Question

You have a deck of 52 playing cards.
(i) How many different 5 card hands can be dealt?
(ii) What is the probability that a hand of 5 dealt randomly contains 3 aces?
(iii) What is the probability that a hand of 5 dealt randomly will have 5 cards of the same

Answer 1

i)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

$C n , k = \frac{n !}{ ( k ) ! ( n − k ) !}$

$C 52 , 5 = \frac{52 !}{ ( 5 ) ! ( 52 − 5 ) !}$ $= 2 , 598 , 960$ ways

ii) To get exactly 3 aces, you need to choose 3 of the 4 aces and 2 of the other 48 cards. The number of ways to do that is

C(4,3)*C(48,2) = $\frac{4 !}{ ( 3 ) ! ( 4 − 3 ) !}$*$\frac{48 !}{ ( 2 ) ! ( 48 − 2 ) !}$ =4512

Hence probability= $\frac{4512}{2598960}$ = 0.00174

iii)Probability of getting a hand that has 5 cards of the same suit (flush, straight flush, royal flush.

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

$\frac{4!}{1!(4-1)!)}\frac{13!}{5!(13-5)!}$ = 5148

The probability of getting a hand with a flush is:

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