Four of a kind combinatorics, why is my logic incorrect? I'm going through a probability textbook and getting stuck on a simple question. From a standard deck of 52 cards, what is the probability of getting a 4 of a kind?
I've seen many of the answers, and I understand how to get there. The straightforward way of answering the question is just to realize there's $ 13$ four of a kinds with $48$ choices for the 5th card so the solution is just $$ \frac {13 *48} {52 \choose 5} = \frac {624} {52 \choose 5}$$
I understand how to get to this answer. My question is, why is the following approach incorrect?
For the first card we can draw any of the 52 cards, we must draw the remaining cards with the same rank. This equates to :
$$52 * 3 * 2 * 1$$
Then for the 5th card we have 48 remaining cards so we have:
$$52*3*2*1 * 48$$
Now, furthermore, these 5 cards can be arranged in any manner. So for the total number of ways we have:
$$ \frac {52 *3 *2 *1 * 48} {5!} $$
Then we simply divide by ${52 \choose 5}$ to get our final probability.
Now it turns out this answer is wrong, but I cannot figure out why?
The mistake is obviously in this step:
$$ \frac {52 *3 *2 *1 * 48} {5!} $$
But for the life of me I cannot figure out why. It seems the correct answer is to divide by $4!$ factorial instead of $5! $, but why? We have 5 cards, they can be arranged in $5!$ ways
 A: You say "for the first card."  Such lines of thinking are dangerous if you are mixing both order matters and order doesn't matter in the same calculation.  For instance, with order mattering, where did you account for the possibility of the first card being the "kicker" (the card not a part of the quadruple)?
So, for order mattering, pick the slot that the kicker could appear in, then pick what it was.  Now, from left to right in the remaining slots pick the first card in the quadruple, then pick the others in the quadruple giving a total of $5\cdot 52\cdot 48\cdot 3\cdot 2\cdot 1$ such hands with order mattering.  Dividing by $5!$ gives the number of such hands if order doesn't matter.  It wasn't that dividing by $5!$ was wrong and $4!$ was correct, it was that you were missing a factor in the numerator corresponding to where the kicker was.
A: When you proceed by counting, you must be quite clear for yourself what set (of possibilities) you are counting, and then ensure that you are counting exactly that set, with no elements left out and no elements counted more than once. Here what matters ultimately is a subset of the set of $\binom{52}5$ hands of $5$ cards, but for a probability question you can also consider the corresponding subset of the set of $52\times51\times\cdots\times 48=\frac{52!}{47!}$ possible ways of successively dealing $5$ distinct cards; this number is $5!$ times larger than the number of hands because on order among the dealt cards is recorded, but the subset will be larger by the same factor than the subset of hands, so the quotient (the probability) will be the same. Since you are dividing at the end by $5!$ to get then number of unordered hands, it is clear that you are counting ways to deal $5$ cards, and in particular the subset of these that result in a hand that is valid as $4$ of a kind.
But given that you are taking order into account, you must count all possible ways in which such a hand can be dealt, and you cannot assume that the first card dealt will be in the group of $4$. In general for counting poker hands in the way, it seems best to separately list the possible groupings among the dealt cards, and then for each count the ways to fill it in. Here the card with a unique rank can be in each of the $5$ positions, giving $5$ distinct groupings. Once this grouping is fixed you can fix the cards in the $5$ positions in any order; choosing those in the other positions first one has $52\times3\times2\times1\times48$ valid possibilities as you correctly computed. But you get the same number for each of the $5$ patterns, so in all $5\times52\times3\times2\times1\times48$ ways to deal a four-of-a-kind. THe probability then is
$$
\frac{5\times52\times3\times2\times1\times48}{52\times51\times50\times49\times 48} 
= \frac{5\times3\times2\times1}{51\times50\times49}
= \frac{13\times48}{\binom{52}5}
$$
in accordance with what is found by the other (easier) method.
