Solving a 13-Card Hand Probability w/ No Pairs

[yoda05378]

hi! i have a homework problem here that I'm stuck on:

What is the probability that a hand of 13 cards contain no pairs?


i know a 13-card hand is C(52, 13) but i have no idea how to represent the probability of that 13-card hand containing no pairs. can anyone give me a hint?

 

[shmoe]

No pairs means at most one of each: Ace, two, ..., ten, jack, queen, king. So how many of each must you have in this case?

 

[OlderDan]

There are probably different approaches, but if you think in terms of the probability of drawing a card that does not match any previously drawn card on every draw I think that will get you there. The first card cannot match any previously drawn card so the probability is 1. The next card has 48/51 probability of not matching the first. The next has 44/50 probability of not matching either of the first two, etc.

The problem can also be done using a multi-type (or multivariate) hypergeometric distribution, which is, I think, the way you would want to go if you were looking for more than one of each type card instead of just one.

 

[yoda05378]

that's 13 different cards and i choose 7 of them: C(13,7)?

so [C(13,7) / C(52, 7)] is the probability?

 

[OlderDan]

that's 13 different cards and i choose 7 of them: C(13,7)?

so [C(13,7) / C(52, 7)] is the probability?

I don't see a 7 in the problem at all. What is 7?

 

[yoda05378]

I don't see a 7 in the problem at all. What is 7?

oops, i meant: [C(13,13) / C(52, 13)]

 

[OlderDan]

oops, i meant: [C(13,13) / C(52, 13)]

That is not it. What has you thinking C(13,13)?

 

[yoda05378]

That is not it. What has you thinking C(13,13)?

umm...13 cards in one hand, so i choose 13 out of the 13 available cards from a suit?

i'm just trying to make some sense out of the examples from my book... I'm so confused :(

 

[OlderDan]

umm...13 cards in one hand, so i choose 13 out of the 13 available cards from a suit?

i'm just trying to make some sense out of the examples from my book... I'm so confused :(

You have 13 different kinds of cards, with 4 of each kind. You want one card of each kind. How many combinations are there of 4 cards taken 1 at a time? How many combinations are there of four things of one kind taken one at a time, combined with 4 things of a second kind taken one at a time, combined with 4 things of a third kind taken one at a time ... combined with 4 things of a thirteenth kind taken one at a time?

 

[yoda05378]

You have 13 different kinds of cards, with 4 of each kind. You want one card of each kind. How many combinations are there of 4 cards taken 1 at a time? How many combinations are there of four things of one kind taken one at a time, combined with 4 things of a second kind taken one at a time, combined with 4 things of a third kind taken one at a time ... combined with 4 things of a thirteenth kind taken one at a time?

4 things of 1st kind combined with 4 things of a 2nd kind combined with...4 things of a 13th kind = 4^13?

so [4^13 / C(52, 13)] is the probability?

 

[OlderDan]

4 things of 1st kind combined with 4 things of a 2nd kind combined with...4 things of a 13th kind = 4^13?

so [4^13 / C(52, 13)] is the probability?

Yes

And if you want to be a bit more formal about it

C(4, 1)^13/ C(52, 13)

Note that 4*13 = 52 and 1*13 = 13

This is a multivariate extension of the hypergeometric distribution for two kinds of things. If you wanted to know the probability of gettng exactly 3 aces in a hand of 7, you would identify aces as one kind, and everything else as another kind and you would have

P = C(4, 3)*C(48, 4)/C(52, 7)

 

[yoda05378]

Yes

And if you want to be a bit more formal about it

C(4, 1)^13/ C(52, 13)

Note that 4*13 = 52 and 1*13 = 13

This is a multivariate extension of the hypergeometric distribution for two kinds of things. If you wanted to know the probability of gettng exactly 3 aces in a hand of 7, you would identify aces as one kind, and everything else as another kind and you would have

P = C(4, 3)*C(48, 4)/C(52, 7)

awesome! thanks OlderDan!