Probability of drawing a pair from a poker hand, unordered with replacement? I am wondering what is the probability with which you can draw a pair in a 5-card hand from a standard 52-card deck, if order does not matter in the context of cards in the hand, and if the cards can be replaced with every draw?
 A: To find the probability of a one-pair hand, it will be easier if we take the order of drawing into account. It is true that all we care about is what hand we end up with. But taking order into account will get us the right answer without the need to look at special cases. 
There are $52^5$ equally likely sequences of $5$ cards. We count the number of $5$-card sequences that satisfy the "one pair" condition. Counting "hands" will not work unless we are very careful, for not all hands are equally likely. 
The two locations in the draw where the  pair will be  can be chosen in $\binom{5}{2}$ ways. Once we have chosen the two locations, the kind we have two of can be chosen in $13$ ways. Once we have done that, the leftmost chosen "pair" location can be filled in $4$ ways, and for each such way we can fill the other pair location in $4$ ways, for a total of $4^2$.
Now it is time to deal with the remaining $3$ positions. There are $48$ ways to fill the leftmost empty position. For each of these, we can fill the next position in $44$ ways, and then the last position can be filled in $40$ ways.
The number of one-pair sequences is therefore 
$$\binom{5}{2}(13)(4^2)(48)(44)(40).$$
Divide by $52^5$ for the probability. 
Remark: We interpreted "the cards can be replaced at every draw" to mean that the drawn card is replaced in the deck before the next draw. So the model is that we draw a card, record its value, replace and shuffle, doing this a total of $5$ times. We want the probability that among the $5$ recorded cards, there are exactly two cards (possibly identical) of one kind, and $3$ useless cards all of different kinds. 
A: "There are 525 equally likely sequences of 5 cards." This is already erroneous. We have 52 possibilities for the 1st card, but this leaves only 51 for the next choice, 50 for the next choice, and so on. So there are 52*51*50*49*48 possible 5-card draws.
Other errors are also present.
