Probability of a drawn card be a diamond given post conditions

  1. Setting is like this: I have 52cards of 4 shapes. that's 13 each.

    I take one card out and put it in a hat. now with 51card I draw 3 consecutive cards.

    It happens to be that I picked 3 diamonds.

    Now I want to find the probability of the card in the hat be a diamond.

    My friend said it should be 10/49 since 3 diamonds are out, but I disagree.

    I think those 3 diamonds with only one draw doesn't give enough information to favor other shapes than diamond at all, since the one card draw happened before any additional information, and 3 consecutive draws cannot affect previous draw in any way making the first draw independent of the second draw.
     
  2. jcsd
  3. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    Try considering another scenario:


    Setting is like this: I have 52cards of 4 shapes. that's 13 each.

    I take one card out and put it in a hat. now with 51card I draw 13 consecutive cards.

    It happens to be that I picked 13 diamonds.

    Now I want to find the probability of the card in the hat be a diamond.​
     
  4. I understand what you're thinking. You're partly right. It is absolutely correct that, before you drew the additional cards, the probability of a diamond in the hat was 1/4. And you're right that the three subsequent draws can't affect the previous draw in any way. But you're wrong to think that that can't affect the probability characteristics of the first draw.

    Probability is a function of information. It's about what you know and don't know. When you draw those three cards, your information changes, and therefore the probability changes.

    Most probability paradoxes are about this. People have a tendency to think that probability exists independent of persons and information. They will naturally think that the probability that the card in the hat is a diamond is a characteristic of that physical setup, the card and the hat, and that if you don't change the card and the hat, the probability can't change. But that's wrong. The probability is about you and what you know. So there is no paradox in the idea that the probability changes when your information changes.

    So, suppose you drew three diamonds, but you didn't show them to your friend, who is also there with you? What's the probability now? You have new information, so your probability is 10/49. But your friend has learned nothing new, so her probability is still 1/4. Which is right?

    You're both right. Since probability depends on information, the same event came have different probabilities for different people with different information. The probability of a diamond in the hat is different for you and your friend. One again, the probability is not just a characteristic of the event.
     
  5. This is the probability for picking 3 cards of diamonds if the one in the hat is Diamond or not:
    [tex]\frac{3*\frac{13}{51}*\frac{12}{50}*\frac{11}{49}+1*\frac{12}{51}*\frac{11}{50}*\frac{10}{49}}{4}[/tex]
     
    Last edited: Jul 16, 2011
  6. Sorry, i didnt read your question correctly. I would say that probability is 1/4
     
  7. 0% after you see the 13 diamonds, 1/4 before you see any diamonds.
     
  8. i understand that information has changed. Nevertheless I'm not convinced that the information affects the original draw, since the population of the first draw doesn't necessarily change before I somehow pick 13 diamonds from 51 cards, which I didn't. I'm aware that if any new information that forces the population of the original event then it influences the probability like having 2 yellow balls and 2 red balls, and trying to find the probability that first ball is yellow WHEN second ball is yellow, which is not 1/2 but 1/3
     
  9. Could you explain to me why this is the case?
     
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    Last edited: Jul 16, 2011
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  13. first single card draw is consisted of equally probable events, isn't it?



    This is completely true, but we're not trying to find number of events concerning first draw as diamond amongst total number of events where 3 consecutive draws after single draw are diamonds. 3 diamonds is not our 'condition'. It is just a result we stumbled upon randomly.

    I'm not saying two random events are equally probable. I'm saying that 3 consecutive card draws turning out to be all diamonds is not a preset condition. If that's the case, answer IS 10/49. It can easily be calculated by finding number of events that has first draw as diamond AND has next 3 draws as diamonds as well which is 13*12C3, and divide this by total number of events that has 3 draws as diamonds but no condition on first draw which is 13*12C3+3*13*13C3.
    However this is NOT what I'm trying to find. Original question never says to find probability that has 3 draws after the first one to be all diamonds. Thus two cases have different number of possible events.
     
  14. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    Er, yes it is your condition. You saw three diamonds, so now all posterior probabilities ought to be conditioned on that fact. You will find that, out of all the experiments you do, less than one quarter of those where you drew 3 diamonds will have a diamond as the first card.

    After drawing 3 diamonds, asking for the unconditioned probability is a purely hypothetical exercise.

    (although it is important to be able to comprehend such hypothetical questions when appropriate)

    Edit: A more blatant example that the posterior probabilities must be conditioned is this:
    I flip over the top card, see that it's a diamond, and put it back. What are the odds the top card is a diamond?​
    Your line of reasoning would say 1/4.
     
    Last edited: Jul 16, 2011
  15. disregardthat

    disregardthat 1,811
    Science Advisor

    Call the event that the card in the hat is diamond A, and drawing 3 consecutive diamonds while one randomly chosen card is in the hat B. We want P(A|B) which is 5/98, or approximately 5.1%. I think.
     
  16. Yes. And after the first draw, the probability of a diamond is 13/52 = 1/4.

    Yes, it is a result that you stumbled on randomly. You seem to think that whenever you use the word "random", that means you can ignore it. But this is not true. Random events still carry information.

    Suppose you were walking around and you saw a $100 bill on the sidewalk. Would you say, "Well, the probability of picking up a $100 bill on the sidewalk is practically zero. And, I was 'not trying to find number of events concerning' being able to pick up a $100 bill from the sidewalk. I 'just stumbled upon it randomly'. So the chance that I can now pick up a $100 bill from the sidewalk is still practically zero. Therefore, I'm not going to bend down and pick it up, because I know that the chance that I can pick up a $100 bill from the sidewalk is not worth the effort."?

    Please, please, PLEASE, I beg you, stop saying, "It's just random". That's meaningless. Everything is "just random", to some degree or other. The idea that you can say "It's just random" and ignore it is WRONG.

    As a matter of fact, that's wrong. The probability of drawing three diamonds after one diamond has been drawn is much less than 10/49.

    No, you're right, the original question doesn't ask you to find the probability of drawing three diamonds after a first draw of a diamond. It asks you to find the probability that the first draw is a diamond, given that the three subsequent draws are diamonds. These are not the same number. But they ARE related (see Bayes' Theorem). It doesn't matter that this was not what you call a "preset condition". You've now drawn three diamonds. Whether you set out to do that or not is totally irrelevant. You now have that information.

    The probability that the first draw was a diamond is multiplied by

    [tex]
    \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was non-diamond}}
    [/tex]

    That's why it's relevant that the two probabilities aren't equal.
     
  17. disregardthat

    disregardthat 1,811
    Science Advisor

    Shouldn't this be

    [tex]
    \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was non-diamond or drawing three diamonds if first was diamond}}
    [/tex]

    ?
     
  18. No -- notice I wrote that this is what the prior probability "is multiplied by". That is, I'm giving the factor by which it changes.
     
  19. 10/49 is the probability of the FIRST draw to be a diamond when three draws after the first are all diamonds.
     
  20. So, here's an experiment. It's a little impractical, but it could be simulated. We agree to play the following game:

    1. Shuffle a deck of cards.
    2. You take the first card.
    3. We turn over the next three.
    4. If those three are not all of the same suit, this round is over: back to step 1.
    5. If the three cards are all the same suit, we look at your card. If your card is the same suit as the three, I give you $3. If your card is of a different suit you give me $0.95. Now, you should be very happy to do this, because you expect a net profit. But I think I'll profit.

    What do you say. Should I run it? I'll show you the code, so you'll know I'm not cheating.
     
  21. This is different from the original question because this process, regarding our net profits, would leave out the cases where first is diamond but next three are of different suit. It would only process our earning only with cases where three diamonds are present.

    Three consecutive diamonds being drawn does not represent the population from which we're trying to calculate our probability. It does represent if we change the question into 'probability of the first card being diamond only when next three cards are also diamond'. I still think these two cases are different, and your game suggests that they are the same. Whether I'm right or wrong I need some explanation on this part to be fully pursuaded.
     
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