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?

[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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383&Forum=probability&Words=inclusion-exclusion&Match=Entire% 20Phrase&Searchpage=0&Limit=25&Old=1year&Main=4169 81&Search=true#Post417383) 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.

[slickpoppa]

dang. you're right.

[iash]

Thanks for the replies!

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

iash

[fnord_too]

[ QUOTE ]
Thanks for the replies!

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

iash

[/ QUOTE ]

In that case you have to compute the probabilities of each individual month not having anyone born in it. (Really, there are only 3 cases: 31 days, 30 days, and 28.25 days, but there are 7 31 dayers, 5 30 dayers, and 1 28.25 dayers).

Call these P31, P30, P28, and do the following:
1 - ( (1-P31)^7 * (1-P30)^5 * (1-P28)) to get the probability that there will be no month that has 0 birthdays in it. (Everything following the first "1 - " is the probability that there is at least one birthday in each month.

I think this is slightly off, too, but close. I think this is off because you would have a non zero probability for a month having someone in it if you only had 11 people, so the above would yield a non zero probability of having no month without a birthday in it for 11 people, which is clearly impossible.

hmmm.... my reasoning is off somewhere, but I do not know exactly where.

Here is another attack, for each person you have a probability of that person being born in each month corresponding to the number of days in the month. You could sum over all combinations that have hat least one person in each month (i.e. you could have 79 in january and one in each of the other months C(90,79) ways, 78 in Jan, 2 in Feb, 1 in each other month, etc), but this is a very arduous approach.

I need to reread this thread to see if someone has the right answer already. I'm sure there is an easy way to do this exactly, but I sure cannot think of it right now.