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Odds: Part 3

by Steve Zolotow |  Published: Mar 18, 2015

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Steve ZolotowThis is the third column on odds and related concepts. The first one dealt with a discussion of what odds are and distinguished between true odds (which focus on the chance of something really happening) and money odds (which relate to the money you can lay or take on it happening). The second one covered the basics of converting the chance of something happening (usually this probability is expressed as a percentage or a fraction) to the odds of it happening. This column will discuss some fundamental points about calculating odds or probabilities for multiple events.

A multiple event, unsurprisingly, consists of more than one event. Here is a hold’em example which illustrates this. A flush draw with one card to come (on the turn) is one simple event, but if the flush draw has two cards to come (on the flop), then it is a multiple event. A card of the key suit may come on fourth street or on the river. Obviously it can also come on both or neither. During March Madness, a basketball team who reaches the final 16 must win four games to be NCAA champion. Finding the odds against their winning requires the calculation of multiple events (four separate games).

When discussing multiple events, the concept of independence also becomes important. When two events are independent, the odds of one happening are not influenced by whether or not the other has happened. If two dice are rolled, the number on one die doesn’t influence the number on the other. If two cards are taken from a deck, the value of the first influences what is available for the second. The first card is an ace four times out of 52. Now when the second card is picked, the second card will only be an ace three times out of 51. Rolls of dice at craps or numbers that appear in roulette are independent. This means keeping track of what has occurred is useless. At roulette, knowing that black has come up four times in a row doesn’t make black any more or less likely to occur. But in blackjack, the cards dealt are dependent on those that have already been dealt. In single deck 21, if four aces have been dealt, there is no chance that the next card will be an ace—they are all gone. This is why counting cards is useful at blackjack, but tabulating results is useless at craps or roulette.

The most straightforward way to calculate the odds for multiple events, especially when they are dependent events, is to use a tree structure. Each event leads to two or more branches. Multiply all the fractions to arrive at the result for a path. It is easier to use probability percentages or fractions to do the calculations, and then work out odds at end. Also, using percentages gives you a chance to check your work. All the percentages must add to one. Let’s work through the case of a flush draw on the flop. On fourth street, there are 47 unknown cards (52 minus your two cards and the three cards on the flop) and nine cards left in the key suit. Note there are nine cards in the suit left when you have two and there are two on the flop or when you have one and there are three on the flop. The odds of making the flush are 9/47, or 19 percent. The odds of missing on fourth are 38/47, or 81 percent (As a check 19 and 81 = 100 percent, everything is accounted for). Next, let’s look at fifth street, the river. Now there are only 46 cards left. If a card of the key suit came on fourth, then only eight of the suit are left. If it didn’t, then nine are still left. So, 19 percent of the time the flush has come in. The other 81 percent of the time, it hasn’t. Of this 81 percent, 9/46, or 19.6 percent of the time, it will come, but 37/46th or 80.4 percent of the time, it won’t. But this branch only occurs 81 percent of the time, so we have to multiple the results by 81 percent. The following chart summarizes the cases:

Event Calculation Percentage
Flush card on 4th and 5th street: 9/47 times 8/46 3.3
Flush card on 4th and not 5th 9/47 times 38/46 15.8
Miss on 4th & flush on 5th 38/47 times 9/46 15.8
Miss on 4th and miss on 5th 38/47 times 37/46 65.0
Total 99.9

Notice that the total is approximately 100 percent, which means we haven’t left anything out. Notice too that the flush comes 35 percent of the time and misses 65 percent. To find the odds against hitting the flush divide 65 by 35 = 1.86 to 1. To make this easier, you can see that the flush hits about one third of the time and is a little less than a two to one underdog. The above chart also shows that when you have a flush draw, 3.3 percent of the time, two of the suit will come. This means that if you make a low flush, with a hand like 8-7 suited, on fourth street, and your opponent has A-K offsuit, but with one of the suit, he will catch you on the river nearly 20 percent of the time you hit on fourth street. There is no need to calculate all of the common situations. Poker nerds have already done the work for you. If you want to verify something, Card Player has a hand odds calculators that is accurate and easy to use. In the next column, I will discuss the odds of a lot of common situations and what they imply. Then pot odds and implied odds will be covered. ♠

Steve ‘Zee’ Zolotow, aka The Bald Eagle, is a successful gamesplayer. He has been a full-time gambler for over 35 years. With two WSOP bracelets and few million in tournament cashes, he is easing into retirement. He currently devotes most of his time to poker. He can be found at some major tournaments and playing in cash games in Vegas. When escaping from poker, he hangs out in his bars on Avenue A in New York City -The Library near Houston and Doc Holliday’s on 9th St. are his favorites.