The fourth and fifth cards cannot be the same as each other nor the first three cards. There are 12 values left, and we need to pick 2: B(12,2) = 12! /2! (12-2)! = 12! /2! 10! = (12×11)/2 = 132/2 = 66 combinations of value. But there are 4 suits for each of the fourth and fifth cards. So there are 66×4×4 = 1,056 ways to choose the fourth and fifth cards.

That makes for (13×4)x(66×4×4) = 54,912 ways to chose a five-card hand that produces three of a kind.

Another way to do this, which is easier to understand, is as follows:

The fourth card in the hand may not be the same as the first three. So we have only (13-1)x4 = 48 cards left. Our total so far is 13×4 x 48.

The fifth card should not be the same as first three nor the fourth, or the latter case will be a full house. (This had me stumped for over an hour.) So there are only (13-2)x4 = 44 cards to choose from for the fifth card.

That suggest that there are 48×44 = 2,112 ways to pick the fourth and fifth cards, but that would be true if order mattered. Since it doesn’t, we have to divide by 2! = 2, producing 2,112/2 = 1,056, as before.

Thus, the total number of 5-card trip hands is 13×4 x 1,056 = 54,912 =~ 1/50 odds.

One Pair

Let’s calculate the number of one-pair hands before two-pair. It’s done similarly to trips (three of a kind). The first card can be any of 13 values. Each value has 4 representative cards, but we just want unique combinations of 2 cards. Pull out the old binomial formula, and we have B(4,2) = 4! /2! (4-2)! = 4×3×2×1/ (2×1) (2!) = (4×3)/(2×1) = 2×3 = 6. So there are 13×6 = 78 ways to get unique pairs for the first two cards.

The third, fourth, and fifth cards of the hand may not be the same as each other nor the same as the first two cards. So there are 12×4 = 48 choices for the third card, 11×4 = 44 choices for the fourth card, and 10×4 = 40 choices for the fifth card. That makes 13×6 x48×44×40 = 6,589,440 possible combinations, if order mattered. Since it doesn’t, we have to divide by 3! = 6. So we have 6,589,440/6 = 1,098,240 =~ 1/2.36.

You could also calculate this using the first method used above: 13 x B(4,2) x B(12,3) x 4×4×4 = 13×6 x 12! /3! (12-3)! x 64 = 78 x (12×11×10/3×2×1) x 64 = 1,098,240, as above.

Two Pair

Let’s use the compact method. The first pair can be selected from any of 13 values. There are B(4,2) = 6 ways to select the first pair of cards. That’s 13×6 = 78 ways.

The second pair can be selected from any of 12 values. Again, there are B(4,2) = 6 ways to do this. So far, there are 13×6×12×6 = 5,616 ways to get two pair for the first four cards – if order mattered. So we have to divide by 2! That means 5,616/2 = 2808

The fifth card cannot be the same value as any of the preceding cards. So we have to pick it from 11×4 = 44 cards. Thus, the total number of unique five-card hands containing two-pair is 2808×44 = 123,552 =~ 1/20. More **cbetcasino.fr**

Full House

A full house consists of a triplet and a pair. There are two ways we can calculate this, but order doesn’t matter, so we’ll just pick one way.

Let’s pick the triplet first. Refer back to the calculations for three of a kind. There are thus 13 x B(4,3) = 13×4 = 52 ways to pick a triplet.

For the pair, we have to pick from 12 values: 12 x B(4,2) = 12×6 = 72 ways.

So the total number of ways to get a full house is (52×72) = 3,744 =~ 1/700 odds.

No pair

The simplest way to calculate the number of five-card hands in poker that aren’t at least a pair is to add up the total number of ways above, then subtract the total from 2,598,960:

2,598,960 – (4+36+624 +54,912 +1,098,240 + 123,552 + 3,744) = 1,281,112 =~ 49% odds of having some favourable hand – if you were the only player.

Understand that these are not only approximate, to make it easier to remember them, but they are not really your odds in a real game of poker. That’s because these odds assume one player.

Adding more players makes calculating the odds far more complex. I’ll try to cover that in the next post.