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Lets say I've drawn 20 cards from a deck and The 20th was an ace. I'm trying to figure out the probability of drawing an ace in the 21st draw.

I'm sure this is just clever combinatorics but all my attemps have led nowhere. The probability of The first event is 4/52, right? And the second one is 3/32?

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    $\begingroup$ Is there no information about the first 19 cards, or do you see each drawn card and the 20th is the first ace ? $\endgroup$ – true blue anil Sep 17 '15 at 15:27
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We can count the ways to draw an ace in the 21st slot, given an ace in the 20th and divide by the ways to have an ace in the 20th slot:

  1. There are ${4 \choose 1} = 4$ aces that can be at the 20th spot;
  2. There are ${3 \choose 1} = 3$ remaining aces that can be at the 21st spot;
  3. The remaining 50 cards may be permuted in any other order around these two spots, so there are $50!$ ways to order the remaining 50 cards;

This gives us $4 \cdot 3 \cdot 50!$ decks in which the 20th and 21st cards are an ace; if we don't require that the 21st card is an ace, then we have $4 \cdot 51!$ ways to permute the cards such that the 20th is an ace, so the probability $P$ is:

$$ P = \frac{4 \cdot 3 \cdot 50!}{4 \cdot 51!} = \frac{3}{51} $$

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    $\begingroup$ What's wrong with the below reasoning - " After it is known that card number 20 is an ace, there are 51 different possibilities for card number 21, of which three are favorable (remaining aces). Hence, 3/51" $\endgroup$ – Deepak Gupta Sep 18 '15 at 20:19
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    $\begingroup$ Nothing. As with almost all discreet math problems, there are multiple ways to arrive at the correct answer. For some people, it will be more intuitive to account for all of the cards rather than just the 20th and 21st; for others the opposite. $\endgroup$ – nben Sep 18 '15 at 21:04
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This problem can be solved with some conditional probability, though the method I will present is tedious and is probably not the most elegant.

We need to find $P(21st\ card\ is\ an\ Ace | 20th\ card\ is\ an\ Ace)$

We know this is equivalent to $\frac{\sum\limits_{i=2}^4 P(21st\ card\ is\ ith\ Ace\ \cap\ 20th\ card\ is\ (i-1)th\ Ace)}{P(21st\ card\ is\ an\ Ace)}$

Now it is an exercise in combinatorics to calculate both the numerator and denominator of this fraction.

I'll help you get started, lets calculate the probability in the denominator of this fraction:

The $21st$ card may be any of the four Aces in a standard deck of 52 cards, thus the probability of drawing $21$ cards such that the $21st$ card is an Ace must be exactly $\frac{4{51\choose20}}{52\choose21}$

Hopefully this helps, I don't think it is too hard now to calculate the numerator of the aforementioned fraction.

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There are 51 possibilities for card number 21, out of which, 3 could be aces. Hence, probability is 3/51.

Note that this probability is independent of the card numbers (20 and 21, in this case)

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