Probability of drawing an ace 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? 

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.
