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Deck of Cards Probability Problem

  1. Feb 1, 2012 #1
    Not sure if this is the correct place to post this. Technically in my school this math class is above calculus but the problem at hand doesn't have any advanced mathematics. Sorry in advance if it is not correct!

    1. The problem statement, all variables and given/known data
    An ordinary deck of 52 cards is dealth, 13 each, at random among A,B,C, and D. What is the probability that (a) A and B together get two aces; (b) A gets all the face cards; (c) A gets five hearts and B gets the remaining eight hearts?


    2. Relevant equations
    Permutation formula.
    [tex]\frac{n!}{(n-k)!}[/tex]


    3. The attempt at a solution
    So the probability of getting an ace is 4/52 and the probability of getting another ace is 3/51. Thus the probability of getting two aces is 12/2652? This doesn't seem right. Wouldn't the two events of drawing an ace be NOT independent? Meaning you cannot just multiply their probabilities? I do not understand how to do this problem. Also don't I need to take into account the random distribution of cards? It says the cards are dealt randomly.

    Can anyone help me out? :(
     
  2. jcsd
  3. Feb 1, 2012 #2
    Hmm, the wording could be interpreted in multiple ways, in my opinion. I think the statement that the cards are dealt at random is there so you don't have to worry about an orderly deal (such as ccw or cw). We have 4 people just so there is a reason for hands of 13 cards. This is my interpretation at least.

    I think part a could also have many interpretations. It could be that between A and B two aces were drawn, your interpretation. Could be that A has one ace and B has one ace. Could be that A has two aces and B has two aces (though I'd think the problem would specify with the word "both", so this is probably not the case, but just wanted to throw this out there because the problem is not very explicit).

    Don't know, maybe someone else will think the wording is less ambiguous. This is my figurative "two cents," that maybe will help you out.
     
  4. Feb 1, 2012 #3
    Ok, let's go over part (a) and see if you can't tackle parts (b) and (c) on your own.

    My basic advice for these combinatorial probability problems is to try and eliminate the time dimension. That is to say, try and make the problem about selecting subsets from some probability space.

    Also, the core of these problems almost always comes down to one of four formulas:
    1. Order doesn't matter and no replacement. (binomial coefficient)
    2. Order doesn't matter with replacement.
    3. Order matters and no replacement.
    4. Order matters with replacement.

    So lets take a look at part (a). We are considering A U B, thus we are looking at 26 cards chosen from 52, no replacement, order doesn't matter. More concisely we are considering half the deck.

    To get some intuition let's first simplify the problem a bit. What if there were just two aces, one black and one red? Well then since A U B and C U D partition the deck into two sets of equal size, and since the cards are dealt randomly, we have our probability space consisting of four disjoint unweighted possibilities:
    1. A U B gets both aces.
    2. A U B gets the red ace but not the black ace.
    3. A U B gets the black ace but not the red ace.
    4. A U B gets neither ace.

    Now to calculate the chance that A U B gets one of the two aces we simply take the number of possibilities in which this occurs and divide by the total number of possibilities, in this case that is 2/4 = 1/2.

    Another way to put this is that since there are two aces, the number of ways for A U B to get exactly one ace is 2 choose 1 = [itex]\binom{2}{1} = \frac{2!}{1!(2-1)!} = 2[/itex]. And the other probabilities will correspond to the associated binomial coefficient as well. For instance the possibility that A U B gets both aces is 2 choose 2 = [itex]\binom{2}{2} = \frac{2!}{2!(2-2)!} = 1[/itex], divided by the total number of possibilities gives us 1/4.

    Now extrapolating back to the original problem, we have four aces and must choose two of them, and then divide this number by the total number of possibilities, given by: [itex]\sum_{k=0}^4\binom{4}{k}[/itex].
     
    Last edited: Feb 1, 2012
  5. Feb 1, 2012 #4

    jambaugh

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    The way I would solve this problem is to count how many total ways you can deal the cards so that your condition is met divided by the total number of ways to deal the cards.

    To count each it is helpful to think about the simplest but most specific way to go about this. For example to count the ways to deal four 13 card hands:
    Shuffle the deck (52! ways to do this.)
    Now take the top 13 and give to player 1, the next 13 and give to player 2, etc. There's no choice here so nothing to count.
    But each player can shuffle his hand and you don't change the hands so take the original 52! and divide it by 13! 13! 13! 13!. That's the number of ways to deal a hand.

    Now to figure something like the probability that A gets all the hearts, imagine actually doing it and count the ways. Pull out the 13 hearts, and hand them to A. (only 1 way to do this). Then shuffle the remaining 39 cards (39! ways) and divide by how the remaining 3 players can reshuffle their hands. The probability of A getting 13 hearts is then:
    ( 39! / 13! 13! 13!) / (52! / 13! 13! 13! 13! ) = 39! 13! / 52!.

    BTW The permutation formula works this way too. For example to figure how many sequences of length 5 you can get out of a set of say 52 cards you shuffle the cards, and divide by the ways to shuffle the remaining 52 - 5 cards once you've laid out the 5.
    52!/(52-5)! or in general cases n!/(n-k)!. If order doesn't matter you also divide by k!, reshuffling the ones chosen.
     
  6. Feb 1, 2012 #5

    Ray Vickson

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    Have you studied the hypergeometric distribution yet? If so, these problem become straightforward; if not, you need to almost develop the hypergeometric distribution on your own, in the context of this special case. Without telling you the answers, I will hint at how to solve these questions using the hypergeometric.

    First, what is the hypergeometric? Given a population of N things, N1 of type 1 and N2 of type 2 (N1 + N2 = N), say we draw a sample of size n without replacement. The hypergeometric probability h(N1,N2,n;k) is the probability that our sample of size n contains k of type 1.

    (a) In the first question consider type 1 = 'ace' and type 2 = 'non-ace'. Now we want the probability of having k = 2 type 1 in n = 26 drawings from a population having N1 = 4 and N2 = 48.
    (b) The second question is similar; can you see how to identify N1, N2, n and k?

    (c) Think of conditional probabilities: A gets 5 hearts, which leaves 39 cards and 8 hearts for dealing to B and C.

    RGV
     
  7. Feb 1, 2012 #6
    Sorry guys but Poopsilon has an answer that is most related to the material I am learning in the class. Thus I can follow it better :p


    Let me see if I can apply that logic for the case of 4 aces as in the problem.
    So the deck is once again partitioned into two subsets and there are now more than 4 possibilities for A U B.

    So the total number of possibilities for 4 aces is..
    [itex]\sum_{k=0}^4\binom{4}{k}=15[/itex]
    So there are 15 cases. The case where A U B has 2 aces is given by [itex]\binom{4}{2}=6[/itex] Thus the probability that A U B together have 2 aces is 6/15 = 0.4

    Would this be correct? I am going to tackle the other parts of the question now! Thanks so much! Ill report back if I run into any problems. :]
     
  8. Feb 2, 2012 #7
    [itex]\sum_{k=0}^4\binom{4}{k} = 16[/itex], so your result for 2 aces is 6/16=3/8.

    Hopefully you can tackle the other two parts with the help of the explanation I gave, although I fear I may have been over-confident in how well my strategy will generalize to parts (b) and (c) of your problem. I think possibly the other two posters' strategies may generalize better, although I haven't given either a thorough read through. Thus be sure to speak up if you need more help.

    For part (b) I'd again try to eliminate the time dimension by first looking at all possible 13 card combinations for A, and then look at how many of those combinations contain all the face cards.

    Thus to look at all possible combinations which contain all the face cards I'd look at how many cards are left over after I've assured all face cards are in A, in this case the number of cards left over makes it fairly trivial, but if it wasn't then I might strategically introduce the time dimension in the following way.

    Say you want to know the number of possible three card combinations which can be dealt from your standard 52 card deck. Well one way you could do this is to say ok there are 52 possible ways to choose the first card, and after that there are 51 ways to choose the next, followed by 50 for the third. Thus you have a 52 x 51 x 50, but now you have over counted (the bane of the combinatorialist) because you've assumed order matters when it doesn't. Thus you look at how many ways there are to rearrange three distinct objects in a row, and discover it's 6 (possibly by simply listing then all out), and so your answer is (52 x 51 x 50)/3!. Now see if you can use this method above.

    How did I know it was 3! ? After all there are other sequences which give 6 for n=3. So I quickly think about just arranging two objects in a row and it's instantly obvious there are 2 ways and now I'm quite sure possible arrangements is modeled by n! and not something like 2n. It's a good idea to learn to instantly recognize the beginning values of sequences like [itex] n!, 2^n [/itex], pascals triangle, etc., that way if your faced with a problem on the test and you don't remember the correct way to model some combinatorial or probabilistic phenomenon, you can look at a few toy examples with small values and see what sequence or pattern is occurring and then match it to what ever function generates that sequence/pattern.

    In part (a) the deck was split in half and so the problem could almost be looked at as the flip of an unbiased coin. That simplification doesn't work in (b).
     
    Last edited: Feb 2, 2012
  9. Feb 2, 2012 #8

    Ray Vickson

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    The hypergeometric distribution gives the probability that A or B get two aces exactly as p = 325/833. Here is the reason. First, convince yourself that of the 26 cards that go to A or B we get the same probability of having two aces in any two positions (out of the 26). So, if A = 'ace' and N = 'non-ace', the two possibilities AANN....NN (2 As, 24 Ns) and NNNANNNNANN ...N have the same probabilities, as do all the other arrangements. The number of such arrangements is the binomial coefficient C(26,2) = number of ways of choosing 2 positions in 26. Now, what is the probability of any one of these arrangements? Look at the first one AANNN...N. Its probability is p1 = (4/52)(3/51)(48/50)(47/49)(46/48)...(25/27), because the prob that the first card is an ace is 4/52, then the probability that the second is an ace is 3/51, then the prob. that the third is a non-ace is 48/50, etc.

    RGV
     
    Last edited: Feb 2, 2012
  10. Feb 2, 2012 #9
    I have not studied the hypergeometric distribution, thus I do not wish to delve into it since it cannot be needed to solve this problem.

    I did part B yesterday and didn't get time to post until now. This is what I did, does it make sense??

    So all the ways to receive a jack OR a queen OR a king is..
    [itex]\Sigma_{k=0}^4\binom{4}{k}[/itex]

    Since receiving a jack, queen, or kind are mutually exclusive the sum of their sets is equal to the union of their sets. So the total number of ways to be dealt a J, Q or K is..

    [itex]3\Sigma_{k=0}^4\binom{4}{k}=48[/itex]

    There is only one way to be dealt all the face cards, so the probability for this to happen is..
    1/48=0.0208

    My concern is that this probability is "to good." This would mean that if I play around 50 games I would be dealt all the face cards. Does this seem correct to you guys? :\

    EDIT: No wait, they wouldn't be mutually exclusive. So this has to be wrong :\
     
    Last edited: Feb 2, 2012
  11. Feb 2, 2012 #10
    So I tried using this approach next.
    There are [itex]\binom{52}{13}[/itex] ways of getting dealt 13 cards. From a hand of 13, 12 of them are the face cards. Thus there are 12! ways of being dealt those 12 cards. There is one extra card with [itex]\binom{40}{1}[/itex] ways of being dealt. Thus the probability is..
    [itex]\frac{40(12!)}{\binom{52}{13}}=0.03[/itex]

    *sigh* what about NOW? :p
     
  12. Feb 2, 2012 #11
    Your problem is the 12!. Think about if A was dealt just 12 cards and they had to be the entire set of face cards, how many ways are there to do this? Is it really 12!, or is it much much less then that?

    Consider two distinct objects, how many ways are there to arrange these two objects in a row? Now consider 3 objects, how many ways are there to arrange these three objects in a row? These are the toy examples I was talking about above, what pattern is emerging? What is the function which describes this pattern? Now conclude what combinatorial phenomenon this function is measuring. Is this phenomenon relevant to our 12 face cards in A?
     
  13. Feb 2, 2012 #12
    Well for 2 objects you could have
    AB or BA. So there are 2 ways
    For ABC you can have
    ABC, ACB, BAC, BCA, CAB, CBA So there are six, this seems to follow n! unless this logic is incorrect. (Which it probably is :p )
     
  14. Feb 3, 2012 #13
    That's good, now follow it to its conclusion: is this relevant to the set of cards in A? For example, are the Texas Hold'em poker hands KJ of spades and JK of spades distinct, or are they the same hand?
     
  15. Feb 3, 2012 #14
    I would say that they are the same hand. So in this case there are 4 of each card. So should I divide by the number of ways to arrange 4 cards? Which would be 12!/4! ???

    I'm getting closer I think! Ha
     
  16. Feb 3, 2012 #15

    Ray Vickson

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    Hints:
    (a) Regard A and B together as a super-player who gets 26 cards. You want the probability that this player gets 2 aces. How many hands are there altogether (i.e., ways of choosing 26 cards from 52)? How many of these hands have 24 non-aces and two aces? Well, how many ways are there to choose 24 cards from 48 (the non-aces). How many ways are there are to choose 2 aces?
    (b) There are 12 face cards and 40 non-face cards. What is the total number of possible hands of 13 cards? In how many ways can you choose all the face cards? In how many ways can you choose one non-face card?
    (c) What is P{A gets 5 hearts}? Well, how many possible hands can A get? How many ways are there of choosing 8 non-hearts? How many ways are there of choosing 5 hearts? Once you have dealt A his 5 hearts and 8 non-hearts, what is the situation faced by B?

    Note: I did not mention the word "hypergeometric" at all in the above, but just posed simple counting issues you have already taken. However, when you answer these questions you are, basically, developing some special cases of the hypergeometric!

    RGV
     
  17. Feb 3, 2012 #16
    Go back to my example of A being dealt only 12 cards where they must be all the possible face cards. Is there really more than one way to do this?

    You've eliminated the time dimension, there is no order in which these cards are dealt to A, they simply materialize in A all at once. Think about the fallacy which led you to believe dividing by 4! was the right move, rather than getting rid of 12! all together. By dividing by 4! you are saying the king of spades and the king of hearts have some relationship that the king of spades and the jack of hearts does not. In a poker hand they certainly might, but here all we care about are face cards, two face cards are either distinct or they are the same, regardless of their rank.

    Thus the correct thing to do would be to divide by 12!, to account for all possible orderings. But that would certainly be a roundabout way to do things. Better to realize that in this problem we are dealing with situation 1. Order doesn't matter and no replacement. And hence not include 12! to begin with.
     
    Last edited: Feb 3, 2012
  18. Feb 3, 2012 #17
    I see now! That makes sense, I guess I got hung on on WHICH face cards he would be getting in certain deals of his hand but it really doesn't matter since there is only one way to get all the face cards.

    I will do part C now. If I feel like what I am doing is way wrong then ill post here once again.

    Thanks for all your help everyone!
    EDIT:
    Ray Vickson: I am using your hint for C and it is helping tremendously!
     
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