Practical Probability — Part X

I ended my last column in this series by mentioning a type of erroneous thinking about probability known as the gambler’s fallacy. Basically, this is the mistaken belief that a series of deviations in one direction will result in the next few results tending to balance out that abnormal sequence. When flipping a coin, have you ever seen a lot of heads in a row and felt that tails was due? If you have, you fell into the trap known as the gambler’s fallacy. Since heads or tails is an independent, random outcome, its probability is unaffected by what has just occurred. This leads us to a discussion of independence.

When the term independence is used in a discussion of probability or statistics, it has nothing to do with the United States breaking away from the British Empire or your ability to pay the bills without any help from mom and dad. It means that the probability of something occurring is completely unrelated to the fact that something else did or did not occur. Independent events are occasionally confused with mutually exclusive events. If two events are independent, the probability that one occurs, given that the other has occurred, is exactly the same when the other event has not occurred. Two independent events may both occur, or only one of them may occur, or neither may occur. If two events are mutually exclusive, both cannot occur. You might finish first or 17th in a tournament, but you can’t finish both first and 17th. Those events are mutually exclusive. Now, let’s get back to independence.

There are several important concepts to be covered. First, it is important to learn to distinguish which events are independent and which aren’t. Second, it is important to learn the rules and mathematical formulas for dealing with independent events. (The rules for dealing with dependent events are a little more complicated, but before this column ends, you’ll have been introduced to both.) In many cases, it should be intuitively obvious that two events are independent. If you roll a die and pick a card, these are independent events. The chance of rolling a 6 on the die is 1/6, and the chance of picking a 6 from a normal deck of cards is 4/52, or 1/13. If you roll a 6, it does not change the probability of your picking a 6, and vice versa. If you roll a 6, and then roll another die, the probability of rolling a 6 the second time is unchanged. It is still 1/6. The second roll is independent of the first. If you pick a 6 from the deck, and then pick again without shuffling the first 6 back into the deck, the odds have changed. Your second pick comes from a deck that has only three sixes and 51 cards. The probability of picking a second 6 is now 3/51, or 1/17. The odds have increased from 12-1 to 16-1. Therefore, picking a second card from the same deck is not an independent event. It has a different probability. That’s why it is possible to benefit from card counting in blackjack. The cards left in the shoe are dependent on the cards that have been dealt. It is also why it is not possible to benefit from counting in dice games or keeping track of rolls in roulette. Each new toss of the dice or new spin of the wheel is independent of the previous ones.

It is frequently necessary to calculate the probability of both of two events happening. (Technically, this is known as the intersection of the two events.) When the events are independent, it is easy. Multiply the probability of one event by the probability of the other.

Let’s look back at our earlier examples. The probability of rolling a 6 on one die is 1/6, and of picking a 6 from a deck of cards is 1/13. So, what is the chance of doing both? We multiply 1/6 by 1/13 and find that it is 1/78. The probability of rolling a 6 twice is 1/6 times 1/6, or 1/36. Nothing could be easier than calculating the possibility of both of two independent events occurring. In fact, if you wanted to do it for more than two independent events, you would just keep multiplying. What is the chance of rolling four sixes in a row? It is 1/6 times 1/6 times 1/6 times 1/6, which equals 1/1,296.

Unfortunately, we are very often interested in the chance of two dependent events occurring, and that is a little more difficult to calculate. Let’s look at the problem of picking two sixes from the deck of cards without replacing the first one picked. The chance of picking the first 6 is 1/13. The chance of picking the second is 3/51, or 1/17. (This is known as a conditional probability. Conditional probability is a fancy way of stating the chance that something will happen, given that we know something else has already happened.) The hard part is figuring out the conditional probability. Once we have that, we go back and use the multiplication rule. Therefore, the chance of picking two sixes is 1/13 times 1/17, which equals 1/221. There is frequently more than one way to calculate the probability of an event. Here is another way to find the chance of getting two sixes. First, we know that there are six ways to get a pair (spade/heart, spade/diamond, spade/club, heart/diamond, heart/club, and diamond/club). Second, we can easily find the number of possible two-card hands. The first card can be any of 52, and the second can be any of the remaining 51. So, there are 52 times 51 ways to get two cards, which equals 2,652 ways. But this number takes into account the order in which the cards are received. That means that the 6 6 and the 6 6 are counted as two different hands. Since we don’t care in what order the cards come, we must divide by 2 to count these as one hand, and we get 2,652 divided by 2 equals 1,326. Six ways to get a pair of sixes divided by 1,326 possible hands is 1/221. As some Eastern guru once said, “There are many roads up the mountain.”

I am going to leave you with a little homework. You will find it instructive to try to calculate the answers. If you decide not to do it, you don’t even have to tell me that your dog ate your answers; just wait till my next column, when I will reveal them to you. In each case, you have to find 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?

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.