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Probability of dealing a higher card

  1. Jan 6, 2012 #1
    From a deck of 52 cards I deal 3 cards each to 4 players. I get the five of spades and two other, non-spades cards. What is the probability that any other spade higher than five is dealt to any one of the other players (ace is the highest card)?

    Back in college it used to be one of the subjects I was good at but after two decades this problem is now above me. Can someone please point me to the right direction?
     
  2. jcsd
  3. Jan 6, 2012 #2

    chiro

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    Hey nickntg and welcome to the forums.

    If this is a homework problem, you need to show an attempt at an answer.

    Regardless of the above, it helps to draw a tree diagram and to break things into disjoint events, as well as independent events.

    As a hint for these kinds of problems, consider the probability of one card, then a hand of cards, and then extend it to all players in the game. In terms of going from a less general to a more general situation, think about any dependencies.

    For example with a spade, the probability of getting another spade is dependent on whether you have existing spades or not.
     
  4. Jan 6, 2012 #3
    Thanks for the time taken to post a reply.

    This is not a homework problem - wish it were, that would mean I'm 20 years younger. I'm coding a skeleton game for a friend and I want to show the odds for the setup I indicated.

    Can't really say I follow your answer but let me try another approach. 3 x 4 = 12 cards are dealt on the table. To get the five of spades dealt there's a 12 x 1/52 = 0,2307 probability, right? 9 higher spades would remain in the deck...and here's where I get stuck. How do you calculate the probability that one of them is being dealt in the remaining 11 cards? I originally thought that for a single card being dealt the probability would be 9/51 but what happens with the rest of the cards?
     
  5. Jan 6, 2012 #4
    you have 3 cards so there are 49 still in the deck and 9 of them are a spade higher than a 5.
    you want to find the probability that an opponent has at least one spade higher than a 9 which is the same as
    1 - the probability that none of the opponents 9 cards are spade higher than a 5

    remember your opponents have 9 cards so the probability that one of them isnt a spade higher than a 5 is 40/49 so what is the probability that none of the opponents 9 cards are spade higher than a 5
     
    Last edited: Jan 6, 2012
  6. Jan 6, 2012 #5
    Are you saying that the probability is 1-40/49 = 0,183?

    That doesn't sound right...I was thinking something along the lines of:

    1 - 40/49 x 39/48 x 38/47 x 37/46 x 36/45 x 35/44 x 34/43 x 33/42 x 32/41 = 0,866.

    What do you think?
     
  7. Jan 6, 2012 #6
    I was saying the same thing you were. 0,866 is the answer... enjoy the game!
     
  8. Jan 6, 2012 #7
    I misunderstood then. I thought you originally indicated that the chance is 1-40/49 = 0,183.

    I've rigged a few hundred thousand simulated runs for this situation and so far the results show a probability around 0,7 of a spade greater than five being dealt. Comments?
     
  9. Jan 6, 2012 #8

    D H

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    Wrong. You, not either of the other two players was dealt the five of spades.
    One way to calculate this is to calculate the probability of not getting the five of spades. This calculation is pretty easy. You are dealt one card that is not the five of spades, then another, then another. That's 51/52 * 50/51 * 49/50, or 49/52. Your odds of getting the five of spades is 1-49/52, or 3/52. In this case, the straightforward calculation is just as easy, but sometimes the contrarian approach (what is the probability that X won't happen?) can make for a much easier calculation.

    However, 3/52 is not the right answer in this case. Per your opening post, "I get the five of spades and two other, non-spades cards." The probability you were dealt the five of spades is exactly one.

    There are 9 remaining cards, not 11. Twelve were dealt out, but you have three of them.

    Here the contrarian approach is very useful. Discounting the three known cards (yours), there are 49 unknown cards, 9 of which are spades higher than the 5. The probability that none of those 9 other dealt cards are one of the 9 higher spades is 40/49 * 39/48 * 38/47 * 37/46 * 36/45 * 35/44 * 34/43 * 33/42 * 32/41 ≈ 0.13309553902. That's the probability no one is dealt one of those higher spades. The probability that at least one person was dealt at least one of these cards is one less this probability, or 1-0.13309553902 = 0.86690446098.
     
    Last edited: Jan 6, 2012
  10. Jan 6, 2012 #9
    That makes sense, thanks for that DH!
     
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