(2/29)(3/29)(4/29)(5/29)(6/29)(7/29)(8/29)(9/29...

The behavior of this sequence depends entirely on where it stops: The sequence will eventually include the term If it goes past

The expression describes the where the numerator increases by 1 for each term and the denominator remains a constant (2/29)(3/29)(4/29)(5/29)(6/29)(7/29)(8/29)(9/29...

people in a room of 29 possible "slots" (like days in February during a leap year) all have unique values, the formula looks like: The behavior of this sequence depends entirely on

This specific sequence often appears in , specifically the Birthday Problem . If you were calculating the probability that (2/29)(3/29)(4/29)(5/29)(6/29)(7/29)(8/29)(9/29...