# Solving a 13-Card Hand Probability w/ No Pairs

• yoda05378
In summary, the conversation discusses the probability of a hand of 13 cards containing no pairs. Different approaches are proposed, including using a multi-type hypergeometric distribution. Ultimately, it is determined that the probability can be calculated by taking the combinations of one card from each of the 13 different kinds of cards, divided by the total number of combinations possible for a 13-card hand. The formula for this is C(4, 1)^13 / C(52, 13). A more formal approach is also mentioned, using the multivariate extension of the hypergeometric distribution.
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?

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?

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.

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

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

yoda05378 said:
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?

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

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

yoda05378 said:
oops, i meant: [C(13,13) / C(52, 13)]

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

OlderDan said:
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 :(

yoda05378 said:
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?

OlderDan said:
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?

yoda05378 said:
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)

OlderDan said:
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!

## 1. What is the probability of getting a pair in a 13-card hand?

The probability of getting a pair in a 13-card hand is approximately 42.26%. This can be calculated by first finding the probability of getting a pair in the first two cards, which is 6/51. Then, for the remaining 11 cards, the probability of getting a pair is (3/50) * (11/49) * (2/48) * (10/47) * (9/46) * (8/45) * (7/44) * (6/43) * (5/42) * (4/41) * (3/40) * (2/39) * (1/38), which is approximately 6.11%. Multiplying these two probabilities gives the overall probability of getting a pair in a 13-card hand.

## 2. What is the probability of getting a specific suit in all 13 cards?

The probability of getting a specific suit in all 13 cards is approximately 0.000000002%. This can be calculated by taking the probability of getting a specific suit in the first card (13/52), then (12/51) for the second card, and so on for all 13 cards. This is equal to (13/52)^13, which is approximately 0.000000002%.

## 3. What is the probability of getting a straight in a 13-card hand?

The probability of getting a straight in a 13-card hand is approximately 4.62%. This can be calculated by first finding the probability of getting a straight in the first 5 cards, which is (4/52) * (4/51) * (4/50) * (4/49) * (4/48), or approximately 0.0032%. Then, for the remaining 8 cards, the probability of getting a straight is (4/47) * (4/46) * (4/45) * (4/44) * (4/43) * (4/42) * (4/41) * (4/40), or approximately 3.85%. Multiplying these two probabilities gives the overall probability of getting a straight in a 13-card hand.

## 4. What is the probability of getting 5 cards of the same rank in a 13-card hand?

The probability of getting 5 cards of the same rank in a 13-card hand is approximately 0.005%. This can be calculated by first finding the probability of getting 5 cards of the same rank in the first 5 cards, which is (13/52) * (4/51) * (3/50) * (2/49) * (1/48), or approximately 0.000000009%. Then, for the remaining 8 cards, the probability of getting 4 cards of the same rank is (3/47) * (2/46) * (1/45) * (44/44) * (43/43) * (42/42) * (41/41) * (40/40), or approximately 0.00499%. Multiplying these two probabilities gives the overall probability of getting 5 cards of the same rank in a 13-card hand.

## 5. How does the probability of getting a full house in a 13-card hand compare to the probability of getting a straight flush?

The probability of getting a full house in a 13-card hand (approximately 0.109%) is significantly higher than the probability of getting a straight flush (approximately 0.000015%). This is because a full house only requires 5 cards to be of the same rank and the remaining 8 cards to be of another rank, while a straight flush requires all 13 cards to be of the same suit and in numerical sequence. Thus, it is much more likely to get a full house in a 13-card hand than a straight flush.

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