View Full Version : Dealing cards
Remclane
Sep 6, 2007, 06:58 PM
What is the probability of being dealt 4 aces and one other card?
worthbeads
Sep 6, 2007, 07:12 PM
52*51*50*49 I would say. I wouldn't count the other card because you will get one other card no matter what. I might be wrong, though.
Jhaynes1988
Sep 6, 2007, 07:19 PM
Worth beads is right. You have a 1 in 6497400 chance of being dealt your desired hand.
Capuchin
Sep 6, 2007, 11:27 PM
I think there's a slight correction you can have a look at if you want to be more accurate, remclane.
If you get the "other card" first, then there's a bigger 1 in 51*50*49*48 chance of getting your 4 aces.
You can also correct for getting the other card 2nd 3rd or 4th, too.
Ken 297
Sep 7, 2007, 03:44 AM
The first ace has four chances in 52, the second 3 out of 51, the third 2 out of 50 and the fourth 1 out of 49.
Not sure how to work in the other card. I guess it depends on when it comes up
ebaines
Sep 7, 2007, 12:10 PM
The number of was that a 5-card hand can be dealt is 52*51*50*49*48. The number of ways that 5 cards satisfy the condition of 4 aces plus 1 other is 4*3*2*1*48*5 - the extra 5 is because the "other" card can be dealt as any one of the 5 cards. So, the probability you're looking for is:
4*3*2*1*48*5/(52*51*50*49*48) = 0.000018469, or 1 in 54,145.
galactus
Sep 7, 2007, 04:37 PM
Yes, I concur with ebaines.
Choose 4 Aces out of the 4 and one other out of the remaining 48.
\frac{C(4,4)C(48,1)}{C(52,5)}=\frac{1}{54145}
s_cianci
Sep 8, 2007, 10:28 AM
Multiply the individual probabilities: P(ace)*P(2nd ace)*P(3rd ace)*P(4th ace)*P(different card)
galactus
Sep 8, 2007, 10:40 AM
Multiply the individual probabilities: P(ace)*P(2nd ace)*P(3rd ace)*P(4th ace)*P(different card)
That'll work too. BUT, don't forget the arrangements. There are 5 cards with for the same
So multiply by 5!/4!=5
(4/52)(3/51)(2/50)(1/49)*5=\frac{1}{54145}
Another good approach.