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Homework Help: Conditional Probability -A Paradox?

  1. Aug 6, 2014 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    Hyper-geometric Probability (badly formatted but bear with me)

    ( n C x )( m C b-x) / ( n+m C b)

    3. 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 cant 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.
  2. jcsd
  3. Aug 6, 2014 #2


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    That seems wrong. Shouldn't the four (conditional) probabilities add up to 1?

    I don't see anything wrong with the first solution.
  4. Aug 6, 2014 #3

    Ray Vickson

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    Let ##A_1##= {first draw is an Ace}, ##A_2##= {second draw is an Ace}. You are asked to compute ##P(A_2|A_1) = P(A_2 \, \& \,A_1)/P(A_1)##.

    Let ##X## be the number of Aces in the top deck initially. For ##k = 0,1,2,3,4## what is ##P(X = k)##? How would you compute ##P(A_1 \, \& \, A_2)## with the help of the probabilities ##P(X=k)##? How would you compute ##P(A_1)## using the ##P(X=k)##? If you refrain from computing probabilities too soon, the expression for ##P(A_2|A_1)## simplifies nicely, in terms of properties of ##X##.

    I get your first answer, but the route to the answer is very different from yours. I am not convinced by the "logic" of your first analysis.

    I cannot understand at all where your "probabilities" come from in your second method.
    Last edited: Aug 6, 2014
  5. Aug 8, 2014 #4
    Well, I am by no means convinced either. Let me try to answer your questions since you seem to be leading me somewhere with them.

    What is P(X=k) for k=0,1,2,3,4 ?

    Well, this is where my "probabilities" are coming from in the second method, or so I think.
    AlephZero, you are correct that they don't add up to one, this is because I excluded the k=0 case. I'm not confident enough to say whether that was a good call or not, but since it isn't working, I'm leaning toward not.

    P(X=k) = (the number of ways to arrange k aces in 26 possible positions) * (the number of ways to arrange the remaining aces in the 26 possible positions of the other half of the deck) / (the number of ways of arranging 4 aces among 52 positions)

    When summed from k=0 to k=4, this is indeed 1.

    (*goes to work*)

    I believe I have fixed it!

    [(26C4)(26C0)]/(52C4) * (4/26) represents A1 union k=4
    [(26C3)(26C1)]/(52C4) * (3/26) represents A1 union k=3
    [(26C2)(26C2)]/(52C4) * (2/26) represents A1 union k=2
    [(26C1)(26C3)]/(52C4) * (1/26) represents A1 union k=1
    A1 union k=0 is zero.

    Summed, these give P(A1) = 1/13

    To find P(A1 union A2), we simply augment each of those terms as follows:

    (46/833)(4/26)(1/27) + (208/833)(3/26)(2/27) + (325/833)(2/26)(3/27) + (208/833)(1/26)(4/27)

    For convenience I simplified the choose notation stuff to simple fractions #/833 in this step.

    This is P(A1 union A2) = 43/5967

    Then, P(A1 union A2)/P(A1) = 43/459

    So, even though my intuition was right in that the k=0 case could be safely ignored, the problem was in my cavalierly eschewing the conditional probability formula. To be fair, the explanation the back of the book gave ignored it as well, giving the abbreviated version (3/51)(26/27)+1(1/27) = 43/459.

    Thanks for the help guys, hopefully this fixed version is correct and made more sense (if not I'll have to save face and return to the problem again . . .)
  6. Aug 8, 2014 #5

    Ray Vickson

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    The random variable ##X## is the number of Aces in Pile I initially, so has the hypergeometric distribution with parameters ##N_1 = 4 \: (\#\: \text{Aces}), \: N_2 = 48 \:(\#\: \text{non-Aces}), n = 26 \: (\# \: \text{sample size})## and pmf
    [tex] P(X=k) = \displaystyle \frac{{4 \choose k} {48 \choose 26-k}}{{52 \choose 26}}, \: k = 0,1,2,3,4 [/tex]
    These values of ##P(X=k)## are the same as yours (that is, your expressions without the factors k/26). Your way is just an alternative (but slightly non-standard) way of writing the hypergeometric pmf. BTW: what you have written are not the probabilities of {X=k} union A1, but of {X=k} intersect A1.

    Anyway, since ##P(A_1|X = 0) = 0## we have
    [tex] P(A_1) = \sum_{k=0}^4 P(A_1 | X=k) P(X=k) =
    \sum_{k=1}^4 P(A_1 | X=k) P(X=k) = \sum_k \frac{k}{26} P(X=k) = \frac{1}{26} EX [/tex]
    Also, when ##X=k## there will be 4-k+1 = 5-k Aces in the new Pile 2. Thus
    [tex] P(A_1 \, \& \, A_2) = \sum_k P(A_1 \, \& \, A_2 | X=k) P(X=k)
    = \sum_k \frac{k}{26} \frac{5-k}{27} P(X=k) = \frac{1}{26 \times 27} [5 EX - E(X^2)].[/tex]
    [tex] P(A_2|A_1) = \frac{1}{27} \frac{5 EX - E(X^2)}{E X}[/tex]

    There are standard formulas for the mean and variance of the hypergeometric distribution: if the two sub-population sizes are ##N_1, N_2## with total population size ##N = N_1 + N_2##, let ##Y## be the number of type-1 in a sample of size ##n##. If ##p \equiv N_1/N##, then:
    [tex] EY = n p\\
    \text{Var} (Y) = n p (1-p) \frac{N-n}{N - 1}[/tex]
    Just apply these to ##N_1 = 4, N_2 = 48, N = 52, n = 26##, then get ##E(X^2)## from ##E(X^2) = \text{Var}(X) + (E X)^2##.
    Last edited: Aug 8, 2014
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