Originally Posted by

**Soroban** Hello, george12345!

There are: $\displaystyle 52!$ possible orderings of the cards.

We have: .4 aces and 48 Others.

Place the 48 Others in row, leaving a space before, after, and between them.

. . $\displaystyle \_\,O\,\_\,O\,\_\,O\,\_\,O\,\hdots \,\_\,O\,\_$

There are 49 spaces.

Select 4 of the spaces and insert the four Aces (in some order).

. . There are: .$\displaystyle _{49}P_4$ ways.

Hence: .$\displaystyle P(\text{no consecutive Aces}) \;=\;\dfrac{_{49}P_4}{52!}$

Therefore: .$\displaystyle P(\text{some consecutive Aces}) \;=\;1 - \dfrac{_{49}P_4}{52!} $

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If the problem meant *exactly* one pair of consecutive Aces,

. . we can modify the above procedure.

We have the 48 Others with spaces before, after, and between them.

. . $\displaystyle \_\,O\,\_\,O\,\_\,O\,\_\,O\,\hdots\,\_\,O\,\_$ . (49 spaces)

We have 4 aces.

Select two of them to be together; there are $\displaystyle _4C_2 \,=\,6$ choices.

Duct-tape the pair together.

Then we have 3 "cards" to distribute: .$\displaystyle \boxed{AA}\,,\:A,\:A$

And they can be placed in: .$\displaystyle _{49}P_3$ ways.

Hence, there are: .$\displaystyle 6(_{49}P_3)$ ways to have one pair of consecutive Aces.

The probability is: .$\displaystyle \dfrac{6(_{49}C_3)}{52!} $