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

Question

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.

Answer 1

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).