it on the river is 1/44, or about 2.3 percent of the time. Since the player who is behind misses quads 98 percent of the time, we need to take 2.3 percent of 98 percent. He will hit quads roughly 4.4 percent of the time.

The final problem is really the first problem in a different setting. You win a seat in the first satellite 20 percent of the time. You will also win your seat in the second satellite 20 percent of the time, but you will play in it only the 80 percent of the time that you fail in the first one. Therefore, you will win a seat in the second one only 20 percent of the 80 percent, or 16 percent of the time. In total, you will get your seat 36 percent of the time.

This type of problem occurs in many “real-world” situations (although for many of us, poker is the real world). For example, you want to buy a new house with your winnings. There are two similar houses that you like. You decide to make a low offer on the first one. You know the owner will accept it only when he really needs to sell the house, and you think this might be about 10 percent of the time. If this fails, you will make a higher offer on the second house. Your real-estate agent assures you that this offer will be accepted more than half the time — say, 60 percent. How often will you get a house? You’ll get the first one 10 percent of the time, leaving 90 percent for you to make an offer on the second. Your total possibility of success is 10 percent plus 54 percent (60 percent of the 90 percent that you missed the first house). You’ll end up with one of the houses 64 percent of the time.

In previous columns, we concentrated on cases in which you needed a favorable outcome on two successive events (think of hitting a runner-runner full house). In this column, the focus was on cases in which you needed a favorable outcome in only one of two events, and being successful twice was no more valuable than being successful once. What about those cases in which being successful once is good and being successful twice is even better?

Every no-limit hold’em cash-game player should be watching High Stakes Poker. If you aren’t watching the show, you should be. It gives you an excellent opportunity to watch some of the world’s strongest players trying to outthink, outmaneuver, and outplay each other in the setting of a cash game with a fairly fast structure (shorthanded, with antes and blinds). After two players are all in, you will frequently hear them discuss “running it twice.” On many occasions, they do so. What does this mean?

First, here’s some history: Standard Vegas practice had been to give insurance to the players in all-in situations. The person selling the insurance almost always got the best of it, and sometimes by quite a lot. For example, when someone insured against a heart on the river, he had folded two hearts.

Negotiating the insurance often took a long time. When the New Yorkers swarmed into Vegas no-limit hold’em and pot-limit Omaha games, we brought with us a technique for these situations that involved dealing it multiple times. Instead of insuring against a heart on the river, we would divide the pot in half and deal two separate river cards, one for each half, or into thirds and deal three separate river cards, one for each subpot. Does this give an advantage to either player? Using our example of a nine-out flush draw on the river, calculate the chances and equity for each player if the river is dealt only once or if it is dealt twice. The answer will appear in my next column.

Steve “Zee” Zolotow, aka The Bald Eagle, is a successful games player. He currently devotes most of his time to poker. He can be found at many major tournaments and playing on Full Tilt, as one of its pros. When escaping from poker, he hangs out in his bars on Avenue A — Nice Guy Eddie’s at Houston and Doc Holliday’s at 9th Street — in New York City.