# In a card game using a standard 52 card deck, 4-card hands are dealt. What's the probability of being dealt 3 diamonds?

$11 , 154$

#### Explanation:

In a standard deck, there are 13 ordinal cards (Ace - 10, Jack, Queen, King) and in each of 4 suits (Hearts, Diamonds, Clubs, Spades) for a total of $13 \times 4 = 52$ cards.

We're asked to find the number of possible 4-card hands containing exactly 3 diamonds. The order of the draw doesn't matter, so we're dealing with a Combinations problem (if the order did matter, it'd be a Permutation problem).

The formula for a Combination is C_(n,r)=(n!)/((r!)(n-r!)) where $n$ is the number of things we have to pick from and $r$ is the number of things we're picking.

So what is it that we're picking and picking from?

First, we need to pick 3 diamonds from a selection of 13, so that is:

C_(13,3)=(13!)/((3!)(13-3)!)=(13!)/((3!)(10!))

There are calculators that will do the math for you, like this one but I'll do the math here:

(13!)/((3!)(10!))=(13xx12xx11xxcancel(10!))/((3!)cancel(10!))=(13xxcancel12^2xx11)/cancel6=13xx2xx11=286

So that's 3 cards down with 1 to go. This last card cannot be a diamond, so we have the remaining suits to pick from. While there are fancy ways to write this, I'll settle for $52 - 13 = 39$.

Now to put this all together. We multiply the two numbers to get the final total of the number of hands possible:

$286 \times 39 = 11 , 154$