1. ## Confused about elementary combinatorics for poker hand calculations

I am quite new to combinatorics, and I seem to be experiencing a discrepancy when attempting to apply common poker-style hand probability calculations on a smaller scale to see what is occurring. Given my newness to the study, and the prevalence of these formulas, I can only assume I am confused. Let it be clear that the style of presentation is only for demonstration so others may find where I have erred. I am not attempting to debunk anything, and I apologize if my notation is sloppy.

Concisely stated, prevalent calculations of poker hand combinations appear, to me, to quantify the the permutations of combinations rather than just combinations. If this is true, it seems to me that the resulting probabilities would be incorrect.

Poker probability calculations divide the product of the quantities of combinations of cards (or the product of the quantities of combinations for their suits and ranks) for a given hand by the quantity of potential hand combinations. What follows is a typical example of calculation of a given hand's probability. In this case, the calculation is for a full house (3 cards of a given rank, and 2 cards of a different rank than the first 3 chosen):
$\displaystyle \\n = 52\\k = 5\\\\\frac{C(13,1) * C(4,3) * C(12,1) * C(4,2)}{C(52,5)}= 3,744 / 2,598,960 \approx 0.14\%$
The difficulty I experience when attempting to follow this calculation of probability is that it would appear the quantity of permutations of the combinations are being calculated, then divided by the quantity of potential combinations for 5 cards drawn. Because the denominator contains only combinations, and not permutations, it appears inconsistent to permute the combinations in the numerator.

The following is an example of a scaled down version of the above calculation--assumed to be analogous--such that it may be easily enumerated:
$\displaystyle \\\mbox{Set R contains all ranks; set S, all suits; and set D, all cards.}\\\\R = \{A,2\}\\S = \{H,D\}\\D = \{AH,AD,2H,2D\}\\\\n = |D| = 4\\k = 4\\\\\frac{C(2,1) * C(2,2) * C(1,1) * C(2,2)}{ C(4,4)} = \frac{\ 2 * 1 * 1 * 1}{1}=200\%$
Why would there be 2 "combinations", yielding a 200% probability of drawing a 4 card hand in 4 draws from a 4 card deck? The only expected combination is: {AH,AD,TH,TD}. One combination. Instead, the calculation presumably finds: {AH, AD, TH, TD}; {TH, TD, AH, AD}. The permutations of suit were found for each rank (or potentially vice versa since both sets are equal in size).

Why does this happen? Consider the case of any five cards chosen from a standard 52 card deck. As with prior cases, this selection of any 5 cards may, presumably, be represented as suits chosen from ranks (or vice versa), though with the added options afforded by prior suits. Presumed equivalencies follow:
$\displaystyle \\C(52,1) = C(13,1) * C(4,1)\\C(51,1) = C(12,1) * C(4,1) + C(3,1)\\C(50,1) = C(11,1) * C(4,1) + 2 * C(3,1)\\C(49,1) = C(10,1) * C(4,1) + 3 * C(3,1)\\C(48,1) = C(9,1) * C(4,1) + 4 * C(3,1)$
Given this understanding, the following should be analogous to the numerator of prior examples:
$\displaystyle \\n = 52\\k = 5\\\\C(52,1) * C(51,1) * C(50,1) * C(49,1) * C(48,1)=\frac{\52!}{47!}=\frac{n!}{(n-k)!}}$
Obviously, this is not the function for calculating quantity of combinations. Again, it appears that multiplying set quantities permutes the combinations. Please tell me where and WHY I am going wrong in all of this.

2. ## Re: Confused about elementary combinatorics for poker hand calculations

Please simply state a problem! Such as "how many ways are there to have a full house in five card poker?"
I have no idea what you posted means.

3. ## Re: Confused about elementary combinatorics for poker hand calculations

Apologies. I am asking what I am doing wrong. I commonly see a full house (3 cards of a given rank/value and 2 cards of a given rank different from the first 3 chosen) resulting from a 5 card draw from a standard deck of 52 cards represented as:

$\displaystyle \frac{C(13,1) * C(4,3) * C(12,1) * C(4,2)}{C(52,5)}= 3,744 / 2,598,960 \approx 0.14\%$

If I attempt to use smaller values for the purposes of viewing the combinations calculated, it appears that the combinations are not calculated correctly. Example:

$\displaystyle \\\mbox{Set R contains all ranks; set S, all suits; and set D, all cards.}\\\\R = \{A,2\}\\S = \{H,D\}\\D = \{AH,AD,2H,2D\}\\\\\frac{C(2,1) * C(2,2) * C(1,1) * C(2,2)}{ C(4,4)} = \frac{\ 2 * 1 * 1 * 1}{1}=200\%$

Is the solution to the first problem correct? Is 2 combinations and a 200% probability an erroneous result for the second problem? Is the second problem not analogous to the first? As far as I can tell, multiplying the combinations results in permutations, not combinations. Combinations are the expected result. I don't know how much clearer I can make my inquiry.

4. ## Re: Confused about elementary combinatorics for poker hand calculations

Originally Posted by a4622
Apologies. I am asking what I am doing wrong. I commonly see a full house resulting from a 5 card draw represented as:

$\displaystyle \\n = 52\\k = 5\\\\\frac{C(13,1) * C(4,3) * C(12,1) * C(4,2)}{{\color{red}C(52,2)}}= 3,744 / 2,598,960 \approx 0.14\%$
Divide by $\displaystyle C(52,5)$

5. ## Re: Confused about elementary combinatorics for poker hand calculations

Originally Posted by a4622
Apologies. I am asking what I am doing wrong. I commonly see a full house resulting from a 5 card draw represented as:

$\displaystyle \\n = 52\\k = 5\\\\\frac{C(13,1) * C(4,3) * C(12,1) * C(4,2)}{C(52,2)}= 3,744 / 2,598,960 \approx 0.14\%$

If I attempt to use smaller values for the purposes of viewing the combinations calculated, it appears that the combinations are not calculated correctly. Example:

$\displaystyle \\\mbox{Set R contains all ranks; set S, all suits; and set D, all cards.}\\\\R = \{A,2\}\\S = \{H,D\}\\D = \{AH,AD,2H,2D\}\\\\n = |D| = 4\\k = 4\\\\\frac{C(2,1) * C(2,2) * C(1,1) * C(2,2)}{ C(4,4)} = \frac{\ 2 * 1 * 1 * 1}{1}=200\%$

Is 2 combinations and a 200% probability an erroneous result? Is the scaled down problem not analogous to the full house problem above? As far as I can tell, multiplying the combinations results in permutations, not combinations. Combinations are the expected result. I don't know how much clearer I can make my inquiry.
In your example with the 4 cards, you are trying the compute the probability of drawing 2 pairs, right? (Which you know will hapen 100% of the time if you draw all 4 cards.)

The problem with your computation is that there is no way to distinguish one pair from another, so you have counted the pairs twice. I.e., you have counted two aces and two deuces, and then counted two deuces and two aces, resulting in an over-count. This situation does not arise in the "full house" computation, because there is no confusing the pair with the three-of-a-kind. For example, 2 aces and 3 deuces is a different hand than 2 deuces and three aces.

6. ## Re: Confused about elementary combinatorics for poker hand calculations

Originally Posted by Plato
Divide by $\displaystyle C(52,5)$
Thank you for catching the typo. It has been corrected.

Originally Posted by awkward
The problem with your computation is that there is no way to distinguish one pair from another, so you have counted the pairs twice.
I believe I see the problem. As written, a subset(either rank or suit; they contain the same number of objects) is selected from in two occurrences instead of one. The computation for drawing the two pairs should be:

$\displaystyle \\\frac{C(2,2) * C(2,2) * C(2,2)}{C(4,4)}$

While that had seemed like an obvious change to make, I could not previously explain why it should be the case. Thank you for the assistance, and sorry about over-complicating the matter with a lengthy post.