Calculating 2 Pair Probability in a Modified Deck of Cards

In summary, the conversation is about finding the probability of obtaining a 2 pair when 5 cards are dealt from a standard deck of playing cards with all the red cards removed. The conversation includes the combination formula and the attempt at a solution, which is later corrected by another individual. The correct answer involves choosing two different pairs out of 13 instead of choosing one pair at a time.
  • #1
walk_w/o_aim
27
0

Homework Statement



A standard deck of playing cards has all the red (heart, diamond) cards removed. Find the probability of obtaining a 2 pair when 5 cards are dealt.

Homework Equations



The combination formula:

[itex]
\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}
[/itex]

The Attempt at a Solution



The way I understood it was that there were 13 pairs of cards. In order to get a two pair, I need to choose 2 pairs from the 13, and choose 2 cards from each pair. For the fifth card, I just pick any pair from the remaining 11, and choose 1 of the pair. Then, that number of combinations I divide by the total number of ways I can pick 5 cards from 26. This is what I came up with:

[itex]
P(2 pair) = \dfrac{\dbinom{13}{1} \dbinom{12}{1} \dbinom{11}{1} (\dbinom{2}{2})^2 \dbinom{2}{1}}{\dbinom{26}{5}}
[/itex]

However, the actual answer is slightly different:

[itex]
P(2 pair) = \dfrac{\dbinom{13}{2} \dbinom{11}{1} (\dbinom{2}{2})^2 \dbinom{2}{1}}{\dbinom{26}{5}}
[/itex]

I can see why the actual answer makes sense, but I can't see why the answer I came up with does not. To me, they seem to be the exact same expression. Am I misunderstanding the meaning behind a combination here? Am I misunderstanding the meaning behind multiplying the number of combinations?

Some light shed on this problem would be much appreciated. Thank you.
 
Physics news on Phys.org
  • #2
Your method counts "A","K" and "K","A" separately. But that's not right; these are combinations, so they are the same. Hence what you want to do is choose two different pairs out of 13, so (13 2). Note that you could just divide your answer by 2: (13 1)(12 1)/2 = (13 2).
 
  • #3
Thanks, hgfalling. I got it now. Picking one at a time then multiplying the total number of combinations with each other makes the order matter -- I should have known that from the largest number n bits can represent or something similar. Thank you.
 

1. What are the chances of getting a Royal Flush in poker?

The probability of getting a Royal Flush in poker is 0.000154%. This means that in a 5-card hand, there is a 1 in 649,740 chance of getting a Royal Flush, which consists of a 10, Jack, Queen, King, and Ace of the same suit.

2. What is the likelihood of getting a Straight Flush in poker?

The probability of getting a Straight Flush in poker is 0.00139%. This means that in a 5-card hand, there is a 1 in 72,193 chance of getting a Straight Flush, which consists of 5 consecutive cards of the same suit.

3. How often does a Four of a Kind occur in poker?

The probability of getting a Four of a Kind in poker is 0.0240%. This means that in a 5-card hand, there is a 1 in 4,165 chance of getting a Four of a Kind, which consists of 4 cards of the same rank and 1 other card.

4. What are the chances of getting a Full House in poker?

The probability of getting a Full House in poker is 0.1441%. This means that in a 5-card hand, there is a 1 in 693 chance of getting a Full House, which consists of 3 cards of the same rank and 2 other cards of the same rank.

5. What is the likelihood of getting a Flush in poker?

The probability of getting a Flush in poker is 0.1965%. This means that in a 5-card hand, there is a 1 in 508 chance of getting a Flush, which consists of 5 cards of the same suit.

Similar threads

  • Calculus and Beyond Homework Help
Replies
31
Views
3K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
678
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
1K
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
10
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
1K
Back
Top