Probability - non gaming related

[BruceZ]

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

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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.

[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]

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Thanks for the replies!

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

iash

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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.