There are 365^730 total ways that 730 people can have their birthdays distributed in a 365 day year.

There are 730!/(2^365) ways to partition 730 people into distinct pairs, each having a different birthday from the rest of the crowd.

Then the probability there are no triples is 730!/(2^365 * 365^730).  Therefore the closed form solution that there is at least one set of birthday triplets is 1 - 730!/(2^365 * 365^730).

Charlie notes that the probability of no triplets can be expressed as the product:

Prod{x=1 to 364}( (1/366-x)^2 * ((365-x)/(365-x+1))^(730-2x) * C(730,x) )

This can be broken into three subproducts:

Prod{x=1 to 364}( (1/366-x)^2 )

Prod{x=1 to 364}( ((365-x)/(365-x+1))^(730-2x) )

Prod{x=1 to 364}( C(730,2x-2) )

The first subproduct evaluates to 1/(365!)^2

The second subproduct evaluates to (364!)^2/(365^728)

The third subproduct evaluates to 730!/2^365

These subproducts multiply together to yield 730!/(2^365 * 365^730), which validates Charlie's numeric answer.