In the standard game of poker, each player gets5 cards and places a bet, hoping his cards are 'better'than the other players' hands.

• Can An Ace Count As A 1 In Poker
• Can You Use An Ace As A 1 In Poker

The game is played with a pack containing 52 cards in 4 suits, consisting of:

13 hearts:
13 diamonds
13 clubs:
13 spades:

♥ 2 3 4 5 6 7 8 9 10 J Q K A
♦ 2 3 4 5 6 7 8 9 10 J Q K A
♣ 2 3 4 5 6 7 8 9 10 J Q K A
♠ 2 3 4 5 6 7 8 9 10 J Q K A

The number of different possible poker hands is found by counting the number of ways that 5 cards can be selected from 52 cards, where the order is not important. It is a combination, so we use `C_r^n`.

The number of possible poker hands

`=C_5^52=(52!)/(5!xx47!)=2,598,960`.

Royal Flush

The best hand (because of the low probability that it will occur) is the royal flush, which consists of 10, J, Q, K, A of the same suit. There are only 4 ways of getting such a hand (because there are 4 suits), so the probability of being dealt a royal flush is

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`4/(2,598,960)=0.000 001 539`

Straight Flush

The next most valuable type of hand is a straight flush, which is 5 cards in order, all of the same suit.

For example, 2♣, 3♣, 4♣, 5♣, 6♣ is a straight flush.

For each suit there are 10 such straights (the one starting with Ace, the one starting with 2, the one starting with 3, ... through to the one starting at 10) and there are 4 suits, so there are 40 possible straight flushes.

The probability of being dealt a straight flush is

`40/(2,598,960)=0.000 015 39`

[Note: There is some overlap here since the straight flush starting at 10 is the same as the royal flush. So strictly there are 36 straight flushes (4 × 9) if we don't count the royal flush. The probability of getting a straight flush then is 36/2,598,960 = 0.00001385.]

The table below lists the number ofpossible ways that different types of hands can arise and theirprobability of occurrence.

Ranking, Frequency and Probability of Poker Hands

Question

The probability for a full house is given above as 0.001441. Where does this come from?

Answer

Explanation 1:

Probability of 3 cards having the same denomination: `4/52 xx 3/51 xx 2/50 xx 13 = 1/425`.

(There are 13 ways we can get 3 of a kind).

The probability that the next 2 cards are a pair: `4/49 xx 3/48 xx 12 = 3/49`

(There are 12 ways we can get a pair, once we have already got our 3 of a kind).

The number of ways of getting a particular sequence of 5 cards where there are 3 of one kind and 2 of another kind is:

`(5!)/(3!xx2!)=10`

So the probability of a full house is

`1/425 xx 3/49 xx 10 ` `= 6/(4,165)` `=0.001 440 6`

Explanation 2:

Number of ways of getting a full house:

`(C(13,1)xxC(4,3))` `xx(C(12,1)xxC(4,2))`

`=(13!)/(1!xx12!)` `xx(4!)/(3!xx1!)` `xx(12!)/(1!xx11!)` `xx(4!)/(2!xx2!)`

`=3744`

Number of possible poker hands

`=C(52,5)` `=(52!)/(47!xx5!)` `=2,598,960`

So the probability of a full house is given by:

`P('full house')`

`='ways of getting full house'/'possible poker hands'`

`= (3,744)/(2,598,960)`

`=0.001 441`

In the last couple of years, as theoretical understanding of poker has galloped forward, an entire new vocabulary has emerged. I mean, when I was playing limit hold'em in San Jose 25 years ago, had you said, 'minimum defense frequency,' they'd have thought you were talking about the 49ers.

One of the words that you hear most frequently now is 'combinations' or 'combos.' Once you accept the concept of an opponent (or yourself) having a range of hands, the next interesting question is, 'Well, how many hands are in that range?' The way you answer that is to figure out how many 'combinations' there are of the hands that make up the presumed range.

Counting Combos: Pocket Pairs

Let's try one. Suppose you are playing $1/$3 no-limit hold'em and before the flop, you raise to $12 in early position with a pair of jacks. A straightforward and uncreative player in late position reraises you to $40. It folds back around to you. Based on your knowledge of this player, you expect her to three-bet with only a pair of queens or better, and all of her ace-kings. So her three-betting range is Q-Q, K-K, A-A, and A-K. How do your jacks fare against her presumed range?

Well, there are six combos of every pocket pair. To determine that, we see that we can randomly pick any of the four queens in the deck, and now have three remaining queens with which to make a pair. So that's 4 x 3 = 12. However, if we pick the first, and then the , that's no different than picking the first and then the . So we must divide by two to get a total of six.

Another way of skinning the same cat: pick the and see that you can then pick the , , or to make a pair. That's three. Now pick the first, leaving just the and to pair. Two more. Finally, the has only the to pair it. One more. 3 + 2 + 1 = 6. Math is beautiful.

So, six combos for each pocket pair. For Q-Q / K-K / A-A, that's a total of 18 combos. So far so good.

Counting Combos: Non-Paired Hands

What about A-K combos? If we give the villain all of the ace-king combos, then she can make one by taking any of the four aces and crossing them with any of the four kings. 4 x 4 = 16 and that's the number of combos.

Of course, if she restricts herself to suited ace-kings, then pretty clearly there are just four of those — , , , .

Calculating Our Equity

We'd agreed that she would three-bet all of her ace-king combos plus queens, kings, or aces, so we conclude she has one of 34 possible hands: 18 pocket pairs and 16 ace-kings.

The 18 pocket pairs are 81-to-19 favorites against us, while we are a 57-to-43 favorite against the 16 ace-kings. To determine our equity against her, we weight each combo by its share of the range pie, compute our equity against that slice, and then sum them up.

Can An Ace Count As A 1 In Poker

For this example, we can compute our equity as follows:

1. Against the higher pairs, we have 0.18 (18%) equity. The higher pairs make up 18 / 34 of the villain's presumed range. So our equity for that piece = 0.18 x 18 / 34 = 0.10 or 10% (0.095 to be closer).
2. Against the ace-king combos, we have 0.57 (57%) equity. The ace-kings make up 16 / 34 of the villain's range. Our equity against that piece is 0.57 x 16 / 34 = 0.27 or 27%.
3. Summing the two, we get 0.10 + 0.27 = 0.37 or 37% equity against her presumed range.

The good news is that there are programs such as Pro Poker Tools and the like that let you ask questions such as 'How much equity does a pair of jacks have against a range of Q-Q / K-K / A-A / A-K?' But it's useful to know how those things are calculated.

Can You Use An Ace As A 1 In Poker

Using Combos to Improve Decision-Making

What to do with that information is beyond the scope of this article, but as an example, if the villain were all in for her $40, we'd know exactly how to proceed.

Setting aside rake for the moment, there's $12 + $40 + $1 + $3 = $56 in the pot. It costs us another $28 to call. Conveniently enough, we're getting exactly 2-to-1 odds to call, so we must have at least 33.3% equity to call her shove. We have a hair above that (37%), so we shrug, slide in the extra $28, and ask the dealer to run out the board.

By the way, I had suggested that we ignore the rake for simplicity. Note that in this case once we take the rake into effect, this could turn into a fold. If you don't see that, subtract the rake ($5 or whatever) from the pot and redo the pot odds calculations, remembering that you still need to call the full $28.

Conclusion

I grant that counting pairs and ace-king combos is relatively simple. But suppose in the heat of battle, a flop comes down and you believe that your opponent could have (among other possible hands) any of the heart flush draws that are two suited Broadway cards, plus all of the ace-high flush draws. How many flush draw combos does she have? (See the answer below.)

Not surprisingly, the best way to get better at this is to practice in the lab (a.k.a. 'your kitchen table'). Go over common situations and learn the arithmetic. Eventually, you'll be as comfortable with the important ones as you're sure that jacks have 37% equity against a range of {QQ+, AK}.

This stuff is not trivial and if you're not used to working with numbers, it can be a bit daunting. But at least some of your opponents are already doing it, and once you get the hang of it, you might even enjoy the mental gymnastics.

P.S. Your villain can have the suited-in-heart combos of A-Q, A-J, A-T, A-9, A-7, A-6, A-5, A-4, A-3, A-2, Q-J, Q-T, J-T, for a total of 13.

Lee Jones can help you count combos and then count your winnings. Go to leejones.com/coaching and schedule a free coaching consultation. Lee specializes in coaching live cash game players.

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