Question

What's the probability of being dealt a full house in 5-card poker?

Answer 1

`P("full house")=3744/(2,598,960)~=.00144=0.144%`

Explanation:

To find the probability of finding a full house (a three of a kind and a 2 of a kind in the same 5-card hand), we find the number of ways we can achieve the full house and divide by the number of 5-card hands possible.

For all of these calculations, we'll use the combinations formula (because we don't care about the order in which we draw the cards, only that we have them):

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

First let's calculate the denominator (picking 5 random cards from a pack of 52 cards):

`C_(52,5)=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))`

Let's evaluate it!

`(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960`

And now to calculate the number of full house hands that can be made.

To solve this, we need to remember that there are 13 ordinal cards (Ace, 2-10, Jack, Queen, King), with 4 suits (Hearts, Diamonds, Clubs, Spades), for a total of 52 cards (`4xx13=52`)

To have 3 of a kind, we select 3 of the 4 cards in one of the 13 available ordinals and then from another of the remaining 12 ordinals, we select 2 of the 4 (and then we multiply the two calculations).

Three of a kind from one of the 13 ordinals...

`C_(13,1)xxC_(4,3)=(13!)/((1!)(13-1)!)xx(4!)/((3!)(4-3)!)=>`

`=>(13xx12!)/(1!12!)xx(4xx3!)/(3!1!)=13xx4=52`

...and two of a kind (pair) from one of the remaining 12 ordinals

`C_(12,1)xxC_(4,2)=(12!)/((1!)(12-1)!)xx(4!)/((2!)(4-2)!)=>`

`=>(12xx11!)/(1!11!)xx(4xx3xx2!)/(2!xx2)=12xx6=72`

Ways to have a full house

`52xx72=3744`

Probability of a full house

`P("full house")=3744/(2,598,960)~=.00144=0.144%`