# Four of a kind and Three of a kind at the same time. What are the odds?

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) ?

• What was your starting hand? – Valentin Grégoire Jul 24 '14 at 14:10
• 3K, that's why i folded it xD. But I wanna know the odds not depending on the starting hand. – ITDE Jul 24 '14 at 15:37
• @ITDE but these odds DO depend on the starting hand. You can't just calculate out a board without knowing which cards you hold (or have otherwise been burned). So you need to provide more clarification on what you're looking for. Otherwise the question is useless. – Unknown Coder Jul 24 '14 at 17:07
• not if I say that i just want to know the odds for an AAAABBB combination, without making a difference if you hava AA, AB or BB on your hand. All the other odds in poker are not dependent on your hand, they are just calculated in general. And there is also no special hand required, this can occure with any starting hand. – ITDE Jul 24 '14 at 19:15
• @ITDE "this can [occur] with any starting hand" shows a fundamental misunderstanding of probabilities. Can't help you. – Unknown Coder Jul 25 '14 at 14:55

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):

• XX -> XX XX YYY There are C(4,3) * 12 = 48 possibilities
• XX -> XX X YYYY There are C(2,1) * 12 = 24 possibilities

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...

• Thanks a lot for this detailed answer, i wonder if i will ever see that hand again. – ITDE Jul 25 '14 at 19:06

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:

``````AAAA222  AAAA666  AAAA101010
AAAA333  AAAA777  AAAAJJJ
AAAA444  AAAA888  AAAAQQQ
AAAA555  AAAA999  AAAAKKK
``````

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.,

A♠A♥A♣A♦ 2♠2♣2♦

A♠A♥A♣A♦ 2♠2♥2♦

A♠A♥A♣A♦ 2♠2♥2♣

A♠A♥A♣A♦ 2♥2♣2♦

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.

• that is not correct, your p is too low... From the moment you have quads "AAAA", there are C(4,3) possible ways to make a set of, say, "222" with a deck which has four deuces. Computing it the way you did (drawing 7 cards from the deck), p is: (13 * 12 * 4) / C(52,7) if I'm not mistaken. I took another approach and gave the two p depending on whether you start with a pocket pair or not. – TacticalCoder Jul 25 '14 at 13:17
• note that (13 * 12 * 4) / C(52,7) is identical to my result. – TacticalCoder Jul 25 '14 at 13:30
• A p of 0.0000046624 instead of 0.00000116605 is still not an excuse for OP to hold his K3 ; ) – TacticalCoder Jul 25 '14 at 13:36
• @TacticalCoder Yep, you're right - I missed the factor of (4 choose 3) = 4. – rphv Jul 25 '14 at 15:26