Practical Probability — Part XIII

I ended my last column with an introduction to the idea of “running it twice.” Running it twice is a way to take some of the randomness out of a poker hand, when one of the players is all in and the betting has been completed. In many high-stakes games, it has come to replace someone giving insurance to the players in all-in situations. The person selling the insurance almost always got the best of it, sometimes by quite a lot. One player might want to insure against his opponent making a heart flush on the river, but the player offering to provide the insurance knows that 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 even into thirds and deal three separate river cards, one for each third of the pot.

For players to accept running it twice, they had to believe that it doesn’t give either player an advantage. I left you with this question: Does this give an advantage to either player? Obviously not, or players wouldn’t be doing it. To illustrate this, I asked you to use the example of a nine-out flush draw on the river, and calculate the chances and equity for each player if the river is dealt only once or if it is dealt twice. Solving this type of problem is a practical way to test your ability to perform probability calculations. For that reason, I am going to go into some detail on this process. You will need to think logically, and multiply and divide, but other than that, no complicated math is required.

Let’s look at a specific case: Player A has the A A and Player B has the 7 6. The board is K Q 3 2. If the river is dealt once, it is fairly easy to calculate the probability of the flush draw being hit, and winning the hand. Since eight cards are known, 44 are unknown. Nine of these 44 are hearts. Player B will make his flush 9/44ths of the time, or 20.45 percent. If there is $100 in the pot and the players decide to split the pot mathematically, B’s equity, the amount to which he would be entitled, is $20.45.

What if it is dealt twice? The first time is the same as dealing it once. The second time, however, is a little more complicated. First of all, there are only 43 cards left. Of these 43, nine are still hearts if the flush missed, but only eight are hearts if the flush hit. There are now four cases: miss, miss; miss, hit; hit, miss; and hit, hit. To make this easier to follow, I am going to arrange these four cases into a table:

There are several points to notice in this table. The odds on the second deal are conditional. The result of what happened on the first deal changes the odds on the second. The first and second are not independent events. The chances of a miss followed by a hit and a hit followed by a miss are the same. Notice that the four cases’ probabilities add up to 1.00. All probabilities can range from 0 to 1. A probability of 0 means that the event is impossible. With a normal deck, what is the probability of being dealt the K K? It is obviously 0. A probability of 1 means that something is sure to happen. This fact is useful in cross-checking probability calculations. In the example above, the fact that column 5’s sum is 1 tells me that I have included all of the possible outcomes and calculated their probabilities correctly. If you ever attempt to find the probabilities for all of the possible outcomes of an event but arrive at an answer that doesn’t add up to 1, you have made a mistake. If you arrive at a figure less than 1, you may have forgotten to include some outcomes. If you arrive at a figure greater than 1, you probably have counted something twice or made some arithmetic error.

A letter I received from a reader by the name of Makya prompts me to review the idea of the gambler’s fallacy. His question, slightly modified, concerns the situation in which a gambler offers you 1,300-to-1 on rolling a 6 on one die four times in a row. (The odds are actually 1,295-to-1.) After you have rolled a 6 three times, you decide to hedge and offer someone else 7-to-1 on your last roll. (The true odds are 5-to-1.) He asks how this last 6 can be both 1,295-to-1 and 5-to-1 at the same time. Each roll of the dice is an independent event. The chance of rolling a 6 is always one out of six. The fact that three sixes have already been rolled doesn’t change that fact. If 20 sixes had been rolled in a row, one might think that, in fact, there is some possibility that the die is not fair (that is, a biased or loaded die), and it may be even more likely that the next roll is a 6. The gambler’s fallacy has been described as assuming that an unusual departure from short-term expected results will be corrected in the short term. A typical case is seeing that black has come up four times in a row on a roulette wheel and thinking that red is “due.”

An even more costly variation of the gambler’s fallacy occurs when it is combined with a mistaken view of the odds. For example, a losing poker player may think he is a favorite to win each time that he plays. After losing steadily for a few hours in a hold’em game with $5-$10 blinds, he decides that he is due to get lucky and win. He therefore moves into a game with blinds of $25-$50. Since his luck in this game is independent of his results in the first game, there is no reason to expect his luck to be better than average. Not only that, but he is probably now playing against much better opponents, so he will need much better than average luck to win.

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.