If you draw two cards, what is the probability that the second card is a queen? We had this question arise in class today and I still don't understand the answer given. We were to assume that drawing cards are independent events. We were asked what the probability that the second card drawn is a queen if we take two from the deck. The answer given was 4/52, which seems counter-intuitive to me. How is the probability still 4/52 if there was a card drawn before it? What if the first card drawn was a queen?
 A: Probablities are a hard thing to wrap your mind around. Let's try a more feeling-based approach:


*

*Take a deck of cards

*Shuffle the cards

*Flip over the second card of the deck and check if it is a queen, while leaving the first card on the deck


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:


*

*Take a deck of cards

*Shuffle the cards

*Draw the first card

*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.
Disclaimer: my proof probably isn't elegant, optimal or nice, but I do believe it to be correct.
A: There are two cases here:
Case 1: First card chosen is a queen
$$\frac{4}{52}*\frac{3}{51}=\frac{1}{221}$$
Case 2: First card chosen is not a queen.
$$\frac{48}{52}*\frac{4}{51}=\frac{16}{221}$$
Adding both the cases, we get $\frac{17}{221}$ = $\frac{4}{52}$ = $\frac{1}{13}$
A: It seems to me that the problem here is not distinguishing between the prior probability and the conditional probability.
If we observed that the first card drawn was a queen, then that would give us a lowered probability of the second card being a queen of $3/51$.
However, this is not the same probability as the one asked of in the problem, but the conditional probability given the first card being a queen.
If we instead had had the first card not be a queen, then the conditional probability would have been heightened to $4/51$.
The prior probability could then be calculated using the probability of both cases as $$\frac{4}{52}\cdot\frac{3}{51}+\frac{48}{52}\cdot\frac{4}{51}=\frac{4}{52}$$
However, the problem can be simplified as what is important isn't the drawing of the card from the deck, but the observation of the value of the card.
As the value of the first card is never considered in the problem, it can simply be considered as not being drawn at all, which brings us to a second point;
If instead of drawing two cards from the deck, you simply fanned out the cards and drew the second card from the top, you would still get the same probabilities as if you chose the top card.
In fact, you would get the same probabilities even if you chose the tenth card or even a card at random.
There is nothing special about the top card of the deck unless you choose to do something with it.
A: Think about it this way: Shuffle a deck of cards randomly. The probability of drawing a queen as your second card is the same as the probability that the second card in the deck is a queen, which is clearly 4/52.
A: A slightly more intuitive way of looking at this:
The probability that the second card is a queen should be the same as the probability that the second card is an ace, and the same as the probability that the second card is a 2 etc. There are $ 13 $ possibilities for the card number/letter, so the answer is $ \frac{1}{13} $
A: Consider a more complete event:
When drawing all 52 cards, what is the probability that the second card is a queen?
The reason of changing the original question to this is because this is an expanded process, in other words, if you do draw all cards you certainly drop the first two ones. To solve the new problem is not difficult. One way is to consider every card of the 52 has an equal chance be at the second place. So the probability is 4/52. 
It is equal to the same probability that the first card, the 3rd card or the 27th card etc. is a queen, or any other type of card.
A: 2 things to consider.  What if the problem stated draw all cards and what is the probability that the last card drawn is a queen then what would your answer be? You might think it is close to 0 because in 51 cards drawn, there is a very good chance that all queens would be already drawn by then.  However, the answer should still be 4/52 because there are 4 queens in the deck and each has an equal chance of being the last card drawn just as all the other cards have.
Also, if you notice on TV when they have Texas Hold 'em, they "waste" the top card when drawing community cards (turn, river...) so if that changed the probability, it would likely be disallowed.
A: Imagine you buy a ticket in a lottery, where a fraction $p$ of the tickets will win. (From the comments I gather that there are some people who are confused by all kinds of newfangled games with different rules, such as being able to compose the number on one's own ticket; I therefore emphasise this is just an old fashioned lottery with a fixed set of tickets established beforehand, all distinct and all sold before the drawing; drawing then determines a subset of the tickets as winners, with the size of that subset being $p$ times the total number of tickets.) What is your chance of winning? It would be hard to argue that it is anything other then $p$.
But now let us add that before buying a ticket, you had to wait in line, and you happen to notice that the person before you also bought a ticket for the same lottery. Of course you don't know whether her ticket will win, but what is the chance that your ticket will win? It is still $p$. But what if her ticket actually wins? Well with that new information your chances will no doubt be less, but you don't have that information. If her ticket actually looses, that will slightly improve your odds, but you don't know that either. In fact you can be sure that there are hundreds of people who also bought tickets, but as long as you know nothing about their results (and you won't until the winning lots are drawn), it will not affect your chances the slightest bit.
The situation you describe is entirely similar to this.
A: put the first card back in the deck; now you've only drawn one card.
the chance it's queen  4/52
A: You can draw a pair of cards by drawing the first card, then drawing the second card. Let's call these cards A and B. You're interested in the probability that card B is a queen.
Now consider a different experiment: draw a pair of cards as before, but this time call the first one card B, and the second one card A. I claim that these two experiments are identical. The reason is that for any two cards X,Y, the probability to draw X then Y is the same as the probability to draw Y then X.
The second experiment makes it clear that the probability that card B is a queen is 4/52, since there are 4 queens out of 52 cards.
A: Can I convince you of this statement:
If I draw two cards, the probability that the first card is a queen is the same as the probability that the second card is a queen.
If you believe this, then your question has the same probability as the following the situation:
What is the probability that, if I draw two cards, the first card is a queen?
It is clear now that the second card pull in this case has no effect on the first, just as how the first card pull has no effect on the second in your case.
So the problem breaks down into:
What is the probability that, if I draw a card from a standard deck, the card is a queen?
which of course is $\frac{4}{52} = \frac{1}{13}$.
A: Let us find out, the $E$ be the event of drawing a queen in the second draw. Let $X_1$ be the event of drawing a queen at the first draw and $X_2$ is a card other than queen in the first draw. So the total probability is $P(E|X_1)+P(E|X_2)= 4/52*3/51+48/52*4/51=4/52$.
A: I would think it would be slightly below 1/13 since 1/13 is the probability for the first pick being a queen then the second round should account for the possibility of there only being 3 queens in the deck.
A: This is basically the Monty Hall problem -- as long as you are talking in advance of the draw, the probability is unchanged, once you draw the odds adjust.
You have two goats and a car, if you open two doors, what is the chance that there is a goat behind the second door?  What is the chance it is a car? 2/3 and 1/3 respectively.
After you open the first door, and find that it is a car, you can say that there is zero chance that there is a car behind the second door.  But at that point, they are no longer independent events, which is why in the Monty Hall problem you always switch.
In this case you are effectively saying that the second card is a queen, and you are doing this before you know what the first card was. The chance that that is correct is 1/13.  It (the chance you were correct) will remain 1 in 13 until either the second card is revealed or the position of all queens have been determined, at which point it is not a matter of probability.
In both cases the odds of you having identified the card is unchanged, what changes is the current odds of that card being a queen (ie is this one of the 1 in 13 times where you were correct).
This is where people go wrong with the Monty Hall problem, they confuse the current odds of the car being behind the door with the odds that they correctly identified the door with the car.  In this instance the confusion is between having identified the card vs the card actually being.
Here's another way of looking at it.  Your odds of correctly predicting card 2 is 1 in 13.  So take 4 decks of cards, divide them into 13 rows by   suit (throwing away the 3 extra suits).  Then pick a value for the 2nd card in each row.  Now start randomly revealing cards until each row has been determined -- do you expect this random reveal to result in you having correctly guessed more than the expected 1 row?  Is it any different from just revealing the 2nd card of each row?  Now, suppose that someone who knows what the cards are does the reveal, revealing each card  until there are only 2 cards left per row -- when you do the final reveal for each row, how many rows would you expect to have guessed correctly?  
Your chance of predicting the card in advance is 1 in 13 for each row, and it remains 1 in 13 -- revealing cards doesn't change the odds of your having guessed correctly.  Your guess is in the past and is now fixed, forever unchanging.  Either it was right or it wasn't.   What changes (assuming random reveal) is my current odds of guessing whether you were right or not.  As with the related Deal Or No Deal game, I get additional information with each random reveal.
