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In Introduction to Probability by Blitzstein & Hwang, Chapter 2 Problem 5:

Three cards are dealt from a standard, well-shuffled deck. The first two cards are flipped over, revealing the Ace of Spades as the first card and the 8 of Clubs as the second card. Given this information, find the probability that the third card is an ace in two ways: using the definition of conditional probability, and by symmetry.

Solution:

Let A be the event that the first card is Ace of Spades, B be the event that second card is 8 of Clubs, and C be the event that third card is an Ace.

$P(C|A,B) = \dfrac{P(A,B,C)}{P(A,B)}$

Numerator: Having first as Ace of Spade, second as 8 of Clubs and third as an Ace, is similar to choosing three cards out of 52 cards without replacement. However, there are 3 ways for the third card to be an Ace since there are three Aces left, Ace of Hearts, Diamonds, and Clubs. $P(A,B,C) = 3\cdot(\dfrac{1}{52})(\dfrac{1}{51})(\dfrac{1}{50})$

Denominator: This is the same as choose two cards out of 52 without replacement. $P(A,B) = (\dfrac{1}{52})(\dfrac{1}{51})$

Therefore, $P(C|A,B) = \dfrac{P(A,B,C)}{P(A,B)} = \dfrac{3\cdot(\dfrac{1}{52})(\dfrac{1}{51})(\dfrac{1}{50})}{(\dfrac{1}{52})(\dfrac{1}{51})} = \dfrac{3}{50}$

Is this solution correct? By the way I don't get it as how to use symmetry to view this problem...

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Symmetry: there are 50 cards left. Each has the same probability so the probability to get an ace is $\tfrac{3}{50}$ as there are 3 aces left.

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  • $\begingroup$ Could you elaborate a bit more the idea of symmetry? $\endgroup$ Jul 20 '20 at 17:19
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    $\begingroup$ If you have several symmetric results, they have the same probability. In this example, you have 50 possible cards to be the 3, all symmetric (no card is more likely to come up next) so each has a probability of 1/50. This is EXACTLY why you wrote "1/52" and "1/51" in your answer. The key is to understand that the problem is like drawing a card from a deck of 50 cards. $\endgroup$
    – YJT
    Jul 20 '20 at 17:23

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