The math of ‘outs’

Jan 6, 2004 2:47 AM

There is often discussion on how to handle outs, so I have written a column which addresses that question.

There are several methods which will work, but the most logical for the beginner or reader with little math experience is the counting method, because it is easy to very that all the cases are covered. The average reader can figure one card outs, because the method is fairly straight-forward. When it comes to comparing two-card outs, the technique differs.

Where most people get confused is in trying to analyze this situation sequentially, i.e. one card at a time, since that it is way one card outs are derived. Before going any further, let’s set some definitions to avoid confusion. A mathematical universe is the total possible events. The odds of an event occurring (usually stated as odds against) is the ratio of good event to bad, or vice versa for odds against.

The first example is a common problem. What are the odds of making a flush when you have two suited cards in your hand and the board flops to of your suit? The two card outs must be treated as events in a two card universe. In this example, there are 47 x 46 total pairs of cards. (Briefly, fourth street can be any one of 47 cards. When you use any one card there are 46 cards left for fifth street.) Since it does not matter if card A comes before card B, e.g. pair AB = pair BA, we can divide 2,162 by 2 for a universe of 1,081 events.

Now, to calculate the number of events that are good versus the number of events that are bad, remove all the good events and count what is left. Nine cards will create a flush in any combination. Set these nine cards aside for a moment. That leaves a deck of 38 cards from which we will draw two at a time randomly with no possible flushes. The first card could be any of 38 cards; the second any of 37 cards and again, pair AB = pair BA. (38 x 37)/2 = 703 events. Subtract 703 events from 1,081 events to discover the number of flush possibilities.

Every two card event using those other nine cards will produce a flush. You should arrive at 378 events which produce a flush. When rounded off, the number should equal 1.86. Those are the odds against making a flush when you have four to a flush with two cards to come in hold’em.

The second example involves the odds of a board pairing in Omaha. Assume that your hand is QH-AS-KH-KC and the board is QD-9D-JS. The two card universe in Omaha is 990 (45 x 44)/2. There are seven cards which pair (a Q, three J’s and three nines) that we subtract from 45. Again we have (38 x 37)/2 or 703 events which are left. Subtracting from 990 leaves 287 favorable events. We are not done yet.

What about all the hands with two running pairs (fourth street is a card other than the seven mentioned pairing on fifth street)? The cards are 2, 3, 4, 5, 6, 7, 8, 10, A. Each number represents six pairs. (Take four cards, A, B, C, D. Combine them AB, AC, AD, BC, BD CD.) Six times nine numbers is 54 events, which we add to 287 for a total of 341.

Is that all the possibilities? Did you find two running kings? It is a slim possibility, but we must include it. With two kings in the hand, there is only one combination of KK because order does not matter. You sum of 342 subtracted from 990 equals 648. 648/342 = 1.89. The odds against the board pairing in this situation would be 1.89 to 1.

As you can see, the amount of math is more than the average player would like to delve into in the heat of battle. There are other methods, but without a solid math grounding mistakes could occur easily because the procedures are not as visible as just counting the cases.

A more sensible solution would be to work out and memorize the more common draws like open-ended straights, etc. Also, try a few variations like two perfect flush cards when you have middle pair to see how much that affects the odds. Then, when you are faced with a decision, you will have a good approximation of your odds.

Poker players, like attorneys, can face a heavy case load also.