How to determine combinations correctly

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The discussion focuses on the correct calculation of combinations for obtaining a full house in poker. The correct formula is 13 nCr 1 * 4 nCr 3 * 12 nCr 1 * 4 nCr 2, which accounts for selecting one rank for the three of a kind and one rank for the pair, along with their respective suits. The alternative formula presented, 13 nCr 2 * 4 nCr 3 * 4 nCr 2, is incorrect because it does not properly account for the order of the denominations. A key point raised is the importance of distinguishing between the first and second denominations when calculating combinations. Understanding these nuances in combinatorics is essential for accurate probability assessments in poker scenarios.
Mathematicize
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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!
 
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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.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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