Question

Consider a standard deck of 52 cards. How many 5 card hands have 2 pairs in them?

Answer 1

123,552

Explanation:

First off, this is a combinations question - we don't care about the order in which the cards are dealt. The general formula for combinations is:

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

Before moving on, let's see how many 5 card hands are possible:

`C_(52,5)=((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`

So of those nearly 2.6 million hands, how many are 2 pair hands?

To achieve 2 pair, we first need to select, from the 13 ordinals (Ace through 10, Jack, Queen, King) 2 of them:

`((13),(2))`

From each of those ordinals, we want 2 suits from the four available (spades, hearts, diamonds, clubs) each:

`((13),(2))((4),(2))^2`

And now we need a last card. It needs to be one of the remaining 11 ordinals and we'll be choosing one card:

`((13),(2))((4),(2))^2((11),(1))((4),(1))=78xx6^2xx11xx4=123,552`

A helpful site for doing poker hand probability and combinametrics is https://en.wikipedia.org/wiki/Poker_probability