Probability - non gaming related

[iash]

Trying to settle an argument with a co-worker.

Office staff of 90 people, what are the odds of nobody having a birthday during any one particular month?

iash

[slickpoppa]

Well, start with the chance of nobody having a birthday on any given day:
(364.25/365.25)^90 = .7813
Then the chance of that happening 31 times in a row is:
(.7813)^31 = 4.76x10^-4 = .000476 ~ 1/2,098
So, the chances are pretty low.

[BeerMoney]

1-(364/365)*(363/365)*(362/365).....
with 86 more terms i think.. The chances would be next to zero.. With about 22 people the chances are about .50

[slickpoppa]

I'm sorry, but that formula is not even close to being correct. Take another look at my answer.

[fnord_too]

[ QUOTE ]
Trying to settle an argument with a co-worker.

Office staff of 90 people, what are the odds of nobody having a birthday during any one particular month?

iash

[/ QUOTE ]

Depends on the month. Take a 30 day month for instance. Then I believe the probability is
(335.25/365.25)^90 =~ .045% (or .00045)

Alternately, treating all months equally
(11/12)^90 =~ .040%

The first calculation takes into account leap years, but not that funky "unless the year is a multilpe of 100 but not a multiple of 400" clause, which impacts the calculations not at all right now (unless you have people born in 1900 on your staff). Damn earth and it's not quite 365 day orbit (and not quite 24 hour rotation).

And of course special care must be taken for Feb, since it is a 28.25 day month.

[BeerMoney]

I skimmed over it, and answered as two people sharing same birthday..

[BeerMoney]

((11/12)^90)*12

[BruceZ]

[ QUOTE ]
Depends on the month.

[/ QUOTE ]

Especially if the month happens to be February. It is more than twice as likely that no one has a birthday in February than for a 31 day month, including the effect of leap year(1400-to-1 vs. 2928-to-1). Try it.

[BruceZ]

[ QUOTE ]
Well, start with the chance of nobody having a birthday on any given day:
(364.25/365.25)^90 = .7813
Then the chance of that happening 31 times in a row is:
(.7813)^31 = 4.76x10^-4 = .000476 ~ 1/2,098
So, the chances are pretty low.

[/ QUOTE ]

The 31 days are not independent, so the probabilities cannot be multiplied this way (except as a crude approximation). If Jan. 1 is not taken, Jan. 2 becomes more likely, and so on.

For 31 days, if we ignore leap year for a moment, the actual probability is:

[ (365 - 31)/365 ]^90 = 0.000339.

Taking leap year into account:

[ (3/4)*(365 - 31)/365 + (1/4)*(366 - 31)/366 ]^90 = 0.000341 or 2928-to-1.

[BeerMoney]

Bruce, wouldn't you have to multiply by 12 choose 1 to account for the different months which this could occur in?