I have come across a problem relating to playing cards, and am having some difficulty figuring out the solution.

Assume you have 1 deck of shuffled cards. You proceed to deal the cards out one by one. I would like to know the probability of getting 2 consecutive aces anywhere in the sequence of 52 cards.

Thus far I have the following:

The chance of getting four consecutive aces anywhere in the deck.

I have calculated as follows:

A sequence can start on Card 1, 2, 3, .... up to the 49th spot. Thus there are 49 possible start positions for the sequence. Further, since there are 4 aces, each sequence can be arranged in 4x3x2x1 ways (4! ways). Thus there are 49x24 possibilities. With these 1,176 possibilites the remaining cards of the deck can be arranged in 48! ways for EACH of the 1,176 possibilities. Thus in total there are 48! x 1,176 ways in which a deck of cards can contain a 4-ace sequence.

There are 52! in which a deck of cards can be arranged.

Thus the probability that a particular card arrangement contains a 4-ace sequence is

(1,176 x 48!) / 52! = 1,176 x 48! / (52x51x50x49x48!)

= 1176 / (52x51x50) = 0.018% (or roughly 1 in 5,525)

I then looked at the likelihood of getting a 3-ace sequence

A sequence can start on Card 1, 2, 3, .... up to the 50th spot. Thus there are 50 possible start positions for the sequence. Further, since there are 4 aces, each sequence can be arranged in 4x3x2 ways. Thus there are 50x24 possibilities. With these 1,200 possibilites the remaining cards of the deck can be arranged in 49! ways for EACH of the 1,200 possibilities. Thus in total there are 49! x 1,200 ways in which a deck of cards can contain a 3-ace sequence.

However this includes the 4-ace sequences which I want to exclude.

49! x 1200 = 7.29 x 10^65

48! x 1176 = 1.46 x 10^64

After the subtraction we are left with 7.16 x 10^65

Thus the probability that a particular card arrangement contains a 3-ace sequence is

(7.16 x 10^65) / 52! = 0.904% (or roughly 1 in 110.5)

Is my approach correct thus far?