# Poker: starting pair and flopping quads probability

I've been looking at charts of probabilities of certain hands occuring in poker and have been trying to work them out myself to see if I get the same answer. I'm having real trouble with having a starting pair and then flopping quads. According to charts, the probability of that happening is 0.25% or 400 to 1. I do not get this answer at all.

I managed to work out the probability the same as in the chart of a starting pair flopping a set (around 12% or 8 to 1). I'm assuming you have to combine the probability first of all of getting dealt a pair ([13C1][4C2]/[52C2]). I have no idea how to do it though. For the actual flop, I've tried numerous ways but it has all been in vain.

Can someone explain and demonstrate how exactly they calculate this as being 0.25%?

In Texas Hold'Em Poker, you are dealt two cards to begin with. After some betting occurs, we see the "flop", which is three community cards dealt into the middle of the table that everyone can use towards making their poker hand. In this example, if you began with, for example, two Aces, and within the three cards in the "flop" there were the other two Aces, you would have "quads" (termed "flopping quads"). This could happen in another way too however. You could start with two random cards (not paired) and then the flop has the three other cards of one of your starting cards (example, you have are dealt an Ace and a King to start, and then there are three Kings in the flop). I'm specifically interested in starting with a pair and then the only other two cards of this rank in the remaining 50 cards coming out in the flop.

So there are 52 cards to begin; after you being dealt your 2 starting cards, there are 50 cards remaining from which three are taken to make the "flop".

After having been dealt your two original cards, if they are a pair, you need 2 of the 3 cards on the flop to have the same value. Since there are 50 cards left to choose from, you have $\binom{50}{3}$ possible flops. How many of these will yield four of a kind? To answer this, notice that such a flop will have one "free" card (doesn't matter what it is) and two that must be there. So, if we restrict ourselves to flops that have the two cards we want, we are left with $\binom{48}{1}$ since we have removed the two matching cards and have to select one of the remaining 48.
Thus, the probability is $\frac{\binom{48}{1}}{\binom{50}{3}} = \frac{48}{\frac{50!}{3!47!}} = \frac{1}{50 \times 49 / 6} = \frac{3}{25 \times 49} \approx \frac{1}{408}$.