Combinations of 5-Card Hands to Make a Full House vs. 4-of-a-Kind

[yellowjack]

I'm running into some confusion about the relative probabilities of finding a full house and 4-of-a-kind.

For a full house, I calculated the number of ways to get a full house to be: C(13,1)*C(4,3)*C(12,1)*C(4,1) = 3744

For 4-of-a-kind, I found it to be: C(13,1)*C(4,4)*C(12,1)*C(4,1) = 624

The concept I am having trouble wrapping my head around is the fact that a full house is 6 times more likely than a 4-of-a-kind (624*6=3744).

Say we know four cards, and three of them are Aces, and the 4th card is a King. One card is still unknown. There is 1 card to catch that will give us 4-of-a-kind, and 3 cards to catch that give us a full house.

[Grizwold]

Hi yellowjack,

I have not checked your combination calculations, but I can see a bias in this statement:

[ QUOTE ]
Say we know four cards, and three of them are Aces, and the 4th card is a King. One card is still unknown. There is 1 card to catch that will give us 4-of-a-kind, and 3 cards to catch that give us a full house.

[/ QUOTE ]

You seem to imply that you believe the chance of a full house should be only 3 times as likely as four of a kind. You calculate the combinations of making any full house and any four of a kind. However, in your final statement, you compare the probability of making a particular full house, aces full of kings, rather than aces full of anything. You should consider it a different way: if we know three cards, and all three are aces, then what are the chances the last two cards will be either a pair, or contain the last ace. Although I have not checked your calculations, they do not seem to represent the example you give.

Clark

[yellowjack]

I see what you are getting at; I'm going to try doing this again but make the basepoint at 3 Aces as suggested.

The number of pairs from two cards are C(12,1)*C(4,2) = 72
However using brute force, 48*3 = 144 (which seems to be the correct answer, therefore C(12,1)*C(4,2) is clearly a mistake)

The number of two cards containing an ace is C(1,1)*C(48,1)/2 = 24

144 / 24 = 6 = happy time

C(12,1)*C(4,2) = 72 is bugging the hell out of me. It's supposed to be the number of ways to make a non-ace pair with two cards.

[jay_shark]

The number of ways of hitting a full house is 3744 but your solution doesn't add up to that number .

It should be 13c2*4c3*4c2*2=3744

There are 13c2 ways to select 2 numbers from 13 . Once we've selected the 2 numbers , there are 4c3 ways to select a 3-of-a-kind and 4c2 ways to select a pair . However , we multiply by 2 since the situation may be reversed .

[jay_shark]

Or, instead of 4c1 as you have written , it should be

13c1*4c3*12c1*4c2 =3744 .

[yellowjack]

Yeah I cheated a bit and just wrote down 3,744 because I had computed it before, made a typo here. Thanks again!