What is the probability of drawing out four aces in a deck of 54 cards (has jokers in it)?

answer is 1/316251

but i just can't figure out how to.

help please!

6 Answers

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  • Anonymous
    8 years ago
    Best Answer

    Assuming you aren't replacing the cards and you're only drawing four times...

    The probability you get an ace on the first draw is 4/54.

    Next, there's one less ace in the deck, so the numerator (top) is 3. There's one less card in the deck, so the denominator is 53. The probability of getting another ace on the second draw is 3/53.

    Similarly, subtract one from the top and bottom again for the third ace. The probability is 2/52.

    Again, subtract one from the top and bottom again for the fourth ace. The probability is 1/51.

    You want to find the probability that the first card AND second card AND third card AND fourth card are aces, so use multiplication (use addition for OR)

    (4/54)(3/53)(2/52)(1/51)=1/316251

  • 8 years ago

    P(1st ace) = 4/54 Remember there are 4 aces in a pack of 54.

    P(2nd Ace) = 3/53 There are now 3 aces in a pack of 53, the 4th ace is kept out of the pack!!!

    P(3rd Ace) = 2/52 There are now 2 aces in a pack of 52, the 3rd & 4th aces are kept out of the pack!!!

    P(4th Ace) = 1/51 There is now 1 ace in a pack of 51, the 2nd, 3rd & 4th aces are kept out of the pack!!!

    P(all 4 aces) = 4/54 x 3/53 x 2/52 x 1/51 = 24/ 7590024

    Since 24 is a common factor to 24 and 7590024, the fraction cancels down to = 1/316251

    This is removing the aces with out replacement.

    However, having drawn an ace, and it is then put back into the pack and the ace is drawn for a second time

    and so on:-

    the probability is 4/54 x 4/54 x 4/54 x 4/54 = (4/54)^4 = (2/27)^4 = 3.01 x 10^-5 = 0.00005

  • david
    Lv 6
    8 years ago

    1 / 54C4

    =4*3*2/(54*53*52*51) should be it

    =1/(27*53*13*17)

    =1/316251

  • 8 years ago

    4/54 = 2/27

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  • DOVE
    Lv 6
    8 years ago

    It is no. of aces/ no of catrds

    = 4/54

    =2/27

  • 8 years ago

    odds against 1st ace = 54/4 to one

    odds against second ace = (54-1)/(4-1) to 1 = 53/3 to 1

    odds against third ace = (54-2)/(4-2) to 1 = 52/2 to 1

    odds against fourth ace = (54-3)/(4-3) to 1 = 51/1 to 1

    Total odds against = (54*53*52*51)/(4*3*2*1) = 316,251

    probability = 1/316,251

    general formula for odds against = n!/((n-m)!*m!) where n = number of cards, m = number of aces.

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