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Deck of Cards in order

  1. Mar 24, 2012 #1
    Hello,

    We were having a conversation about odds, probability and so forth... and we were shuffling some cards.

    All of sudden, a question was asked, and we have no idea how to figure it out.

    After shuffling a deck of cards repeatedly (long after it has come out of the box), what would be the mathematical odds of turning the deck over to reveal that every card is in order - ace, ace, ace, ace, two, two, two, two and so on.

    To take it a step further, what would be the odds of them showing up in numerical order, AND in the same order of suit - hearts, diamonds, spades, clubs for each set?

    Thanks in advance!

    Mandy
     
  2. jcsd
  3. Mar 24, 2012 #2
    Since there are 52 unique cards, there are 52! possible orders (permutations) of the deck. If every card has a "correct" place in that order, then the probability of randomly finding that particular ordering is 1/52!. (52! is a large number, equal to 1*2*3*......*51*52 [itex]\simeq[/itex]8.06581751709439e+67)
     
    Last edited: Mar 24, 2012
  4. Mar 24, 2012 #3
    That's awesome. Much simpler than we were making it out to be.

    Thanks so much!

    So according to our calculations, if it's NOT suited, and it's just in order, then the odds are 4/52 * 3/53 * 2/52 * 1/51 * 4/50 * 3/49 * 2/48 * 1/47.... which is 1.08667018142645000000E-50.

    Sound about right?

    Thanks again!

    Mandy
     
  5. Mar 24, 2012 #4
    My calculation is for every card having a particular place in the deck. If you allow any variability, the probability increases. From the above, it looks like you are specifying a place in the order for every card
     
    Last edited: Mar 24, 2012
  6. Mar 25, 2012 #5
    Sorry. I think I misunderstood your question. If you want to ignore suits, then I believe the correct approach is (52C4)13!. That is, the number of ways to choose 4 cards from 52, times the number of permutations of 13. The reciprocal of this should give the probability of randomly getting the order you specified in post 1 when ignoring suits. I leave the calculation to you.
     
    Last edited: Mar 25, 2012
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