Probability (1 Viewer)

[Linda N]

Hey.. this question below looks simple but got trouble getting the answer.

A game of poker uses a deck of 52 cards with 4 suits (hearts, diamonds, spades and clubs). Each suit has 13 cards, consisting of an ace, cards numbered from 2 to 10, a jack, queen and king. If a person is dealt 5 cards find the probability of getting:

a) four aces
b) a flush (all ccards the same suit)

 

[withoutaface]

a)
find the number of combos with four aces:

AAAAN

so you have four aces and one non-ace

the non-ace can be selected from anyone of the other 48 cards

so 48x5!

is the number of 4 ace combs

then you divide this by the total number of combos:

48x5!/52P5

b) This is a bit easier:

you have 13 cards from each suit to start, then 12, then 11 etc it is irrelevant which card is selected first, so long as the others are all of the same suit, so the answer is

(12/51)*(11/50)*(10/49)*(9/48)


I have a feeling im wrong using permutations for the first one, so check with the answer in the book before you accept that one

 

[Estel]

I think combinations is the better way to go.

a) 4 Aces (1 way) + one of any other of the 48 cards
48C1/52C5

b) 5 of one suite. (4 different suites)
13C5*4/52C5

 

[Linda N]

The first part is right but the second part is wrong.

b) 3 * 33/16660 = 0.00594

I think the answer is wrong in the text book.