Don't be intimidated by the polysyllabic title of this article -- it really can be fairly easy stuff -- if I can do a good job in explaining it. The methods that I'll describe can be used to figure such things as the chances of there being three or more cards of one suit on the table in Hold'em (or Omaha), etc.-- it's not too difficult, once you understand a couple of basic principles.
To create an example, let's say you're to choose a meal from a 7-item menu of "steak," "chicken," "shrimp," "egg roll," "pizza," "lobster," "chocolate cake" -- and you're to choose three different items from it (I'll take pizza, steak, and chocolate cake). How many different combinations of choices is it possible for you to make? The answer is 35. If you want to convince yourself of this, write down a list of every combination of three of these items that you can think of. If you get less than 35, you left some combinations out, or if you get more than 35, you counted some combinations more than once. The actual mathematical expression for how many different ways you can choose a number of things (3 here) from a set of things (7 here) is:
(7 x 6 x 5) / (3 x 2 x 1) = 35
The numerator (7 x 6 x 5) starts with the number of menu items (7), and has the other numbers counting downward, until you have as many numbers in it as the number of choices you will be making (3). This reflects the fact that you can choose any one of the 7 items first, then any one of the other 6 items next, and then for your third choice, you can choose any one of the 5 items that were not chosen first or second. The numerator in this example (7 x 6 x 5), when the three numbers in it are multiplied together, equals 210. The denominator (3 x 2 x 1) has all the numbers, starting with the number of choices (3), down to 1. The denominator in this example (3 x 2 x 1), when the three numbers in it are multiplied together, equals 6. This reflects the fact that there are 6 orders in which you can make the same 3 selections (that is, if I had steak, chocolate cake, and pizza, in that order, it would amount to exactly the same thing on my bill -- though maybe not so in my stomach!). And finally, when you divide 210 by 6, the result is 35.
This mathematical method has many useful applications -- such as with how many different 5-card poker hands can be made from a deck of 52 cards. That would be (better use your calculator for this one):
(52 x 51 x 50 x 49 x 48) / (5 x 4 x 3 x 2 x 1) = 2,598,960
All well and good, so there are 35 different combinations of 3 choices that we can make from a 7-item menu, and 2,598,960 different 5-card poker hands -- but how do we apply this method in a meaningful practical way? Let's work on the question of how many ways there are for there to be 3 spades in the 5 Hold'em (or Omaha) up-cards. There are 13 spades in the deck, so just think of it as choosing 3 items from a 13-item menu; and there are 39 non-spades for the other 2 up-cards, equivalent to 2 items from a 39-item menu:
(13 x 12 x 11) / (3 x 2 x 1) = 286 and (39 x 38) / 2 x 1) = 741
And to take into account that any of the 286 combinations of 3 spades can occur with any of the 741 combinations of 2 non-spades, we multiply these two numbers together, 286 x 741 = 211,926 ways for there to be 3 spades in the 5 up-cards. How about 4 spades then, with 1 non-spade?
(13 x 12 x 11 x 10) / (4 x 3 x 2 x 1) = 715
And we then multiply this by the 39 different possibilities for the 1 non-spade up-card, 715 x 39 = 27,885 different ways to have 4 spades out of the 5 cards on the table. And 5 spades?
(13 x 12 x 11 x 10 x 9) / (5 x 4 x 3 x 2 x 1) = 1287
So, we have 1,287 ways to have 5 spades. I hope you're still with me in this, as we're now ready to start getting "practical" here. We can add these numbers together, 211,926 + 27,885 + 1,287 = 241,098 ways to have 3 OR MORE spades in 5 up-cards. Remember, from way back toward the beginning of this article, we calculated a total of 2,598,960 different 5-card combinations -- well, 241,098 of those combinations (nearly 10%) contain 3 or more spades. And the numbers are the same for all 4 suits, so we multiply, 241,098 x 4 = 964,392 different ways out of 2,598,960 to have 3 or more of any one suit -- so, a little over
37% of the time, there will be possible flushes in Hold'em (or Omaha).
How about the chances of having 2 cards of the same rank (a pair), and 3 others of different ranks on the table? Well, there are 13 different ranks to choose from, and to choose 2 of the 4 cards of any rank, we have:
(4 x 3) / ( 2 x 1) = 6
And from the other 12 ranks (that is, those ranks other than that of the pair), we have to have 3 of them for the 3 side-cards:
(12 x 11 x 10) / (3 x 2 x 1) = 220
And from each of the 3 side-card ranks, we can have any one of the 4 cards of that rank. So, to figure the number of 5-card combinations including exactly one pair, we multiply, 13 x 6 x 220 x 4 x 4 x 4 = 1,098,240, which is a little over 42% of the 2,598,960 total 5-card combinations. So, in Hold'em and Omaha, there will be one pair on the table (thus possible full houses) a little more than 42% of the time. By similar application of this method, you can work out the chances of two pair, three of a kind, etc., coming up on the table.
Here's a little test for you to see if you have mastered the method described in this article. If I write down the names of 3 of the 50 states of the U.S., and seal those names in an envelope, and I then charge you $1 for a chance for you to guess what 3 states I've written down, how large a prize (in dollars) should you receive (if you guess all three correctly), so as to make this a fair proposition? Incidentally, if your guess is Wisconsin, New Mexico, and Rhode Island--don't call me, I'll call you...
