the birthday puzzle

how many people in a crowd do i need so that i have a 99% chance of finding at least a pair of individuals who shares the same birthday every year?

the answer, surprisingly, is 57.

read more about this classic birthday conundrum in this NYTimes post, as i can't fully explain to you the math behind it as succintly as that author did.

to fully test this theory, i got hold of UPCM 2015's class list, isolated the first 57 people on that list, and took note of their birthdays on a separate spreadsheet.

January 21, 1989
January 23, 1991
January 28, 1988
February 17, 1989
February 27, 1989
March 10, 1988
March 12, 1989
March 26, 1989
March 26, 1989
March 26, 1990
March 29, 1988
March 5, 1992
April 1, 1989
April 10, 1989
April 12, 1988
May 17, 1988
May 7, 1990
May 8, 1990
June 11, 1992
June 18, 1989
June 23, 1989
June 25, 1989
June 29, 1984
July 1, 1987
July 13, 1986
July 16, 1990
July 20, 1990
July 26, 1989
July 27, 1988
July 31, 1989
July 8, 1990
August 12, 1992
August 2, 1991
August 25, 1989
August 26, 1989
August 4, 1988
August 7, 1989
September 10, 1989
September 11, 1990
September 2, 1987
September 2, 1988
September 25, 1988
October 11, 1988
October 13, 1990
October 26, 1988
October 28, 1989
October 31, 1991
October 6, 1989
November 17, 1989
November 2, 1992
November 24, 1988
November 4, 1988
November 5
December 11, 1989
December 17, 1991
December 30, 1990
December 9, 1989

it turns out that the theory holds water!

coincidentally i was taking a module on probabilities and variances last week, and i only got 57.1% in that week's quiz. guess i need to review my math.