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I have a homework assignment on probability that goes like this:

Two cards are selected one after the other, at random and without replacement, from a well-shuffled, standard deck of 52 playing cards. Let A1 be the event that the first card selected is an ace, and let A2 be the event that the second card selected is an ace. What is the probability that the second card selected is an ace?

The solution argues that because we don't know what we picked first, the probability that the second card selected is an ace is still $\frac{4}{52}$. However, I am very confused by the idea that not knowing what was picked, is equivalent to assume that an ace was not picked.

Also, this leads to the following sort of paradox. Given two observers, one who knows what is on the first card, the other who does not, the probability of picking an ace on the second card changes from conditional to unconditional.

Instead, I would see the experiment as a case of conditional probability in both cases, which cannot be estimated unless we know what is on the first card.

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    $\begingroup$ What possible mechanism could cause an ace to have any more or less probability of being selected for the second card than any other denomination? Since there isn't one, the answer must be the proportion of aces, or $1/13$. Your solution suggests you have a different question in mind, perhaps one about a chance of an event conditional on the outcome of the first card. Are you sure you quoted the question correctly? $\endgroup$ – whuber Sep 27 '17 at 21:43
  • $\begingroup$ @whuber Thanks for pointing that out, the quoted part was correct but not the solution as you suggested. My point of confusion is that what comes out on the first card should affect what I could possibly pick on the second. The fact that I ignore what came up in the first, should not make the probability of the second card unconditional. $\endgroup$ – Dambo Sep 27 '17 at 21:49
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    $\begingroup$ You might find it intuitively helpful to contemplate a generalization of this question in which all cards are drawn, in succession: what is the chance that the last card is an ace? It might be easier in this case to see that although (obviously) the first 51 cards completely determine the last one, the fact of drawing them provides no information that should change your answer from $1/13$. $\endgroup$ – whuber Sep 27 '17 at 21:55
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    $\begingroup$ Let's bet, then. I will shuffle the deck, remove 51 cards without showing them to you, and will allow you to make one guess concerning the rank of the card that remains. What will you guess and, if I charge you one dollar to make that guess, how much should I pay you if you're right? $\endgroup$ – whuber Sep 28 '17 at 13:59
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    $\begingroup$ If we were to ask the same questions about the first card--what would you bet, etc.--it seems that your entire set of comments would still apply: you cannot possibly observe the other 51 cards when you're considering only the first card. The only difference between the two situations is an irrelevant physical one: in one case the other 51 cards were moved from the deck into a discard pile and in the other they were not moved. If your intuition is "so what," it's correct. $\endgroup$ – whuber Sep 28 '17 at 16:24
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The solution, as you've stated it here, seems to be incorrect. Perhaps you've misunderstood what it was saying?

A nice way to think about problems of conditioning like this is with the law of total probability: \begin{align} \Pr(A_2) &= \Pr(A_2 \mid A_1) \Pr(A_1) + \Pr(A_2 \mid \overline{A_1}) \Pr(\overline{A_1}) .\end{align} That is, we need to both consider both the cases that an ace was picked first, and that an ace was not picked; then we need to combine those two by the probability of the respective first events. From this formula, you can easily work out the probability: \begin{align} \Pr(A_2) = \frac{3}{51} \times \frac{4}{52} + \frac{4}{51} \times \frac{48}{52} = \frac{1}{17 \times 13} + \frac{16 \times 4}{17 \times 13} = \frac{1}{13} \end{align}

Another way to think about it is as whuber suggested in the comments: the second card, given that we don't know what the first card was, is simply going to be completely random out of the deck. How could any card be more likely than any other? Thus, more directly, we have $$\Pr(A_2) = \frac{4}{52} = \frac{1}{13}.$$

Your "paradox," though, is not really a paradox. The observer who saw whether $A_1$ happened now has more information about whether $A_2$ will happen or not. Thus, from their perspective, the probability of $A_2$ has changed.

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