Curious observation

[--Pete]

Hi,

Yes, of course. The probability that someone has a birthday on any particular day is just 1 - (probability that no one has a birthday on that say). The probability that at least one person has a birthday on every day of the year is just that last number to the 365.25 power. So, the probability that there exists at least one day when no one has a birthday is just one minus this last number. Or, at least that's my zeroth order answer without much thought.

For the case of 500 people, the "answers" are
Probability of no birthday on a given day, P1 = (364.25/365.25)^500 = 0.2539
Probability of a birthday on a given day, P2 = 1 - P1 = 0.7461
Probability of a birthday on every day of the year, P3 = P2^365.25 = 3.455e-47
Probability of one or more days without a birthday, P4 = 1 - P3 = virtually 1

If we increase it to 2000 people, we get
P1 = 0.004156
P2 = 0.9958
P3 = 0.2185
P4 = 0.7815

(all calculations done on the Windoz calculator, no intermediate rounding, and answers *not* checked).

All such calculations can be fun. But they are nothing but counting problems. Sampling problems are much more interesting, IMO ;)

--Pete