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Cards Probability Question

  1. Mar 7, 2006 #1
    From a well shuffled deck of 52 playing cards(no jokers) ,a person draws out cards one by one until he picks all 13 hearts in the deck.
    On which turn is he most likely to pick the last heart?
    Don't give me 52 as the answer bcos the question also includes the probability that he reaches that turn to pick the 13th heart.
    Last edited: Mar 7, 2006
  2. jcsd
  3. Mar 7, 2006 #2
    Can't any gurus answer this question
  4. Mar 7, 2006 #3
    I would venture that the final heart is most likely to be drawn in either 49th, 50th, 51st, or 52nd position (each occurring with equal probability).

    In an infinite serious of trials, probability will ensure that after 12 hearts are drawn, 12 of each of the remaining suites (clubs, spades, diamonds) are drawn as well. This gives us 48 cards out and 4 remaining. It is not possible to determine whether or not the case heart will necessarily be drawn in the next card but at this point there is a 25% chance that it will be drawn in 49th, 25% for 50th, 25% for 51st, and 25% for 52nd.

    Without hard math, common sense would suggest that the final heart will most often be found in this 4-card grouping (remaining 4) when compared with similiar groupings occurring before 48 cards are selected.

    Any thoughts?
  5. Mar 8, 2006 #4


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    Probability doesn't work like this. You can't guarantee in any way that the last 4 cards will be from 4 different suits. Your conclusion of 25% for the 52nd, 51st, 50th, and 49th can't possibly be correct either, it would mean that all the other possibilities have zero probability.

    arunbg, what have you tried? If you show some work, you are much more likely to get responses. By the way what do you mean by "Don't give me 52 as the answer bcos the question also includes the probability that he reaches that turn to pick the 13th heart."? Specifically the bold part, I don't see what you're getting at.
    Last edited: Mar 8, 2006
  6. Mar 8, 2006 #5
    I do think that HokieBalla34 is on to something. Doesn't it seem reasonable to assume that over many draws the probability is that the hearts come off in a regular pattern and are evenly distributed?

    If we wind up with one heart in the last four cards we have: 1/4 to get it the first time, but 3/4 of the time we miss giving: 3/4x1/3=1/4 on the second draw, and so forth: 1/4 + 3/4(1/3)+1/2(1/2) + 1/4(1)= 1 or the total probability in the four draws.

    In fact, as a general case, if we have one heart among n cards, we have 1/n chance of picking it up immediately, and (n-1)/n chances of not, but we might get it on the second try: [tex]\frac{n-1}{n}*\frac{1}{n-1}=1/n[/tex] and so on down the line, [tex]\frac{n-2}{n}*\frac{1}{n-2}=1/n [/tex] etc... So that the probability is the same, 1/n, for each of the n possible cases.

    Or as Laplace would see it, probability consists of sorting things into equally probable cases. For cards, that could consist of all permutations. If we have 1 heart in n cards, then there would be n permutations which consist of putting the one heart in one of the n choices.
    Last edited: Mar 8, 2006
  7. Mar 8, 2006 #6
    What I mean is that the person always stops when he picks the 13th heart.For eg, if he reaches the 52nd turn ,the fact that he picks up the 13th heart in that turn is sure.But you have to also take into account the probability that he reaches the 52nd turn to pick the card and so the probability changes.That is he might have obtained all 13 earlier and stopped.
    As for the other replies regarding an infinite no of trials, I am compelled to think that there might be a lead there but can someone give me concrete mathematical backing for this?
    Thanks guys
  8. Mar 8, 2006 #7


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    I absolutely agree with this given the "if" at the start, but you would need to work out the probability that after 48 cards you had drawn 12 hearts. This you can do though, but this direction might get you thinking in an overly complicated way (specifically I mean if you've found this probability you should focus on the probability that the 49th is your last heart. If you wanted the probability the 50th is your last heart, find the prob that you have 12 hearts picked after 49 draws, etc.).

    Let's call P(m) the probability your mth draw is the 13th heart, so P(1)=P(2)=...=P(12)=0. We can work P(m) out directly in the other cases. In order to draw the last heart at pick m, at m-1 you must have picked 12 hearts. Find the probability of picking 12 hearts with m-1 picks (this is a little simpler as we don't care where in these m-1 picks these 12 happen). Given 12 hearts in hand after m-1 picks, you can find the probability the next pick is a heart.

    there's absolutely no reason to consider an "infinite" number of trials or some kind of limit as there are a finite number of outcomes.
    Last edited: Mar 8, 2006
  9. Mar 8, 2006 #8


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    Allow me to posit that the opposite question might be easier to study.
  10. Mar 8, 2006 #9
    What u have said was exactly what I was doing initially.
    I was thinking about maximising the obtained probability function for nth turn to get the most probable turn.
    what do u think.
  11. Mar 8, 2006 #10


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    I think if you find what I called P(m) it's not hard to find where it's max occurs, you could always just calculate P(m) for m=13, 14, ... 52 if you can't see any other way of maximaizing P(m). What did you get for P(m)?
  12. Mar 8, 2006 #11
    There is a way to get an exact answer without regard for probability. Just construct the rectangular row. Say in the case of n=6, looking for two hearts, we proceed:

    (An x represents a heart and a blank representing a space, and where I had to put some s's for space, since it condensed it otherwise.)

    In this particular case we will have to continue for 6C2 = 15 rows within our 6 columns. So, therefore you can do 15 lines, say using Excel, and determine the total matrix for all outcomes based on an equal probability for each case. Then just add up the cases.
    Last edited: Mar 8, 2006
  13. Mar 8, 2006 #12


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    Enumerating all possible outcomes is not reccomended. You would have to look at all possible ways to distrubute 13 hearts in a deck of 52, i.e. 52C13=635,013,559,600, which will take up alot of space.
  14. Mar 8, 2006 #13
    Once you get into these things, there is an easy way to see the diagram. (It does help to do some drawing.) We are going to have 52 slots available for 13 hearts. We put one heart in the left most, or first, slot. Then we distribute 51C12 rows of hearts all over the remaining necessary spaces.

    To get the next case, we leave the first slot blank, put a heart in the second place, and distribute 50C12 hearts over the remaining cases. We continue on down until we have 12 hearts to put in 12 slots, or 12C12=1. Now we turn the situation around and call the first slot the last.

    Now lets look at the actual situation, which seems to work out: [tex]\frac{\sum xC12, x=12...51}{52C13} =1[/tex]

    This gives the final ratio in the 52 slot as the highest at: [tex] \frac{51C12}{52C13}=1/4.[/tex]
    Last edited: Mar 8, 2006
  15. Mar 8, 2006 #14
    I didn't mean that the probability, from the start, was 25% for it to be selected in positions 49-52. I was (perhaps incorrectly) considering the probability after 48 cards have been drawn (believing that 1 of each suit would likely remain). You are just as likely to have drawn 'n' amount of hearts as 'n' amount of the other suits in 4*n draws. Therefore after 48 cards are drawn, you are most likely to have 12 hearts, 12 clubs, 12 spades, and 12 diamonds, no?

    I do not believe that you can dissect which position the 13th heart is the most likely to occur in. Instead, I believe that it would only be possible to determine the GROUP of cards it most likely appears in--and my conclusion is that it is most likely to occur in the group that consists of positions 49,50,51,52.
  16. Mar 8, 2006 #15


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    I'll make my hint more explicit!

    It's much easier (but still not trivial) to deal from the bottom of the deck, and ask when you find the first heart.

    Why not? Each position has a corresponding probability. There are finitely many of them. Just compute them all and take the one that's largest.

    It's a perfectly well defined question, with a perfectly straightforward (if tedious) solution.

    The only interesting part is to find a good shortcut to speed things up, or a good bit of theory to promote understanding.

    And yes, the 13th heart is more likely to be at the bottom of the deck than at any other particular position.

    (Of course, it's more likely not to be at the bottom of the deck than it is to be at the very bottom)
    Last edited: Mar 8, 2006
  17. Mar 8, 2006 #16
    Hurkyl: It's much easier (but still not trivial) to deal from the bottom of the deck, and ask when you find the first heart.

    What we could have done is constructed a computerized robot that would endlessly shuffle and deal. Then we could search its memory banks and determine the hearts in the 52nd place. BUT, we could have had the robot deal from right to left and then considered the first card as the last.

    BETTER YET, we could ditch the robot, and not check anything. Just ask yourself what is the heart probability for the first card to be dealt?
    Last edited: Mar 8, 2006
  18. Mar 9, 2006 #17


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    This has probability (13C12)^4/(52C48), the number of ways to select 12 cards from each suit divided by the total number of ways to select 48 cards. This is about .1055. In other words you are far more likely to have an uneven distribution of suits.

    nCr here means the usual number of ways to select r objects from n.
  19. Mar 9, 2006 #18
    I think this is the right way to go.
    So for P(n) we can get a series like

    P(n)= 39/52 * 38/51 * 37/50 ..... (n-1 terms) * 13/(52-n+1)
    where n <= 40
    Here P(n) refers to picking up the first heart when the deck is dealt from the bottom.
    Note that the last term becomes 1 when when n=40 ,ie if he reaches the
    40 th turn he is sure to pick the first heart. But the overall probability is the least.
    I think the maximum comes when n=1 where P(n) = 1/4 but I am not sure.
    I'll try working it out in C++.

    Thanks for the feedback guys
  20. Mar 10, 2006 #19
    To be more exact on this, I have already shown the number of cases divided by all cases ending at any place from 13th to 52nd. For example if the last heart is at the 51st place, the ratio is

    50C12/52C13 =19%.

    Let us suppose we put the final heart at the 13th place. Then how many ways can we fill 12 spots with 12 hearts? The answer in combinations is only 1. Now if the last heart was placed at the 14th spot, then we have 12 hearts to place in 13 places or this can be done in 13 ways.

    Looking at it this way it is completely obvious that the more viable "space" that we have to maneuver the hearts in, the greater is the ratio of that case to all possibilities, referred to as the probability of that outcome. (Once we place a heart at position X then only the "space' before this placement is viable, we can not place anything after this point by definition of what we are supposedly doing.) Remember that all these ratios have to add up to covering all posibilities, and so sum to 1.

    So that the question of what we are actually trying to do, or WHAT IS THE SAMPLE SPACE is extremely important. We have to define the characteristics of this sample space. That in itself, can be become confusing, and is the reason why it is best to try and clearly see how this sample space, i.e. all the combinations, can be ordered.
    Last edited: Mar 10, 2006
  21. Mar 11, 2006 #20
    I get your point sir.

    Thanks a lot
  22. Mar 19, 2006 #21
    Well, just for academic interest this is the result of my program that calculated the probability for every turn ,starting from the 52nd turn.


    card 52 = 0.25
    card 51 = 0.191176
    card 50 = 0.145294
    card 49 = 0.109712
    card 48 = 0.082284
    card 47 = 0.061275
    card 46 = 0.04529
    card 45 = 0.033213
    card 44 = 0.024155
    card 43 = 0.017414
    card 42 = 0.012439
    card 41 = 0.008798
    card 40 = 0.006159
    card 39 = 0.004264
    card 38 = 0.002917
    card 37 = 0.001971
    card 36 = 0.001314
    card 35 = 0.000864
    card 34 = 0.000559
    card 33 = 0.000356
    card 32 = 0.000222
    card 31 = 0.000136
    card 30 = 8.172414e-05
    card 29 = 4.790725e-05
    card 28 = 2.737557e-05
    card 27 = 1.520865e-05
    card 26 = 8.189274e-06
    card 25 = 4.258422e-06
    card 24 = 2.129211e-06
    card 23 = 1.018318e-06
    card 22 = 4.62872e-07
    card 21 = 1.983737e-07
    card 20 = 7.934949e-08
    card 19 = 2.923402e-08
    card 18 = 9.744674e-09
    card 17 = 2.866081e-09
    card 16 = 7.165201e-10
    card 15 = 1.43304e-10
    card 14 = 2.0472e-11
    card 13 = 1.57477e-12

    Thanks you guys
  23. Apr 2, 2006 #22
    Imagine a whole pack of 52 in a random order in your head. What is the location of the last heart? Now reverse that deck. What is the location of the first heart?

    13/52 times the first heart will be the first card. 13/52 times it'll be the last card too.
  24. Apr 4, 2006 #23
    pokerponcho: 13/52 times the first heart will be the first card. 13/52 times it'll be the last card too.

    This is not what the problem was about, i.e., what is the probability that the last heart is at position n? Well, in that ordering there can not be the last heart in the first 12 places.

    We might as well say,going along with you, that at any position the probability of a heart is 1/4.
  25. Apr 9, 2006 #24
    if you consider it the other way and find the FIRST heart, it cannot be in the LAST 12 places. the problems (as Hurkyl suggested) are identical, but it is easier to consider where the first is.
  26. Apr 12, 2006 #25
    If we have not uncovered any hearts, the probability is 1/4 for a heart in any position. However, if the first heart is in position 1, then it's 12/51 that a heart is found at another position. We might look at this as being asked to choose a card from somewhere in the deck.
    Last edited: Apr 12, 2006
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