Question

You have a deck of 52 playing cards (check online for the full composition of a 52 card deck if you are unsure).

(i) How many different 8 card hands can be dealt?

(ii) What is the probability that a hand of 8 dealt randomly contains (exactly) 2 aces?

(iii) What is the probability that a hand of 7 dealt randomly will have 7 cards of the same suit? 

Answer 1

$i)$

Using combinations, we can find the number of 8 card hands that can be dealt as follows,

number of ways =$\binom{n}{x}={n!\over(x!(n-x)!)}$

From the question, $n=52, x=8$

We now have,

$\binom{52}{8}={52!\over(8!\times44!)}=752538194$

Therefore, there are 752538194-8card hands that can be dealt

$ii)$

To get exactly 2 aces, we need to choose 2 out of 4 aces and 6 of the remaining 48 cards. The number of ways to do this is

$\binom{4}{2}\times\binom{48}{6}={4!\over(2!*2!)}*{48!\over(6!*42!)}=6\times12271512=73629072$

The probability is, ${73629072\over752538194}=0.0978(4dp)$

Therefore, probability of getting exactly 2 aces is 0.0978.

$iii)$

Since there are 13 cards in a suit, we choose 7 cards from the 13 cards using combinations as follows, $\binom{13}{7}={13!\over(7!*6!)}=1716$

There are 4 suits but we only want 1suit. Therefore, the number of ways for choosing 1 out of a total of 4 suits is, $\binom{4}{1}={4!\over(1!\times3!)}=4$

Therefore, total number of ways is 1716*4=6864

Hence the probability is, ${6864\over752538194}=9.121132e{-6}$

Therefore, the probability that a hand of 7 dealt randomly will have 7 cards of the same

suit is 9.121132e-6.