Probability ( counting principles)

I just need an explaination to a question I saw in a statistics book. I understand the concepts but I don't really understand what the question is asking. This question asks '' Find the probability of being dealt at random and without replacement a thirteen card bridge hand consisting of a)thirteen cards of the same suit b) 2 clubs, 3 diamonds, 5 hearts and 3 spades. A thorough explaination will assist me.

Part a) is easy: There is a total of $52\choose 13$ hands that can be dealt. Out of these hands, there are $4$ of the same suit, hence the answer is $4\over{52\choose13}$.

For b) the number of "good" possibilities is counted by first selecting 2 out of 13 clubs, then 3 out of 13 diamonds, then 5 out of 13 hearts, and finally 3 out of 13 spades. Thus, the answer is ${{13\choose2}{13\choose3}{13\choose5}{13\choose3}}\over{52\choose13}$.

Do we have to worry about mixing the suits to accommodate for the fact that these cards might be dealt in different orders of suits (e.g. in order "a spade, a heart, a club, ...") No! Since we count unordered hands with $52\choose13$ in the denominator, we must do the same in the numerator - which means that we may assume any specific order we wish, as if after dealing we sort our hand by suits.

Edit: Adapted solution to fixed typo in the problem statement.

• @ Hagen, thank you for your assistance. for the b) I did the same way but wasn't that sure. Thank you... – Eugene Mettle Sep 3 '12 at 5:08
• Hagen, I think its suppose to be 13 cards consiting of 2 clubs, 3 diamonds 5 hearts and 3 spades. It was a typo but i get it now.. – Eugene Mettle Sep 3 '12 at 5:24

For a: There are four hands that are all the same suit. There are $52 \choose 13$ total hands.

For b: 2+3+5+7=17

• @ Ross, can you pls explain it explicitly? What you wrote is kinda someway.. I just want you to explain it to me.. A little detail will do.. – Eugene Mettle Sep 3 '12 at 5:01
• D'oh, unlike Ross I hadn't noticed that b) describes a hand of more than 13 cards. Is there a tipping error in the problem statement or are the rules of bridge not what I think they are (13 out of 52 cards in a hand)? – Hagen von Eitzen Sep 3 '12 at 5:07
• @HagenvonEitzen: I don't know what the problem setter was thinking, but my answer would be the probability is 0. The rules of bridge specify a 13 card hand. The problem seems to help you this way, also specifying a 13 card hand. – Ross Millikan Sep 3 '12 at 5:13
• @ Hagen and Ross, I think its suppose to be 13 cards consiting of 2 clubs, 3 diamonds 5 hearts and 3 spades. It was a typo. – Eugene Mettle Sep 3 '12 at 5:23
• Although providing solutions to a) and b) explains the question, all the OP asked for was an explanation of the question which would simply be to find the probability of getting all thirteen cards in the same suit (i.e. all the clubs, hearts, diamonds or spades) and in b) that exactly 2 cards are clubs, 3 are diamonds, 5 are hearts and 3 are spades. – Michael Chernick Sep 3 '12 at 13:06