- #1
Fictionarious
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Homework Statement
We've got a standard deck of 52 cards. We shuffle the deck well, then cut it into two piles of 26 each, a top pile and a bottom pile. We reach in and pull a random card out of the top pile, observing that it is an Ace. We then put it into the bottom pile, shuffle the bottom pile, and reveal a card at random from it. What is the probability that it is an Ace?
Homework Equations
Hyper-geometric Probability (badly formatted but bear with me)
( n C x )( m C b-x) / ( n+m C b)
The Attempt at a Solution
I have found two ways of solving this problem, one of which I think is correct, and one which I believe is oversimplified. I will show the one I believe is oversimplified first.
This is to observe that there is initially a uniform probability distribution of 4/52 for each card being an Ace, but after we reveal one, the probability goes to being 3/51. This applies to every card except the one we KNOW is an ace. Since there are 27 cards total in the bottom pile when you pick one from it,
(3/51)(26/27)+1(1/27) = 43/459 = 0.0936819172
However, another approach is to use choose notation (permutations and combinations):
Consider that there were 4 possibilities. The top half originally contained one of the following number of Aces x={1,2,3,4}, with the bottom half containing y=4-x. Note that the top could NOT have originally contained none, given that the card we observed and then moved was an Ace. We might then wish to express each of these cases' respective probabilities.
[(26C4)(26C0)]/(52C4) , [(26C3)(26C1)]/(52C4) , [(26C2)(26C2)]/(52C4) , [(26C1)(26C3)]/(52C4)
In the first case, there is one Ace in the bottom 27 cards when we select at random from it (the one we just moved there from the top), so we would multiply that term by (1/27). In the second, third, and fourth, we would multiply by (2/27), (3/27), and (4/27), respectively. Then we would sum them. This method yields 0.1008847984
I don't see any explicit error in the second method or the first, but both of them can't be right. My tentative belief currently is that the second one is the correct method. If anyone could prove me right or wrong, and shed further light on this problem, it would be very much appreciated.