Probability of flopping four of a kind

A pocket pair flops four of a kind 0.24% of the time, 1 in 408. An unpaired hand flops it 0.010% of the time, 1 in 9,800.

  • 0.24%Pocket pair
  • 0.010%Unpaired hand
  • 19,600Flops counted

What is the chance of flopping four of a kind?

A pocket pair makes quads when both remaining cards of its rank are on the flop, with any third card: C(2, 2) x 48 = 48 flops of 19,600, 0.24%, or 1 in 408. An unpaired hand needs the flop to be all three remaining cards of one of its hole ranks, which happens in 2 of the 19,600 flops, 0.010%.

Four of a kind on the flop
EventChanceOddsFlops out of 19,600
Pocket pair: four of a kind0.24%1 in 40848
Unpaired hand: four of a kind0.01%1 in 9,8002

How rare is it, in plain terms?

With a pocket pair you flop quads about once in 408 deals of that pair. An unpaired hand does it about once in 9,800 deals of that hand. Compare that with the chance of a set, 11.76%, to see how much the second card changes the picture.

Where do these numbers get used?

The chance to flop a hand is the first number in a plan. The pages below turn it into prices: what a call can cost, how many outs a draw has, and how many ways a hand can be dealt.

How were these numbers computed?

After two hole cards are dealt, 50 cards are unseen and the flop is any 3 of them, so there are C(50, 3) = 19,600 equally likely flops. For each starting hand we deal all 19,600, classify the five cards with our hand evaluator and count. Suits are symmetric, so one combination stands for every combination of the same hand. The closed forms on this page use binomial coefficients and give the same counts.

Source: GeorgeLiPoker's own exact count of the 52-card deck.

Questions

What is the probability of flopping quads with a pocket pair?
0.24%, or 1 in 408.
Can an unpaired hand flop four of a kind?
Yes, when the flop is all three remaining cards of one hole rank. It happens on 2 of 19,600 flops, 0.010%.
What is the chance of quads by the river with a pocket pair?
0.82%, or 1 in 122: both remaining cards of the pair must be among the five board cards.