Results 1 to 6 of 6

Math Help - hard probability question

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    249

    hard probability question

    i had this probability question on my test and i had absolutely no idea how to do it. please help me.

    suppose you have a standard 52 card deck of playing cards and 1 joker. you spread these cards face down over a table. now one by one, you flip over cards and you can only flip over each card once. What is the probability that you will flip over all 4 aces before you flip over the 1 joker?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2009
    Posts
    108
    probability:

    numerator: ways to get 4 aces before 1 joker in a deck of 53 cards
    denominator: ways to deal 53 cards


    A successful dealing will look like this:

    aAbAcAdAeJf where the sum of a through f is 48.

    So the question is really asking you how many ways you can have 6 integers sum to 48, then divide by the number of ways to deal 53 cards.

    Show us the progress you've made so we can help you find your mistakes.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Let A denote the event that in first (j-1) draws 4 aces have been drawn and let B denote the event that the j'th card is the joker.

     <br />
P(A\cap B)=P(A)P(B|A)=\sum_{j=5}^{53}\frac{\binom{4}{4}.\b  inom{48}{j-5}}{\binom{53}{j-1}}.\frac{1}{54-j}<br />

     <br />
=\frac{1}{5}<br />
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, oblixps!

    I'd like to see your teacher's solution to this problem.
    I don't think it can be solved during a timed exam.


    Suppose you have a standard 52-card deck of playing cards and 1 joker.
    You spread these cards face down over a table and flip them over, one by one.
    What is the probability that you will flip over all 4 aces before you flip over the joker?
    There are 53! possible permutations of the cards.
    Then I have a long long list of cases . . .


    If the Joker (J) is in one of the first four positions {#1, 2, 3, 4},
    . . it is impossible to find the four Aces first.

    J at #5
    The four Aces are among the first four cards: . 4! ways.
    The other 48 cards have 48! possible permutations.
    Hence, there are: . (4!)(48!) ways for J at #5.

    J at #6.
    The four Aces are among the first five cards
    . . along with one of the other 48 cards: . _{48}C_1 choices.
    These 5 cards can be ordered in 5! ways.
    The other 47 cards can have 47! permutations.
    Hence, there are: . (_{48}C_1)(5!)(47!) ways for J at #6.

    J at #7.
    The four Aces are among the first six cards
    . . along with two of the other 48 cards: . _{48}C_2 choices.
    These 6 cards can be ordered in 6! ways.
    The other 46 cards can have 46! permutations.
    Hence, there are: . (_{48}C_2)(6!)(46!) ways for J at #7.

    . . . \vdots

    J at #51.
    The four Aces among the first 50 cards
    . . along with 46 of the other 48 cards: . _{48}C_{46} choices.
    These 50 cards can be ordered in 50! ways.
    The other 2 cards can have 2! permitations.
    Hence, there are: . (_{48}C_{46})(50!)(2!) ways for J at #51.

    J at #52.
    The four Aces are among the first 51 cards.
    . . along with 47 of the other 48 cards: . _{48}C_{47} choices.
    These 51 cards can be ordered in 51! ways.
    The last card can have only 1! order.
    Hence, there are: . (_{48}C_{47})(51!)(1!) ways for J at #52.

    J at #53.
    The rest of the deck occupies the first 52 positions.
    . . And it has 52! permutations.
    Hence, there are: . 52! ways for J at #53.


    Task: add these numbers and divide by 53! . . .


    I'll wait in the car . . .
    .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by oblixps View Post
    i had this probability question on my test and i had absolutely no idea how to do it. please help me.

    suppose you have a standard 52 card deck of playing cards and 1 joker. you spread these cards face down over a table. now one by one, you flip over cards and you can only flip over each card once. What is the probability that you will flip over all 4 aces before you flip over the 1 joker?
    This is not as hard as you think.

    Ignore all the other cards (they don't matter) and consider only the aces and joker. All 5! sequences of the 5 cards are equally likely. Now, how many of the sequences end in a joker? 4!. So the probability is

    \frac{4!}{5!} = \frac{1}{5}.

    Or, even more simply, one of the 5 cards must be the last turned over. They are all equally likely. So the probability the last card is the joker is 1/5.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Thank you 'awkward'.

    I knew there had to be some simple explanation but couldn't give much thought to it in the morning as there was an early morning class.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hard Probability Question
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: August 3rd 2010, 08:15 AM
  2. Probability Question- could be simple/could be hard??
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 7th 2010, 04:56 AM
  3. I'm stuck with this hard probability question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 11th 2009, 12:35 PM
  4. probability qn(not too hard!!)
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 10th 2008, 07:52 PM
  5. A rather hard probability question.
    Posted in the Statistics Forum
    Replies: 5
    Last Post: September 25th 2007, 11:35 AM

Search Tags


/mathhelpforum @mathhelpforum