# Probability of getting 3 of a kind by at least the river (assuming unpaired hand)

Here's the question: Assuming an unpaired hand of 2 cards in poker (2 & 10, 3 & Q, NOT A&A), what is the probability of having 3 of a kind after all 5 community cards have been dealt?

My work: The probability can be calculated as:

outcomes where you have a triple / all possible outcomes

We know that order does not matter.

So, numerator can be calculated as (3C2) * (48C3) * 2. Assume we have 1 ace. Then, there are 3C2 ways to choose the next 2 aces to complete our triple. We then have 48C3 ways to choose the other 3 cards that make up the 5 community cards. we multiply by two since we have two cards in our hands, and therefore 2 ways to make a triple.

The denominator is just the number of ways to choose 5 cards from 50. so 50 C5.

This returns 4.9%. Is this correct? I've seen a few posts about flopping three of a kind. But haven't seen one about getting three of a kind by the river. If there is a pre-existing post, please direct me to it if you are aware of one.

• The question is not clear. Do you mean '3 of a kind' or '3 of a kind or better'? For example, what if you have a full house? Or a flush as well as 3 of a kind? Or just a flush? What if there are 3 of a kind on the board - does that count? Also, note that your "multiply by two" step is conceptually incorrect because you will double-count the boards where both of your hole cards make 3 of a kind. May 14 at 8:30
• Apologies - I asked the question in an unclear manner. I meant 3 of a kind only using the two cards that are in your hand, assuming they are not paired. Additionally, this would include any combination of cards that has 3 of a kind (with 1 card in the 3 being in your hand). So a full house would be counted, as would a flush and 3 of a kind, and quads. But not just a flush. In terms of correcting for overcounting - this is what I would subtract from the numerator: 3C2 * 3C2 * 46. 3C2 ways to choose the two sets of two cards that match the ones in your hand, and 46 remaining cards to choose from May 14 at 17:20

Requirement

The clarification in the comments confirms that this question relates to all hands which include three of a kind (including any better winning hands such as a flush or full house) where at least one of the triple in the original (unpaired) starting hand. Hands where both hole cards make three (or four) of a kind are also included.

Overcounting is not (as suggested in the comments) as simple as just subtracting `3C2 * 3C2 * 46`. The problem is that when you use `46C2` or `46`, those 46 cards include an Ace and a King, so boards with three Aces or three Kings will still be counted twice.

The exact result is quite fiddly to calculate accurately - you need to be very careful to avoid double-counting - but one approach is given below.

Approach

Without loss of generality we can assume the original hand is AK (meaning that any other unpaired hand will yield the same result). Suitedness is not relevant as flushes do not matter.

First, sum the following:

1. The number of boards with exactly two Aces (and any three other cards).
2. The number of boards with exactly three Aces (and any two other cards).
3. The number of boards with exactly two Kings (and any three other cards).
4. The number of boards with exactly three Kings (and any two other cards).

Note that this will double-count the following boards (where `x` indicates a card which is not an Ace or King): `AAKKx` (in #1 and #3 above); `AAAKK` (in #2 and #3); and `AAKKK` (in #1 and #4). We therefore need to calculate the number of each and subtract those numbers. (Note that these three board textures are themselves distinct, i.e. do not overlap, so there is no double-counting here.)

Lastly, use the result as the numerator and the total number of possible boards as the denominator (as per the question).

As you say, we can ignore the order of the board, so by "number of boards" I mean the number of distinct combinations of five cards, ignoring their order.

Calculation

1. The number of boards with exactly two Aces is `3C2 * 47C3`. (We use 47 because we exclude four Aces plus the King in your hand.)

2. The number of boards with exactly two Kings is also `3C2 * 47C3`, by equivalent logic.

3. The number of boards with exactly three Aces is `47C2`. (Again, use 47 because we exclude four Aces plus the King in your hand.)

4. The number of boards with exactly three Kings is also `47C2`, by equivalent logic.

And for subtractions (for double-counting):

1. The number of `AAKKx` boards is `3C2 * 3C2 * 44` (44 because there are 44 cards which are not an Ace or a King).

2. The number of `AAAKK` boards is `3C2` (choosing two of the three Kings not in your hand).

3. The number of `AAKKK` boards is also `3C2`, using equivalent logic.

Putting all of this together, the total number of unique boards containing at least two Aces and/or at least two Kings is:

`(3C2 * 47C3 * 2) + (47C2 * 2) - (3C2 * 3C2 * 44) - (3C2 * 2)`

The total number of boards (again, up to ordering) is `50C5`, so divide the above result by `50C5` to get 0.046749 or 4.67% (to 2dp).

As expected, this is very close to your original answer, but slightly lower because the original calculation included some double-counted boards.

Finally, I should point out that this calculation includes hands where your hole cards do not actually contribute. For example, suppose you hold 32 and the board is AAA22. You have made "three of a kind" with your 2, but this doesn't improve upon the board itself which is Aces full of 2's - in other words, you can do no better than "play the board". This seems slightly at odds with the requirement to include only three-of-a-kind hands where one of the triple is in your hand, but maybe that's ok (and of course this only makes a small difference to the overall result).