All four cards of a kind (like all four sixes) between two pockets and flop

I have a question in relation to a rare kind of hand. I have been playing poker for ten years and remember only two hands like this so I am guessing the probability is one in maybe a gazillion.

This is where, between two players and the flop, all four of a kind are out. (In each case standard 9 handed holdem)

I shall explain by two examples, real world:

Hero: pocket 87 Villain: 88 Flop: 877
Hero: pocket 86 Villain: A6 Flop: 664

These are two hands I had IRL, many years apart. I had the worst of both and donked off a lot of doughnuts.

This seems to happen almost never, I should dearly love to know the odds of this.

It is quite a different situation where one player flops quads: normally not too much action, guy with quads has too much of the board. Also different where, for example, 3 players have an ace and there is an ace high flop. Here too all four cards of a kind are out between pockets and flop though the action tends to be more restrained. In contrast, when all four are out between two players and the board, it is going to be all in on the flop.

Many thanks if you have time for this one ;)

Quote: anglodonk

I shall explain by two examples, real world:

Hero: pocket 87 Villain: 88 Flop: 877
Hero: pocket 86 Villain: A6 Flop: 664

These are two hands I had IRL, many years apart. I had the worst of both and donked off a lot of doughnuts.

This seems to happen almost never, I should dearly love to know the odds of this.

The first thing that you want to determine is the probability of a three of a kind in four cards. This is as easy as determining a specific three-of-a-kind in four cards and just multiplying it by thirteen ranks.

(4/52 * 3/51 * 2/50 * 48/49) + (4/52 * 3/51 * 48/50 * 2/49) + (4/52 * 48/51 * 3/50 * 2/49) + (48/52 * 4/51 * 3/50 * 2/49) = 0.00070920675

0.00070920675 * 13 = 0.00921968787

Now that we have done that, all we have to do is determine the probability of the other card showing up on the flop. That probability is simply 1 - the probability of that card NOT showing up, which is easier to do:

1 - (47/48 * 46/47 * 45/46) = 0.0625

Okay, so even with the other three out, there's still a probability of 6.25% that the fourth one is one of the three flop cards. That checks out, as you'll see:

(47/48 * 46/47 * 1/46) + (47/48 * 1/47) + (1/48) = 0.0625

All that remains is to combine the probabilities, so the first thing we want to happen is three out of four of the player's first two cards to be the same, and then we want to factor in the 6.25% probability of the fourth showing up on the flop:

0.00921968787 * .0625 = 0.00057623049 or 1/0.00057623049 = 1 in 1,735.42*****

*****This assumes playing heads up. With more players, it becomes more likely to happen because the wired pair as compared to the other players off-card have multiple things that they could be. It could also happen to more than two players in one hand!

******Also, don't get upset with yourself for, "Donking off," especially not in the first instance. How do you not bet a Full House?

Quote: anglodonk

I have a question in relation to a rare kind of hand. I have been playing poker for ten years and remember only two hands like this so I am guessing the probability is one in maybe a gazillion.

This is where, between two players and the flop, all four of a kind are out. (In each case standard 9 handed holdem)

I shall explain by two examples, real world:

Hero: pocket 87 Villain: 88 Flop: 877
Hero: pocket 86 Villain: A6 Flop: 664

I'm confused by what you mean by "all four of a kind are out." In the first one, doesn't the hero have four 7s if the Turn or River is a 7, and in the second one, what if the Turn or River is a 4 and one of the other seven players has pocket 4s?

I believe he meant that all four of the cards, with one player having them in the pocket, are out by the time the flop drops.

Quote: Mission146

The first thing that you want to determine is the probability of a three of a kind in four cards. This is as easy as determining a specific three-of-a-kind in four cards and just multiplying it by thirteen ranks.

(4/52 * 3/51 * 2/50 * 48/49) + (4/52 * 3/51 * 48/50 * 2/49) + (4/52 * 48/51 * 3/50 * 2/49) + (48/52 * 4/51 * 3/50 * 2/49) = 0.00070920675

0.00070920675 * 13 = 0.00921968787

Now that we have done that, all we have to do is determine the probability of the other card showing up on the flop. That probability is simply 1 - the probability of that card NOT showing up, which is easier to do:

1 - (47/48 * 46/47 * 45/46) = 0.0625

Okay, so even with the other three out, there's still a probability of 6.25% that the fourth one is one of the three flop cards. That checks out, as you'll see:

(47/48 * 46/47 * 1/46) + (47/48 * 1/47) + (1/48) = 0.0625

All that remains is to combine the probabilities, so the first thing we want to happen is three out of four of the player's first two cards to be the same, and then we want to factor in the 6.25% probability of the fourth showing up on the flop:

0.00921968787 * .0625 = 0.00057623049 or 1/0.00057623049 = 1 in 1,735.42*****

*****This assumes playing heads up. With more players, it becomes more likely to happen because the wired pair as compared to the other players off-card have multiple things that they could be. It could also happen to more than two players in one hand!

******Also, don't get upset with yourself for, "Donking off," especially not in the first instance. How do you not bet a Full House?

Actually, I just noticed that one of your example hands has the pair in with the flop. It's actually easier for me to reverse-engineer that one as of the flop, so let's take a look at the probability of getting a pair out of three cards:

https://wizardofodds.com/games/three-card-poker/

Already done. 0.169412

Okay, so now we want each of the other cards to be one of the cards of each player's hand. That's easy:

(2/49 * 47/48) + (47/49 * 2/48) = 0.07993197278

AND:

(1/47) + (46/47 * 1/46) = 0.04255319148

Okay, so let's multiply all that together:

0.04255319148 * 0.07993197278 * 0.169412 = 0.00057623129 or 1/0.00057623129 = 1 in 1735.41

You see that it's actually the same as the other way, except for a rounding error somewhere...probably because of the order in which we did things.

Okay, so if we combine that order:

One Player-One Villain-Two Flop

With what was figured earlier

One Player-Two Villain-Two Flop

0.00057623049 + 0.00057623129 = 0.00115246178 or 1/0.00115246178 = 1 in 867.71

Most importantly, that actually ignores the fact that the person we're identifying as, "Player," could also be the one to get the pair. That doesn't change the math other than to add another of the first number:

0.00057623049 + 0.00057623129 + 0.00057623049 = 0.00172869227 or 1/0.00172869227 = 1 in 578.47

At the end of the day, not that it would matter in your example, but the flop could also have three of the 4OaK cards. Ultimately, the probability of a 4OaK in seven-card stud is .00168 or 1/.00168 = 1 in 59.524

Either way, I'm pretty confident that, with two players, the combined probability of:

4OaK/4Oak---4Oak/x (Either order)---4OaK/x/x (Any Order)

4Oak/x (Either order)---4OaK/x (Either order)---4OaK/4Oak/x (Any order)

AND

4Oak/x (Either Order)---4OaK/4OaK---4Oak/x/x (Any order)

Is about 1 in 578.47 and does not account for many other possible 4OaK combinations.

In other words, this is not at all rare...in terms of it being dealt out. Keep in mind that this is also for heads-up, or if you prefer, randomly picking any two players at a nine-seat table. When you start adding in more people, it becomes more likely and could even happen multiple times in one hand.

The only reason it seems more rare than it is is because you probably don't average 578 hands every day of your life, or anything, and the 4OaK with the off card, as a practical matter, probably gets folded a lot. Hell, the wired pair may not even always play.

wonderful analysis, thank you ;)