Anyone good in math? Solve this (poker) question.

    • ragney
      ragney
      Bronze
      Joined: 02.08.2010 Posts: 2,417
      Lets say there is 40 cards left in the deck:
      11 of them :spade:
      7 of them :heart:
      13 of them :diamond:
      9 of them :club:

      You draw exactly 4 cards.

      Questions: (when we draw 4 cards)
      1) What's the chance that at least one of the cards is spade?
      2) What's the chance that at least two of the cards is heart?
      3) What's the chance that at least three of the cards is diamond?
      4) What's the chance that [B]at least one card is spade and one card is heart[/b]?

      Can you guys explain how to calculate these stuffs?
  • 6 replies
    • sentin1ty
      sentin1ty
      Bronze
      Joined: 20.03.2009 Posts: 123
      Ok i'll try. So lets say to pick one spade is 11/40, one heart 7/40 and so on.
      Sooo 1) Atleast one spade is so there can be 4 spades x 3 spades and heart, 3 spades and diamond and 3 spades and a club x 2 spades .... and 1 spade and 3 different suit. I hope you get it, if not ill try to explain more clearier
    • VorpalF2F
      VorpalF2F
      Super Moderator
      Super Moderator
      Joined: 02.09.2010 Posts: 8,910
      Hi Ragney,

      You need to work with the Combination Function

      I sidestep that and use Excel.
      Excel has a neat function =Combin(Total,Number_chosen)

      For small numbers it is easy to see:
      Combin(4,2) = 6 (the number of possible pairs of a given rank)

      etc.

      I'll illustrate the process with your first example, because it is the easiest, and I have a shot at getting it right :D

      In your 40 cards there are:
      Combin(40,4) = 91390 combinations of 4 cards.
      There are Combin(29,3) = 3654 3-card combos that have NO spade.
      There are 11 possible spades to add to each 3-Card combo, so the total number of combos containing EXACTLY 1 spade is 3654 x 11 = 40194
      There are 406 2-card combos where neither of them is a spade.
      There are 55 two-card combos using the 11 spades, for a total of 22330 combos of EXACTLY 2 spades in the 4 card group.
      There are 165 3-card combos of spades.
      There are 29 remaining non-spade cards, so there are 4765 combinations with exactly 3 spades
      There are 330 combinations where all of the cards are spades.

      So...
      40194 + 22330 + 4765 + 330 = 67619 combinations of cards with AT LEAST one spade

      67619 / 91390 = 74%


      The next two are done the same way.

      For the last one do you mean:
      a) at least one spade and exactly one heart
      b) at least one spade and at least one heart

      In any case it just a wrinkle on the first method.

      Cheers,
      --VS
    • Saka000
      Saka000
      Bronze
      Joined: 21.07.2010 Posts: 136
      1)11/40 X 11/39 X 11/38 X 11/37

      2)7/40 X 6/39 X 36/38 X 35/37

      3)13/40 X 12/39 X 11/38 X 34/37

      4)11/40 X 7/39 X 36/38 X 35/37
    • VorpalF2F
      VorpalF2F
      Super Moderator
      Super Moderator
      Joined: 02.09.2010 Posts: 8,910
      Originally posted by Saka000
      1)11/40 X 11/39 X 11/38 X 11/37

      2)7/40 X 6/39 X 36/38 X 35/37

      3)13/40 X 12/39 X 11/38 X 34/37

      4)11/40 X 7/39 X 36/38 X 35/37
      Good idea, but I think that you need to do it backwards.

      In other words, what are the chances that NONE of the 4 cards are spades?

      for the first card:
      29/40
      2nd: 28/39 etc
      so 29/40 * 28/39 * 27/38 * 26/36 = 0.26
      1 - 0.26 = .74

      I always do things the hard way...
      <sigh>

      --VS
    • TinoLaan
      TinoLaan
      Bronze
      Joined: 12.10.2011 Posts: 6,411
      I'd like to add my two cents to this as well. At the above, Saka000, unfortunately it's not quite that simple, although you're on the right track. I'll do some rough rounding up/down here so these numbers won't be exact, but they'll be incredibly close.

      Like Vorpal mentioned, there's 40 c 4 = 91390 total combinations of 4 cards. This is a very basic calculation:

      40! / (4! * (40 - 4)!) = 91390


      However, you really don't need to know this figure, or any other exact numbers. All you need to know are probabilities.

      1) At least 1 spade
      To calculate the number of hands which have at least one spade, we simply have to calculate the number of hands that do not have a spade. For this, Saka's calculation is actually almost correct, except you have to subtract that number from 91390. So the chance of not getting a spade is:

      29/40 * 28/39 * 27/38 * 26/37 = ~26%.

      For every non-spade we draw, there's one less non-spade in there and one less card in the deck. That's how you get to the above calculation, and this applies to all calculations below as well.

      Therefore the chance to get at least 1 spade is roughly 1 - .26 = 74%.

      2) At least 2 hearts
      This one is already a bit more complicated. However, this can also be done in a simpler way. You're looking for 2, 3 or 4 hearts. The chance to get 0 hearts can be calculated the same as above:

      33/40 * 32/39 * 31/38 * 30/37 = ~44.8%

      The chance that you will get exactly one heart is

      4 c 1 * 7/40 * 33/39 * 32/38 * 31/37 = ~41.8%

      4 c 1 = 4, the reason we add this multiplier in there is because the heart can be drawn in 4 different cases.

      Therefore the chance that you'll get 2 hearts or more is roughly 1 - .448 - .418 = 13.4%

      3) At least 3 diamonds
      The calculation for 3 or 4 diamonds is basically the same as the above, except we can now just use the calculations for the chances of getting 3 diamonds and 4 diamonds.

      3 diamonds: 4 c 1 * 13/40 * 12/39 * 11/38 * 27/37 = ~8.4%
      4 diamonds: 13/40 * 12/39 * 11/38 * 10/37 = ~0.8%

      Therefore, the chance of getting at least 3 diamonds is roughly 0.084 + 0.008 = ~9.2%

      4) I'm not exactly sure which of the two you mean here (same as Vorpal). However, I'm running short on time and the same principles as above apply. :)

      Hopefully that helps!
    • ragney
      ragney
      Bronze
      Joined: 02.08.2010 Posts: 2,417
      Originally posted by VorpalF2F
      For the last one do you mean:
      a) at least one spade and exactly one heart
      b) at least one spade and at least one heart

      In any case it just a wrinkle on the first method.

      Cheers,
      --VS
      Hey thx! I'll read after dinner. For the last I mean we need to draw at least one spade and one heart. The other two cards can be anything. So option b I guess.