# Suppose five cards are drawn from a deck. What is the probability of obtaining a flush (five cards of the same suit)?

Probability of getting a hand that has 5 cards of the same suit (flush, straight flush, royal flush) $= \frac{5148}{2598960} \cong .00198$.

Probability of getting a flush (and so excluding straight and royal flushes) $= \frac{5108}{2598960} \cong .00196$

#### Explanation:

To find the probability, we need to find the fraction where the numerator is the number of ways to have a flush and the denominator is the number of 5 card hands possible.

Each of these numbers will be found using Combinations (we don't care about the order of the draw, merely of what ends up in our hand). The general formula for Combinations is:

C_(n,k)=(n!)/((k)!(n-k)!) with $n = \text{population", k="picks}$

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 let's calculate the numerator.

We're going to calculate all hands that involve a flush (so not just a flush but also a straight flush and royal flush) so that we'll be looking at any hand that has five cards of the same suit (with a suit having 13 cards in total). We can express getting this with:

${C}_{13 , 5}$

Keep in mind that there are 4 suits this can happen in, but we only want 1, and so we need to multiply by ${C}_{4 , 1}$. Putting it together then, we get:

C_(4,1)xxC_(13,5)=(4!)/((1!)(4-1)!)xx(13!)/((5!)(13-5)!)=(4!13!)/(3!5!8!)

Let's evaluate this!

(cancelcolor(red)(4!)xx13xx12xx11xxcancelcolor(brown)10xxcancelcolor(blue)9^3xxcancelcolor(orange)(8!))/(cancelcolor(blue)3xxcancelcolor(brown)(2xx5)xxcancelcolor(red)(4!)xxcancelcolor(orange)(8!))=13xx12xx11xx3=5148

(keep in mind that we've just calculated all hands that have a flush element to them, including straight flushes and royal flushes!)

The probability of getting a hand with a flush is:

$\frac{5148}{2598960} \cong .00198$

~~~~~

To exclude straight and royal flushes (these are hands that have 5 consecutive value cards in the same suit, such as 3, 4, 5, 6, 7 of hearts), we can deduct those possibilities from the 5148 flush hands. There are 10 different ways to get a straight (A-5, 2-6, 3-7,..., 10-A) and 4 suits, so we can deduct $4 \times 10 = 40$ hands, so that's

$5148 - 40 = 5108$ hands, giving:

$\frac{5108}{2598960} \cong .00196$