Practical Probability — Part XI

In my last column, I left you with a little homework. I hope that you tried to solve the problems, since it is enlightening to see if you can calculate the correct answers. If you didn’t do them, try them now. Even if you don’t actually do the calculations, try to think through the process that leads to the solution and approximate the answer. In both problems, you had to determine the chance of someone hitting a perfect card on both the turn and the river. (Hint: These are dependent events.)

1. You are way ahead after the flop in a hold’em hand. You get knocked out of a tournament when your opponent sucks out, as usual, hitting a heart on the turn and a heart on the river to make a flush. Assuming that you don’t have any hearts, what is the chance of your opponent making a flush?

2. Later, as you sit at the bar, you meet a cute girl who was at your table earlier. It is springtime, and love is in the air. So, naturally, you talk about your tough beats. When you tell her your tale of woe, she says that hers is much worse. She claims that she flopped a set of aces against an opponent who had only a pair of sixes. Her opponent also sucked out, by hitting a 6 on the turn and making quads on the river. What are the chances of this happening?

In the first problem, there are 45 cards left in the deck after the flop. (You and your opponent each have two, and there are three on the board.) Of these cards, 10 are hearts. (Of the 13 original hearts, three are gone, leaving 10 in the remainder of the deck. Your opponent might have two and there is one on the board, or he has one and there are two on the board.) So, he will hit a heart on the turn 10 times out of 45. If he does, he will then hit one of the remaining hearts nine times out of 44. He will do both 10/45 times 9/44. This is equal to .0455 (just less than 5 percent). You are about a 21-1 favorite.

In the second problem, there also are 45 cards left after the flop. Two of these cards are sixes. Her opponent hits a 6 on the turn two times out of 45. If he does this, he will then hit another 6 only once out of the remaining 44 times. He does both 2/45 times 1/44. This is equal to .0010, or about one-tenth of 1 percent. If she was telling the truth, she lost as nearly a 1,000-1 favorite. Since this column is part of a series on practical probabilities, I must point out that the practical chance that a cute girl in a Vegas bar is lying is a lot less than nearly 1,000-1.

These problems dealt with cases in which you had to calculate the possibility that both of two events would occur. In technical terms, this is known as finding the joint probability of two events. It is also called finding the intersection of two events. This type of problem is solved for independent events by simply multiplying the chance of one event occurring by the chance of the other one occurring to get the chance that both will occur. If they are dependent events, as they were in the problems above, you multiply the chance of one occurring by the chance that the other will occur, given that the first one has already occurred.

The other common situation involves calculating the possibility that either of two events will take place, technically called the union of two events. You have to calculate the possibility that either one of two events (or both) will happen. This type of problem is very common in gambling situations. For example, a team is ahead 3-2 in a seven-game series. What is its chance of winning? You need to win a seat in either of two supersatellites to play in a tournament. What is your chance of winning a seat?

I recommend using a sequential method for this type of problem. First, calculate the chance of both a favorable and an unfavorable result in the first event. Then, calculate the chance of a favorable result in the second event. Remember that the second event is relevant only when you had an unfavorable result in the first event, so you need to multiply their probabilities. Let’s assume that the Lakers need to win one of the final two games to win the NBA championship. They will win at home against this opponent 60 percent of the time. So, they will win the first game .60. The remaining 40 percent of the time, they will play a game seven on the road, and will win that 45 percent of the time that it is played. This will happen .4 times .45, which equals .18. Therefore, they will win the championship .60 plus .18, or 78 percent of the time. They are more than a 3.5-1 favorite.

The sequential method often works well for poker problems. To illustrate this, let’s modify the two problems with which we started. In the first one, your opponent flops a four-flush. In the second, her opponent also flops a set. This leaves us with two new problems in which we need to calculate the union of two events.

The following three problems are your homework for next time. Calculate only the possibility that the opponent hits a flush in problem No. 3 or quads in problem No. 4. Don’t worry about other ways that these hands might be won or lost. The easiest approach is to (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.

3. 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 on the river to make a flush. Assuming that you don’t have any hearts, what is the chance of your opponent making a flush?

4. 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?

5. You win tournament seats via 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 in order to get some rest.)

The solutions 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.