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52 cards split four ways

  1. Sep 8, 2015 #1
    1. The problem statement, all variables and given/known data
    If you have a 52card deck and split it evenly to 4 people (a,b,c,d) then what is the probability that persons a and b have 7 spades and person c has 3 spades.

    2. Relevant equations
    P(A) = |A|/|S|

    3. The attempt at a solution
    I divided up the decks 4 ways. So Persons a and b should have 26 cards total. There's no requirement that one person have more spades than the other.
    So I collected them together.

    The chance that they draw 7spades out of the top of the deck should be
    13/52 * 12/51 * 11/50.... *7/46

    Then the chance that they draw non spades should be:
    39/45 * 38/44... * 20/26

    For there are still 6 spades in the deck and 19 carded must be drawn.

    I would do this again for person C, he would be allowed to draw:

    6/25 * 5/24 * 4/23 *19/22 * 18/21... * 11/14.

    I may have messed up that last fraction.

    Does my logic make sense..?
     
  2. jcsd
  3. Sep 8, 2015 #2
    ***edit, persons A and B should end at 21/27

    And

    Persons C should be:
    6/26...4/23 * 20/23...* 11/14

    I went one too far
     
  4. Sep 8, 2015 #3

    Ray Vickson

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    How can a and b both have 7 spades, when there are only 13 spades altogether? Or, did you mean that a and b together have 7 spades?
     
  5. Sep 8, 2015 #4
    Within both of their combined 'decks' are 7 spades. So A could have 0-7 so long as B had the remaining count.
     
  6. Sep 8, 2015 #5

    andrewkirk

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    The above analysis implicitly assumes that the a+b 'team' gets all its 7 spades in its first seven cards. That will dramatically understate the probability of the team getting seven spades.

    These problems are best analysed by counting the number of different possible combinations (D) and the number (N) of those combinations that satisfy the criteria (7 spades for a+b and 3 spades for c). The probability is then N/D.

    Imagine that the cards are dealt first to the a+b team (we don't need to distinguish between a and b in this, so we treat it as a single player who gets double the number of cards), next to player c and last to player d. We can ignore the deals to player d because once c has been dealt, d's hand is fully determined.

    Do you know how to calculate the number of different combined hands the a+b team can get? Call that number Dab.
    Next calculate the number of different ways that 13 cards can be chosen for c out of the remaining 26. Call that Dc.
    Then the number of possible, equally-likely, arrangements we are interested in is D=Dab * Dc.

    The process to calculate N is similar but with extra steps. For a+b, number the deals (a 'deal' is giving a card to the player) they receive from 1 to 26. First you need to calculate the number of different ways Nwhenclubs1 we can select 7 different numbers from 1 to 26 - those being the deals in which the team gets its clubs. Then you calculate the number Nwhichclubs1 of different combinations of 7 clubs that can be dealt in those seven deals. Then you calculate the number Nwhichnonclubs1 number of different combinations of nonclubs that can be dealt to a+b in their remaining six deals. Then the number of possible different hands of a+b that meet the criteria is Nab = Nwhenclubs1 * Nwhichclubs1 * Nwhichnonclubs1.

    Then you do the same thing for player c to get Nc the number of different possible hands with three clubs that c can get, given that the hand for a+b has already been dealt and has seven clubs. Then N=Nab * Nc.
     
  7. Sep 8, 2015 #6
    This is the portion I struggle with... Seeing which becomes what. At best I start throwing factorials without any logic.
    For finding D (my sample set) I said
    Dab = 26!/(7!19!)
    Dc = 13!/(4!9!)
    I believe this to be the way to count these two sets with respect to their given information, but my professor put a lot of doubt in this explanation when I discussed it with him. He keep repeating himself despite my questions, and so I do not know if that 7! Adequately distinguishes between spades or not.

    (Btw, we are dealing with spades, not clubs)

    As for the next section, the best I can do is something similar

    Nwhenspades1 = 26!/(7!19!)
    For they can have 26! Combinations, and we wish 7 spades and 19 not spades.

    Nwhichspades1 = 7!
    Nwhichnospades1 = 19!


    This is starting to look very incorrect as I'm canceling out my own denominators here... But I'll stay the course.

    NCwhenspades1 = 13!/(3!10!)
    This is the combinations of 13 where 3 are spades and 10 are not.

    NCwhichspades1 = 3!
    NCwhichnospades1 = 10!

    So D= 26!13!/(7!19!4!9!) N= 26!13!

    D/N = 1/(7!19!4!9!)

    I now can confirm I really do not know what I am doing.
     
  8. Sep 8, 2015 #7
    I got the fraction wrong, but flipping that over gives me a probability greater than 1 so i still know I am wrong. I never would have thought that learning to count would be this difficult...
     
  9. Sep 8, 2015 #8

    andrewkirk

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    This is where it starts to go off track.

    Dab is the number of ways that 26 cards can be chosen from 52, which is a 'Combination', written ##C^{52}_{26}## or sundry variations on that, and spoken '52 choose 26'. Usually in this type of course they teach you the formulas for Combinations and Permutations (permutations are where the order of choosing matters, Combinations are where it doesn't). Have they taught you the general formula for combinations ##C^n_r##? If they haven't you can work it out by first principles, by first working out the number of Permutations ##P^{52}_{26}##, which is done by observing that there are 52 possibilities for the first card, 51 for the second, and so on down to the 26th card. You then need to divide that permutation number by the number of different ways the 26 chosen cards can be ordered, which is 26!.

    You then do the same trick for Dc, but this time choosing 13 cards from the remaining 26, ie ##C^{26}_{13}##.
     
  10. Sep 8, 2015 #9

    andrewkirk

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    The answer is correct, although I don't understand your reason. The number is just 26 Choose 7.
    No. This step is to choose seven spades from the available 13, ie 13 choose 7.
    No. This is another Combination. How many cards are you choosing, out of how many?
    The calc is correct, but we need to call this NCwhenspades2, as 1 as has already been calculated above, and is different. My plan was to use 1 to refer to the a-b team and 2 to refer to c.
     
  11. Sep 9, 2015 #10
    Here's what I gathered from your help. I'm sorry, I still have trouble visualizing these tactics. I understood that when counting (and I gather that order does not matter because we are only using their suit and not their identity or their location in the deck) then we must use combinations. What confuses me most I suppose is the scaling of the sample space as we define the these combinations. It made a little more sense when I did 52C26. That split the suit in half seeking out all combinations of 26 cards where 26 of them were the ones we wanted, I wrote comments justifying what I did, but i'm not the least bit confident in that final answer. It doesn't calculate well and I think I still lost it at the last second when I did Nc.

    To me, I had 13 possible cards to choose from (otherwise my approach would not be consistent with what i did for Nc) and then I sought out the possible remaining spades and the possible remaining nonspades from their respective remaining wholes. This is why I used the numbers I did.

    Then I just attempted N/D using N * 1/D
     

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  12. Sep 9, 2015 #11

    andrewkirk

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    I'm afraid I can't read that attachment.
     
  13. Sep 9, 2015 #12
    I'm sorry, I will type it in the morning. We went over it in class and I believe it is clearer, but counting is surprisingly difficult to wrap my head around. I'm trying to comprehend my professors idea of seeking first out the tuple and then identify ways to partition it. I think this is where you were headed as well.
     
  14. Sep 10, 2015 #13
    I would do it this way. I am also learning probability, so I ask the people who know this stuff for further verification. The best way to do this is by using combinations of a+b with 7 spades and c with 3 spades. If you use the total number of outcomes in the sample space which is 52! then, it is going to be a little bit difficult finding the outcomes of the event.

    The sample space would be basically the whole possible number of combinations of (7+3) that can be made of 52 possible spots. Think about this like if you have 52 boxes in a bag. How many groups of ten can you make?. But hence there are 13 spades, first we have to find the number of groups of 10 spades that can be made of 13 spades. " We don't care about the way how the other spaces are filled".

    Sample space =nCr(13,10)*nCr(52,10)

    For the event that you are trying to find, we are going to do a similar thing that we did to the sample space, but this time we are going to compute the 7 and 3 spades separately. Let's start with the 7 spades for a and b. a and b have in total 26 spots, so if you were to find the whole possible number of combinations of 7 that can be made from 26 you would do nCr(26,7), but again we have 13 spades, so you have to find the total possible groups of 7 spades that can be made of 13 spades. so one part of the probability of the given event would be nCr(13,7)*nCr(26,7), but now we have another part of the problem to complete which is the 3 spades for c. c has 13 spaces so I am not going to repeat the same thing that I did previously. The outcomes in this case would be nCr(13,3), but there are 6 spades left. Why 6? Because we already put 7 in a and b., so you have to find the total possible combinations of 3 spades that can be made of 6 spades, so putting the whole pieces together, we get

    Outcomes of the event=nCr(13,7)*nCr(26,7)*nCr(6,3)*nCr(13,3)

    then P= Outcomes of the event/Sample space

    think that this process may look a little bit different to yours, but I think that this works, because in the sample space and in the event we are just considering what happens with spades and not what happens with the rest of the cards. Sorry for posting multiple times, but it just doesn't give me the option to delete posts. I hope that someone who has more credibility checks if this is right.
     
    Last edited: Sep 10, 2015
  15. Sep 10, 2015 #14
    mmmm why did you put 39 choose 19, you are saying that it is 52-13, but I think that this is wrong because you are making groups of 7 spades not 13, so it would be 45 choose 19 and the sample space is also a little bit weird to me. In this case you are finding the sample space by distributing 52 cards among the 4 players. If you o 52 choose 26 and
     
    Last edited: Sep 10, 2015
  16. Sep 10, 2015 #15
    I think that this process may look a little bit different to yours, but I think that this works, because in the sample space and in the event we are just considering what happens with spades and not what happens with the rest of the cards.
     
  17. Sep 10, 2015 #16

    Ray Vickson

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    A far, far safer approach is to do the problem using conditional probabilities. Let A = {7 spades to a & b}, B = {3 spades to c}. We can join a & b to get a larger group ab that receives 26 cards. We want the probability that group ab gets 7 spades and 19 non-spades. This is a hypergeometric probability:
    [tex] P(A) = h(7|13,39,26) = \frac{C(13,7) C(39,19)}{C(52,26)}, [/tex]
    where ##C(a,b) =## "a choose b".

    Given that event ##A## occurs, that leaves 26 cards, of which 6 are spades and 20 are non-spades. We draw 13 cards, and want the probability of getting 3 spades (and 10 non-spades). This is a hypergeometric probability
    [tex] P(B | A) = h(3|6,20,13) = \frac{C(6,3) C(20,10)}{C(26,13)}. [/tex]
    Altogether, we have
    [tex] \text{Answer} = P(A \cap B) = P(A) \cdot P(B|A) = \frac{C(13,7) C(39,19) C(6,3) C(20,10)}{C(52,26) C(26,13)}. [/tex]

    The hypergeometric distribution just "automates" the processes that you (and other respondents) have done herein; rather than having to re-think each problem, I personally prefer to do it once-and-for-all and apply the results over again to new situation (provided, of course, that the scenarios fit the underlying "hypergeomtric" assumptions). Briefly: the hypergeometric distribution refers to the following: (i) we have ##N_1## objects of tkype I and ##N_2## of type II; (ii) we withdraw ##n## objects randomly, without replacement, from the population of ##N = N_1 + N_2## objects. The probability of the sample having ##k## type-I objects in it is
    [tex] P\{ k \; \text{type I} \} = h(k|N_1,N_2,n) = \frac{C(N_1,k) C(N_2, n-k)}{C(N,n)} [/tex]

    See, eg., https://en.wikipedia.org/wiki/Hypergeometric_distribution or
    http://mathworld.wolfram.com/HypergeometricDistribution.html or
    http://www.math.utah.edu/~lzhang/teaching/3070summer2008/DailyUpdates/jun23/sec3_5.pdf
    for more material on the hypergeometric distribution.
     
  18. Sep 10, 2015 #17
    Hi Ray , thank you for your help. Why mine doesn't work?. I already computed the prob and yeah, it doesnt make any sense.
     
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