Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook