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.

[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?

[BruceZ]

[ QUOTE ]


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

[/ QUOTE ]

I'm computing the probability of no one having a b-day in one particular 31-day month (specified in advance). That is how I interpret the original question because of the word "particular", though I suppose you can read it differently. The probabilities for different months are different depending on the number of days in the month. If you wanted the probability that no one had a b-day in some month, then you would start by adding the probabilities for each of the 12 months. That would give a close approximation, but to be exact we would have to use the inclusion-exclusion principle to take into account the small probabilities of more than one month not having b-days since these are not mutually-exclusive.

In your solution, assuming each month has an equal 11/12 probability is not a good approximation since the difference between 28, 30, and 31 days gets very significant when the probabilities are raised to the 90th power. Even if we could assume each month is equally likely, multiplying by C(12,1) would be an approximation since we would really need the inclusion-exclusion principle to combine these.

EDIT: Actually, for the problem you were solving, that is "at least 1 month with no b-day", your method turns out to be quite close. This is because 11/12 is right in the middle of the probability you get for 30 and 31 days. It turns out that your method gives an overall probability of 0.004766623 or 209-to-1, and adding the exact probabilities for each month gives 0.004890898 or 203-to-1.

[iash]

Thanks for the replies!

What I meant in my original post was at least one month without birthdays...any month.

iash

[slickpoppa]

dang. you're right.

[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.

[BeerMoney]

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