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.