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Counting poker hands ?

  • Thread starter cragar
  • Start date
  • #1
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Homework Statement


A poker hand conatins 5 cards chosen from a deck of 52.
1. How many hands are possible
a. 4 of a kind
b.3 of a kind
c. flush
d. full house
e. straight
f. straight flush

The Attempt at a Solution


When i write (5,2) I mean 5 choose 2 .

1. (52,5)
a. I have 13 different choices from 2 .... king, ace .
so (13,1)*(4,4)*(12,1)*(4,1) then my last card has to come from the remaining 12 and then of the 4 of those cards I have to pick 1 .
b. (13,1)(4,3)(12,1)(4,1)(11,1)(4,1)
When I pick my last cards I have to pick from different rows to avoid getting 4 of a kind or a full house.
c. I have 4 different suits to pick from so I pick 1 of the four then pick 5 of the suits 13 cards. so it will be (4,1)(13,5)
d. full house will be (13,1)(4,3)(12,2)
e. Well there are 9 ways to get a straight with 5 cards numerically in a row.
and their are 4 copies of an ace so i think the answer is 4^5(9)
f. well there are 9 ways to get a straight an 4 suits so 9*36 but it would be (9,1)(4,1)
 

Answers and Replies

  • #2
vela
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I think you have the wrong answers for b and d.

With three-of-a-kind, you're double-counting since you count XXXab and XXXba separately.

With the full house, the factor of (12,2) isn't right because you want the final two cards to be a pair, and you left out the factor because of the possible suits for the pair.
 
  • #3
2,544
2
yes your right. part d. should be (13,1)(4,3)(12,1)(4,2)
and on b. I am off by a factor of 2 .
for part b. (13,1)(4,3)(12,2)(4,1)(4,1)
 

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