What is the probability that you will draw a queen from a standard deck of cards then draw a second queen without replacement?

Perhaps you can see how the chance is the exact same ($\frac{4}{52}$) if you had picked the top card instead of the second to top. Chance, after all, does not play favorites between the two cards.

Then, consider the following set of actions:

1. Take a deck of cards
2. Shuffle the cards
3. Draw the first card
4. Draw the second card and check if it is a queen

Now, you should be able to see that the chances for the two cases are the exact same. After you shuffle the deck, it doesn't matter if you draw two cards and check the second or you just check the second card without looking at the first; if the second card is a queen, it's a queen, if it's not, it's not.

The flip side of the feeling-based approach is the math to back it up. Let's start with the simple case: you have 52 cards, and want a queen on the second draw. (This has also been done by other answers, but I'll repeat it here.)

You draw a queen on the second draw if:

• You draw a queen on the first draw and one on the second $$\frac{4}{52}*\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}$$
• You draw something other than a queen on the first draw and a queen on the second $$\frac{48}{52}*\frac{4}{51}=\frac{192}{2652}=\frac{16}{221}$$

So in total, the chance is:

$$\frac{1}{221}+\frac{16}{221}=\frac{17}{221}=\frac{4}{52}$$

Now, let's up the ante a little bit. Rather than wanting to know something about queens in a complete deck, I want to know about the more general case. I have a pile of $n$ shuffled cards. In that pile, I know there are $p$ cards that I "like". What I want to know is: what is the chance I draw a card I like.

For the first card, it's simple. The chance simply is $\frac{p}{n}$.

For the second card, we once again have two options:

• I like both the first and second card $$\frac{p}{n}*\frac{p-1}{n-1}=\frac{p^2-p}{n^2-n}$$
• I like the second card, but not the first $$\frac{n - p}{n}*\frac{p}{n-1}=\frac{pn-p^2}{n^2-n}$$

Adding the two, you get:

$$\frac{p^2-p}{n^2-n}+\frac{pn-p^2}{n^2-n}=\frac{p^2+pn-p^2-p}{n^2-n}=\frac{pn-p}{n^2-n}$$

Moving things around a bit more:

$$\frac{pn-p}{n^2-n}=\frac{p(n-1)}{n(n-1)}=\frac{p}{n}$$

Which is the same as the chances for the first card. So, now I can say that no matter the deck size or the amount of cards that represent "success", it doesn't matter if I look to the first or second card to determine success. (Of course, if I look at the second card, it's important that I don't care what the first card is at all.)

I could actually repeat the experiment for each different card in the deck, and then I could draw the conclusion that in general: it doesn't matter if I look at the first or second card, the chances for the card to be a specific one are equal.

The next step could be to proof that the the other cards (third, fourth, etc) have the same chance as well, but I'll leave that as an exercise for the reader.

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