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  1. #16
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    5. Cards from an ordinary deck are turned face up one at a time (w/o replacement). Compute the expected number of cards that need to be turned face up in order to obtain 2 aces.


    Captainblack, this was the way you taught me the last time for a question: Draw 2 cards without replacement from a deck of playing cards. What is the expected number of aces?

    Let random variable x1 = 1 if an ace turns up on the turn, 0 otherwise.
    E(x1) = 1/13

    Let random variable x2 = 1 if an ace turns up on the 2nd turn, 0 otherwise.

    E(x2) = 1/13

    E(x1 + x2) = 2/13

    Now, instead of # of ace turning up, it's finding the expected number of cards that needed to be turned to obtain 2 aces. Can I use the same way?

    Can anyone please advise? Thank you very much
    Mmmm.. There must be a way of looking at this which simplifies the
    problem, but I can't see it at present.

    I have run a Monte-Carlo experiment for this problem and the result
    from that is the average number of cards that needs to be turned over
    to find two aces is 21.21+/-0.02 (2 sigma).

    OK, I can confirm that the expectaion is 21.2, but the computation is a
    but long winded and fiddly to type here at present.

    RonL
    Last edited by CaptainBlack; December 6th 2006 at 08:24 AM.
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  2. #17
    Junior Member Ruichan's Avatar
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    It's doesn't help anything even if I do it this way, does it?

    P(1 ace) = 4/52
    P(2nd ace) = 3/51
    Total prob = 4/52 * 3/51?

    Doesn't make much sense. Geez... I hope my prof isn't going to put such questions in the finals otherwise I'll never be able to finish the paper.
    Last edited by Ruichan; December 6th 2006 at 09:58 AM.
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  3. #18
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    It's doesn't help anything even if I do it this way, does it?

    P(1 ace) = 4/52
    P(2nd ace) = 3/51
    Total prob = 4/52 * 3/51?

    Doesn't make much sense. Geez... I hope my prof isn't going to put such questions in the finals otherwise I'll never be able to finish the paper.
    As I said in another post I have a way of evaluating this but without some
    more work on its not of much use to you as it still involves some heavy duty
    calculation which I think you cannot be expected to do. If I find a way of
    simplifying it that reduces the amount of calculation I will post it.

    It involves calculating the probailities P(k,n) of the k-th card being an ace
    and the n-th card being an ace, n>k, and all the cards 1,2, ..k-1,k+1, ..n-1
    not being aces.

    The the probability that the n-th card is the 2-nd ace is:

    P(n)=sum(k=1..(n-1)) P(k,n)

    then:

    E(number of cards to turn to find second ace)=sum(n=2..,50) n.P(n)

    RonL
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  4. #19
    Junior Member Ruichan's Avatar
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    Another card question

    Fourteen playing cards are dealt (w/o replacement).
    a) What is the probability that the first ace occurs in the fourteenth card?
    b) What is the probability that the fourteenth card dealt is an ace?


    I did it the following way. Can someone check if I did it correctly or point out that I'm wrong? My prof is too busy to help.

    a) Total # of playing cards = 52
    # of aces = 4
    Probability of picking an ace = 1/13
    Geometric distribution:
    (1/13)(12/13)^13 = 0.0272
    geo(14, 0.0272) = (0.0272)(1-0.0272)^13 = 0.019
    Therefore probability that first ace occurs in the fourteenth card = 0.019?

    b)P(fourteenth card an ace) =(1/13)^14
    = 2.5 X 10^-16????
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  5. #20
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by Ruichan View Post
    5. Cards from an ordinary deck are turned face up one at a time (w/o replacement). Compute the expected number of cards that need to be turned face up in order to obtain 2 aces.


    Captainblack, this was the way you taught me the last time for a question: Draw 2 cards without replacement from a deck of playing cards. What is the expected number of aces?

    Let random variable x1 = 1 if an ace turns up on the turn, 0 otherwise.
    E(x1) = 1/13

    Let random variable x2 = 1 if an ace turns up on the 2nd turn, 0 otherwise.

    E(x2) = 1/13

    E(x1 + x2) = 2/13

    Now, instead of # of ace turning up, it's finding the expected number of cards that needed to be turned to obtain 2 aces. Can I use the same way?

    Can anyone please advise? Thank you very much
    My prof just changed the question to with replacement. But I still don't know how to go about finding the expected number of cards.
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  6. #21
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    My prof just changed the question to with replacement. But I still don't know how to go about finding the expected number of cards.
    That makes it simpler.

    Consider the probability P(n,k) that n>k, and that n-th card is an A and the
    k-th card is an A, and that all the other cards from 1-n are not A's

    P(n,k)=(4/52)(4/52)(48/52)^(n-2)

    that is the product of the probability that card n is an A, the probability that
    card k is an A and the probability that the other n-2 cards from 1 upto n
    are not A's.

    Then the probability that the 2-nd A occurs at position n is:

    P(n)=sum(k=1..n-1) P(n,k)=(n-1)P(n,1)=(n-1)(4/52)(4/52)(48/52)^(n-2)

    (as the P(n,k)'s are independent of k)

    So the expected position of the second A is:

    E(n)=sum(n=2..52) nP(n)=(4/52)(4/52)(sum(n=2..50) n(n-1)(48/52)^(n-2)

    RonL
    Last edited by CaptainBlack; December 7th 2006 at 12:42 PM.
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  7. #22
    Junior Member Ruichan's Avatar
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    P(n,k)=(4/52)(4/52)(48/52)^(n-2)


    Why do you multiply (4/52) twice? Isn't once enough since I don't know which card is the ace turning up at?


    Thank you very much.

    One question is what has the expected position of Ace has to do with the expected # of cards that need to be turned?
    So that I find the position and then count the # of cards?
    But in this case, no numbers have been given so there's no way to find the expected # of cards needed to be turned in numerical value since I don't know the position of the Aces?
    Last edited by Ruichan; December 7th 2006 at 12:26 PM.
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  8. #23
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    P(n,k)=(4/52)(4/52)(48/52)^(n-2)


    Why do you multiply (4/52) twice? Isn't once enough since I don't know which card is the ace turning up at?
    There are two aces in the claculation one at position k the second at position n (the others we don't care about as if they occur will be later in the sequence). Each of these aces gives a 4/52, just as the non-aces give
    facors of 48/52.

    RonL
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  9. #24
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    One question is what has the expected position of Ace has to do with the expected # of cards that need to be turned?
    So that I find the position and then count the # of cards?
    But in this case, no numbers have been given so there's no way to find the expected # of cards needed to be turned in numerical value since I don't know the position of the Aces?
    The expected number of cards turned is by definition:

    sum(n=2 .. 52) nP(n)

    where n is the number of cards turned to reach the second ace.

    But n is the position of the second ace in the sequence.

    (note this sum is to 52 I may have used 50 earlier if I did it was a hang over
    from the non-replacement case I will correct it when I finish this).

    RonL
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  10. #25
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    There are two aces in the claculation one at position k the second at position n (the others we don't care about as if they occur will be later in the sequence). Each of these aces gives a 4/52, just as the non-aces give
    facors of 48/52.

    RonL
    Oh ok, i understand now. Initially I was thinking that I could apply just one time 4/52 since I do not know where the ace will turn up, so it doesn't matter at position k or at position n since it will be one calculation of 4/52(48/52)^n-2.

    Thank you very much.
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  11. #26
    Junior Member Ruichan's Avatar
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    5. Cards from an ordinary deck are turned face up one at a time with replacement. Compute the expected number of cards that need to be turned face up in order to obtain 2 aces.

    Captain Black, my prof did on the whiteboard the solution but he didn't explain. I'm wondering if you could help me with it.

    Solution:
    Let X1 denote the number of nonace card before the first ace. X2 denote the number of nonace cards before the first 2 aces.

    P{X1=i, X2=j} = P{X1=j, X2=i}
    ===> X1 and X2 have the same distribution

    E[X1] = 48/5

    E(2 + X1 + X2) = 106/5.

    How did he get the values for E[X1] and E[2+X1+X2]?
    He just wrote down the solution to this question then erased it away to do another question as we were short on time.
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