# Thread: Probability of getting a card hand dealt in a particular order.

1. ## Probability of getting a card hand dealt in a particular order.

52 cards, dealt 5 cards
P(1st 4 are aces, last is face card)
how do i do it (besides the [4/52][3/51][2/50][1/49][12/48])?
shouldn't 4 nCr 4 x 12 nCr 1 all over 52 nCr 5 work?????!?!?!

2. Originally Posted by mairjuanaman
52 cards, dealt 5 cards
P(1st 4 are aces, last is face card)
how do i do it (besides the [4/52][3/51][2/50][1/49][12/48])?
shouldn't 4 nCr 4 x 12 nCr 1 all over 52 nCr 5 work?????!?!?!
That would work, if it were not for the fact that you are asked to determine the probability of being given those 5 cards in that particular sequence: 4 aces first, and only then one single face card. Also, if you are interested in the probability for a particular sequence of aces, you cannot use 4 nCr 4 to count the number of possible sequences of 4 aces either.

3. Originally Posted by mairjuanaman
52 cards, dealt 5 cards
P(1st 4 are aces, last is face card)
how do i do it (besides the [4/52][3/51][2/50][1/49][12/48])?
shouldn't 4 nCr 4 x 12 nCr 1 all over 52 nCr 5 work?????!?!?!
The order of the cards is important.

4. 52C4 = 270725

Leaves 48/52 = 1/4

1/270725 * 1/4 = 1/1082900

5. Hello, mairjuanaman!

52 cards, dealt 5 cards
. . P(1st 4 are aces, last is face card)

The answer is: . $\frac{1}{1,\!082,\!900}$

How do i do it? . . . besides: . $\frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}\cdo t\frac{1}{49}\cdot\frac{12}{48}$ . . This is correct.

Shouldn't . $\frac{(_4C_4)\cdot(_{12}C_1)}{_{52}C_5}$ . work? . . . . . no

That is the probability of getting 4 Aces and 1 Face Card in any order.