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Five cards are dealt from a standard deck of 52. What is the probability that the 3rd card is a Queen?

What I dont understand here is how to factor in when one or both of the first two cards drawn are also Queens.

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You can just ignore the first two cards. Think of dealing five cards, the move the first to third position. You are in the same situation as your procedure. As long as you don't look at the first two, they don't matter.

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The usual way to do this is to use the total probability theorem.

Let $A_i$ denote the event that the $i$th draw is a queen (for $i=1,2,3$). Then

$$\Pr(A_3) = \Pr(A_3|A_2 \cap A_1)\Pr(A_2 | A_1) \Pr(A_1)+ \Pr(A_3|A_2^c \cap A_1)\Pr(A_2^c | A_1) \Pr(A_1) + \Pr(A_3|A_2 \cap A_1^c)\Pr(A_2 | A_1^c) \Pr(A_1^c) + \Pr(A_3|A_2^c \cap A_1^c)\Pr(A_2^c | A_1^c) \Pr(A_1^c)$$

That is

$$\Pr(A_3) = \dfrac{2}{50}.\dfrac{3}{51}.\dfrac{4}{52} + \dfrac{3}{50}.\dfrac{48}{51}.\dfrac{4}{52} + \dfrac{3}{50}.\dfrac{4}{51}.\dfrac{48}{52} + \dfrac{4}{50}.\dfrac{47}{51}.\dfrac{48}{52} = \dfrac1{13}$$

Looks like $\Pr(A_1) = \Pr(A_3)$. This rather puzzling behavior can be explained by staring closely at the sample space.

In terms of the sample space, assigning a sequence of draws $(x_1,x_2,x_3)$, to the permutation $(x_3,x_2,x_1)$ defines a bijective map.

To understand the "bijective" map: Consider two parallel universes $U$ and $V$, where $A$ and $B$ are drawing cards (three times) simultaneously in $U$ and $V$ respectively. The randomness in the universes $U$ and $V$ are coupled in such a way that $A$'s first draw in $U$ is the same $B$'s third draw in $V$, and the second draw is the same for both.

Now if drawing a King of Hearts, Nine of Spades and Queen of Hearts represented as the vector $($KH,9S,QH$)$, then $($KH,9S,QH$)$ for $A$ will be the draw $($QH,9S,KH$)$ for $B$.

If you think carefully, you will realize that all the questions about the third draw in $A$'s universe is really the same question about $B$'s first draw.

In our case, the question transforms to 'what is the probability that $B$ will draw a queen on the first draw'. But that is easy to answer: $$\Pr(A_3) = \dfrac{4}{52} = \dfrac1{13}$$

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This is correct, but it seems unnecessarily complicated. You deal some cards onto a table face down. You point at one of them and wonder whether it's a queen. Well, there are $52$ possible values for that card and $4$ of them are queens, so the probability is $\frac{4}{52} = \frac{1}{13}$. – Pete L. Clark Dec 19 '12 at 20:19
That's true. In retrospect, the idea is complicated. However, until I write the sample space and see the bijective map, I somehow feel unconvinced that the probabilities are actually the same! – Isomorphism Dec 19 '12 at 20:28
Upon reflection, part of what I am objecting to is this: your answer seems to treat as a "black box" the assumption that the probability that the first card is a queen is $\frac{4}{52}$. I think that if you reflect on this you can see that whatever justification you have for that is equally well a justification for the probability of the third card being a queen. Reducing the second case to the first one via a bijection of the sample space is better but still, paradoxically, slightly asymmetrical. – Pete L. Clark Dec 19 '12 at 20:40

All orderings of the $52$ cards in the deck are equally likely. So the probability the third card in the deck is a Queen is exactly the same as the probability that the $17$-th card in the deck is a Queen, or that the first card in the deck is a Queen: They are all equal to $\dfrac{4}{52}$.

The fact that $5$ cards were dealt is irrelevant.

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