I recently got 3333KKK at Texas Holdem in one round. I know that only the Quads 3 count, but I still wonder what the probability of such a "Big Full House" is. Haven't found anything about it on the internet and wasn't able to figure it out myself. So can anyone tell me the odds of Four of a kind and Three of a kind in the same round (in general, not only for 3 and K) ?
p is 0.0000046642
Here are two different ways to arrive at that result...
If you have XY, there is one way to make "quad X" and C(3,2) ways to make "set Y".
If you have XY, there is also one way to make "quad Y" and C(3,2) ways to make "set X".
C(3,2) is 3, so starting from XY, there are 6 possibilities to make what you called the "big full house" (very nice hand btw) if the deal goes to river.
You know your two holecards, so there are 50 cards left (as far as you're concerned, you have no idea what your opponents have). There are C(50,5) ways to choose 5 cards out of 50, which gives 2 118 760 possibilites.
So if you do not start with a pocket pair, you have: p = 6 / 2 118 760 (or approximately 0.000002831845)
If you have XX the probability is actually higher, for you can make a hand with any Y (there are 12 Y left):
So there are 72 possibilities out of 2 118 760 boards to come.
So if you start with a pocket pair, you have: p = 72 / 2 118 760
Note that I did write some piece of code to try this and found both 6 possibilities when you start without a pair and 72 possibilities when you start with a pocket pair, while trying all the possible boards of five cards to come.
If you want to "combine" these two, there are 3/51 (i.e. 1/17) that you're dealt a pocket pair and 48/51 (16/17) that you're not, so:
72 / 2118760 * 1/17 + 6 / 2118760 *16/17 = 0.0000046642...
Which can be compared to user "rphv"'s method (drawing 7 cards out of 52) formula:
(13 * 12 * 4) / 133784560 = 0.0000046642...
There are (52 choose 7) = 133784560 total possible hands in 7 card poker. Of these, 12 * 4 * 13 are a "Big Full House." To see why, consider the 12 seven-card hands with four aces and three of a kind, disregarding suit:
For each of these 12 hands, there are (4 choose 3) = 4 ways to draw the "three of a kind" taking suit into account e.g.,
Finally, repeat this process for each of the 13 possible "four of a kinds".
Thus, the chances of drawing a "Big Full House" are (12 * 4 * 13) / 133784560 = 0.00000466421, or roughly 1 in 220,000.
Don't use this as an excuse to hold your K3, though.