[Buzz]

“can you please break down the numbers:
"(5*4/2+5*42+2*3+9*6)/1081 = 280/1081 = 0.259 is the true probability ". What do the numbers represent; where do you get them from? I like math and would like to understand what you're trying to show me above.”

Domit - Okay, I’ll break down the numbers for you. But first let me make it clear that I agree with Al Mirpuri. Get yourself a copy of Winning Concepts in Draw and Lowball by Mason Malmuth.

Second, I have to warn you that I’m not a mathematician and advise you that you might get a better explanation from one of the mathematicians who post over on the 2+2 probability forum.

Third, you should understand there are a variety of ways to solve probability problems. I might (likely would) set up the problem differently the next time I did it. But here goes.

Think of some particular holding in a setting where you might want to keep an ace kicker. Note that I’m not advocating keeping an ace kicker. But suppose, for example, you were dealt JhJdAc2s3h in a very tight game and decided to keep JhJdAc and draw two cards.

You know the whereabouts of five cards, the three you are holding and the two you just mucked. In a 52 card deck there are 47 cards you haven’t seen. When you draw two cards, you could get any two of these forty seven cards.

The order in which you are dealt the two cards for your draw is immaterial. This allows calculation of two-card “combinations” rather than two-card “permutations.”

The total possible number of two-card combinations of forty seven cards, the total number of possible two card draws is given by the expression 47 choose 2. (There are several ways mathematicians write this, but 47 choose 2 will do). 47 choose 2 turns out to equal 1081. Let me try to explain.

Take a deck of cards and remove JhJdAc2s3h. You should have 47 cards left. Now take the first of these cards, whatever it is. How many cards can you put with it to make some two-card combination? The answer is 46. If the order in which the cards are combined (which comes first and which comes second) is unimportant, then there are 46 ways to combine the first card with some other card. Take a minute and make sure this concept makes complete sense to you. Okay? Now we’re done with that first card. There is no other way to make a two-card combination with that first card. Take it out of the pack and put it with the cards you have already mucked.

How many cards are left? The answer is 46. Next take the first of these 46 cards, whatever it is. How many cards can you put with it to make some two-card combination? This time the answer is 45.

Perhaps you can see ahead. There are going to be 46+45+44+43+42+41+40+
39+38+37+36+35+34+33+32+31+30+
29+28+27+26+25+24+23+22+21+20+
19+18+17+16+15+14+13+12+11+10+
9+8+7+6+5+4+3+2+1 two-card combinations of the 47 cards. All those numbers added together equal 1081.

Probably no one would calculate 47 choose 2 by actually adding all those numbers together. 47 choose 2 also equals (47*46)/(1*2).
47 choose 2 also equals 47 factorial divided by 2 factorial and divided by 45 factorial.
In other words, 47 choose 2 also equals (47!)/(2!*45!) or 47!/2!/45!.
And any of these equals 1081.

At any rate, when you know the whereabouts of five cards, and are drawing two of a possible 47 cards, there are 1081 possible outcomes for you.

That’s where the denominator, 1081, of the expression “5*4/2+5*42+2*3+9*6)/1081” comes from. 1081 is the number of possible two card combinations in 47 unknown cards. Now for the numerator.

If you have kept JhJdAc and discarded 2s3h, next we need to know how many of these 1081 two-card combinations would improve your hand. There are four ways a two card draw combination could improve JhJdAc to a hand better than a pair of jacks with kickers. You could make quad jacks, aces full, jacks full, trip jacks, two-pairs-aces-over-jacks, or two-pairs-but-not-aces. That’s the full extent of it.

(1) A jack or an ace plus a jack or an ace makes either quad jacks or a full house. (2) A jack or an ace plus any other card makes either trip jacks or two pairs, aces over jacks. (3) Any pair (other than aces or jacks) makes two pairs.

(1). 5*4/2 represents the number of possible two-card combinations that have a jack or an ace plus a jack or an ace.

(2). 5*42 represents a jack or an ace plus any other card.

(3). 2*3+9*6 represents any pair (other than aces or jacks).

Breaking category (3) down a bit further, there are three ways to be dealt a pair of deuces, since you have discarded the deuce of spades. They are 2h2d, 2h2c, and 2d2c. Similarly there are three ways to be dealt a pair of treys. Thus 2*3.

Continuing breaking category (3) down a bit further, there are six ways to be dealt a pair of fours. They are 4s4h, 4s4d, 4s4c, 4h4d, 4h4c, and 4d4c. Similarly there are six ways to be dealt a pair of fives, sixes, sevens, eights, nines, tens, jacks, queens, and kings. There are nine ranks of cards here. (Fives are one rank, sixes another, sevens another, etc.). Thus 9*6.

Thus 5*4/2 + 5*42 + 2*3 + 9*6 = 280 represents the total number of two-card draw combinations that will improve JhJdAc to a hand better than a pair of jacks with kickers.

Lastly, divide the 280 by 1081 to get the probability of improving JhJdAc to a hand better than a pair of jacks with kickers.

Hope this makes it clear.

Buzz