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Simple Probability

  1. May 27, 2009 #1
    ok long story short I have a card game I play with friends and there is a circumstance that happens way to often. I don't know how to calculate it so I am asking you guys...

    You flip 4 cards over to start the game. Out of those 4 cards I am wondering what the probability is for 1 to be an Ace, and for 2 of them to be an Ace.

    I am more curious about 1 and 2 and if that happens how many hands should go by before it happens again. If you want to do 3 and 4 that is great, but that rarely happens.

    Thanks guys
     
  2. jcsd
  3. May 27, 2009 #2
    The chance of an ace is around 1 in 4, and the chance of 2 aces is around 1 in 40. Unless this is a homework question, that should be close enough (neither answer is exact).
     
  4. May 28, 2009 #3
    No it is not a homework question...but I would like to be as precise as possible please...like your numbers are they estimations or did you round off?
     
  5. May 28, 2009 #4
    Also, how would you arrive at the solution??? I understand that the probability of drawing 2 Aces is 1/221 but I don't know how the two extra chances would factor into affecting the formula
     
  6. May 28, 2009 #5

    HallsofIvy

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    There are 4 aces in a 52 card deck and 48 "non-aces". The probability that the first card will be an ace is 4/52. Now there are 51 cards left. The probability that the second card is NOT an ace is 48/51. The probability the third card is not an ace is 47/50 and the probability the fourth card is not is 46/59. Writing "A" for "ace" and "N" for "non-ace", the probability of ANNN (first card is an ace and the next three is not) is (4/52)(48/51)(47/50)(46/49). Now what about NANN? The probability the first card is not and ace is 48/52, the probability the second is an ace is 4/51, then not-an-ace is 47/50 and 46/49 for the third and fourth cards. That probability is (48/52)(4/51)(47/50)(46/59). It is easy to see that those are the same numbers- we just have the numbers in the numerator rearranged- and so the products are the same. That is, the probabilities of ANNN, NANN, NNAN, and NNNA are all (4/52)(48/51)(47/50)(46/49). Since there are 4 of them, the probability of a single ace in 4 cards is 4(4/52)(48/51)(47/50)(46/49).

    The same kind of argument works for two aces. The probability of AANN is (4/52)(3/51)(48/50)(47/49) while there 4!/2!2!= 6 different orders. The probability of two aces is 6(4/52)(3/51)(48/50)(47/49).
     
  7. May 28, 2009 #6
    Alternatively... first choose 2 aces (C(4,2) ways) then choose 2 non-aces (C(48,2) ways) and divide the result by C(52,4).
     
  8. May 28, 2009 #7
    that is awesome...thank you so much
     
  9. May 29, 2009 #8
    i used to play poker quite regularly and after reading a few books about it, the decisions that are made become quite mechanical. sort of took the fun out of the game for me.
     
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