What is the probability of being dealt 4 aces and one other card?

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.

Worth beads is right. You have a 1 in 6497400 chance of being dealt your desired hand.

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.

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

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.

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}

Multiply the individual probabilities: P(ace)*P(2nd ace)*P(3rd ace)*P(4th ace)*P(different card)

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.