I would love it if somebody could hold my hand through this one. My brain turns to mush when attempting to account for the first of the quad giving cards being chosen and manipulating the probability of the second card being chosen.

Any related literature to what I'm sure seems to be inane to most of you is appreciated

Cheers!

I think you are asking if you are holding a pocket pair in NLHE what is the probability of making quads of that particular rank after the 5-card board is dealt out?

To make quads of that particular rank, the remaining two of your pocket pair rank must appear on board, of course. There are 48 cards of a differing rank in the deck from which the other 3 board cards will come. And there are 50 total cards (the original 52 cards minus your 2 hole cards) in the deck from which the 5 board cards are chosen.

So the prob of making quads of your rank when you hold a pocket pair is given by:

[C(2,2)*C(48,3)] / C(50,5) = (1*17,296) / 2,118,760 = 0.816%

where C(X,Y) reads "X choose Y" denotes the number of combinations of choosing Y items from X total items when choosing without replacement and when the order of the choices does not matter.

I hope that is what you are asking and I hope that I answered it correctly.

I don't know if this will help or hurt, but you can also derive the answer via "direct" probabilities (I think that is what your original question referred to).

Since the order of the 5 board cards does not matter, we will calculate the probability of making quads of our rank in a specific order, and then multiply by the total number of relevant orderings of the board cards that would give us our desired quads.

Tackling this second point first. You need 2 specific cards to appear out of 5 cards. Denote (X,Y) as the cards you seek appearing as cards X and Y of the board. Then you make your quads if (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), or (4,5). So by counting we see that there are 10 ways for our two desired cards to appear out of 5 card orderings. Meaning that we will need to multiply the probability of any one "quad-giving" ordering by 10 to get the overall "quad-making" probability below.

Of course, we could have derived this 10 figure in other ways. Using the Choose notation from the earlier post, there are C(5,2) = 10 ways for your two cards to appear when the specific order does not matter. In case anybody does not know, C(X,Y) = X! / [Y! * (X-Y)!] where X! is the product of the integers from 1 to X. So C(5,2) = 5!/(2!*3!) = 120/(2*6) = 120/12 = 10.

Okay, now we seek the probability of getting quads of our rank when we get to specify which board card positions in which we desire our ranked cards to appear. For simplicity let's have our two cards appear first and second. We would get the exact same probability no matter which two specific positions we chose here.

Prob of the first board card being our rank = 2/50 since there are 2 cards in the 50 which are of our rank.

Prob of the second board card being our rank (given that the first card to appear was of our rank) = 1/49 since there would now be 1 card left in the remaining 49 cards which is of our rank.

Note that we don't need to carry on any further here since the other board positions can be filled with any remaining cards and we would still make our quads. That is, the probability of the three remaining positions are all equal to one.

So, we can put this all together by saying that the overall probability of making quads of our rank when the complete 5-card board is dealt out when we hold a pocket pair is:

= 10 * (2/50) * (1/49) = 20/2450 = 0.816% as we found in the prior post.

Thank you, guys!