How to determine combinations correctly

  • Context: Undergrad 
  • Thread starter Thread starter Mathematicize
  • Start date Start date
  • Tags Tags
    Combinations
Click For Summary
SUMMARY

The correct method for calculating the number of distinct ways to receive a full house in a five-card draw is represented by the formula: 13 nCr 1 * 4 nCr 3 * 12 nCr 1 * 4 nCr 2. This formula accounts for selecting one rank for the three-of-a-kind, three suits from that rank, one rank for the pair, and two suits from that rank. The alternative formula, 13 nCr 2 * 4 nCr 3 * 4 nCr 2, is incorrect as it fails to properly account for the designation of the ranks as first and second, leading to a discrepancy in the total combinations.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations (nCr).
  • Familiarity with poker hand rankings and the concept of a full house.
  • Knowledge of the multiplication rule in probability.
  • Basic understanding of card suits and denominations.
NEXT STEPS
  • Study the principles of combinatorial mathematics, focusing on combinations and permutations.
  • Learn about the different poker hands and their probabilities, particularly full houses.
  • Explore the multiplication rule in probability and its applications in card games.
  • Investigate alternative methods for visualizing combinations beyond subsets, such as tree diagrams.
USEFUL FOR

Mathematicians, poker enthusiasts, and anyone interested in combinatorial probability and its applications in card games.

Mathematicize
Messages
11
Reaction score
0
Hey everyone,

I have a question regarding combinations and poker! Could someone explain to me why the number of distinct ways to receive a full house from a five card draw is:

13 nCr 1 * 4 nCr 3 * 12 nCr 1 * 4 nCr 2
and not,
13 nCr 2 * 4 nCr 3 * 4 nCr 2

I usually make small mistakes here and there with combinatorics and I can never find a good answer from anyone. Also, I am more interested in the not in this case. Intuitively they seem to say the same thing to me. (choose two distinct denominations, choose 3 cards for the first denomination and 2 cards for the second denomination). Also does anyone have a good way at viewing combinations other than the number of subsets of size r from a set of size n?

Thanks!
 
Physics news on Phys.org
Hey Mathematicize and welcome to the forums.

It might help the readers here if you show your thought process on how you derived your answer because doing this will help us understand where the thinking was wrong or even if the answer that has been provided is wrong if this is the case.
 
The first answer is correct after computing the probabilities and checking with various sources. So I will explain my thought process on the second answer in which I obtained. First I did 13 nCr 2 because in a full house we need 2 distinct denominations where order does not matter. Next, we must pick the suit for the first denomination, and there are 4 suits and 3 that we must choose in which order does not matter, so I did 4 nCr 3. Next I needed 2 suits for the second denomination. So with the same reasoning as my last step I get 4 nCr 2. Then since each step is independent of each other, I invoke the multiplication rule to get the total number of ways to receive a full house.

The different between the answers is off by a factor of 1/2. I have a hard time visualizing this or any real way at checking my work for correctness which is a big problem.
 
Last edited:
Mathematicize said:
(choose two distinct denominations, choose 3 cards for the first denomination and 2 cards for the second denomination).

You have to account for the number of ways that one of the denominations can be designated as "the first" deonomination. When you enumerated picking the denominations, you didn't enumerate it as picking a "first" and "second" denomination.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K