I perform 5 "independent" card draws from a deck w/ replacement. All drawn cards are queens, what is the probability of queen of spades drawn twice? So, drawing a queen of spades on the first draw has probability $\frac{1}{52}$.
As there is replacement involved the next card, assuming the card drawn is again, the queen of spades, must be of probability $\frac{1}{52}$. As i need exactly two queen of spades, the rest of the draws must be queens but not the spade suit. So the probability of drawing a queen, but not a spade is $\frac{3}{52}$. Hence the answer is $\frac{1}{52}$$\frac{1}{52}$$\frac{3}{52}$$\frac{3}{52}$$\frac{3}{52}$.
I am absolutely new to the idea of probability, or even if I have learnt it before I have completely forgotten anything about it. The answer I have calculated seems too absurd to be true, so kindly help me out with this problem. Thanks.

Looking at all the replies and hints in this post, this is what i came up with, albeit incompletely.
Assuming all cards are queens (-Don Thousand),probability of drawing the card (queen of spades) is 1/4. The rest of the cards can be drawn in 3/4 ways. Hence the joint(?) probability of 5 independent card draws is $\frac{1}{4}$$\frac{1}{4}$$\frac{3}{4}$$\frac{3}{4}$$\frac{3}{4}$. After this point in reference with John's reply there are $5\choose2$ ways of drawing the 5 card hand (edit: with respect to the positioning of the two queen of spades), as there is no order(?) constraint. 
The answer being $5\choose2$*$\frac{27}{1024}$. The answer seems to be correct, but did the way I go about finding it logically sound? Thanks a lot for the help guys!
