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.
 A: Pretending that the deck has only the four queens is great advice.
Let's assume that you need exactly two queen of spades.
Choose which two of the five draws you get your queen of spades. How many ways?
How many ways can you get three non-queen of spades in the other three slots?
Divide by the total number of ways to draw the five cards.
Spoiler:

 There are $_5C_2 = 10$ ways to choose the particular draws for the two queen of spades. There are $3^3 = 27$ ways to draw the other three non-queen of spades, and $4^5 = 1024$ ways to draw five queens. So the probability is $10 \times 27 / 1024 = 270/1024.$

A: 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!
