Re: The matching birth-month formula by Afonso
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Fri, 07 Apr 2006 10:01:36 EDT
Let c(n) , n=1, 2, 3, …, 12 the cumulative days of
the year
____c(1) = 31
____c(2) = c(1) + 28.25
____c(3) = c(2) + 31
____c(4) = c(3) + 30
...
____c(11) = c(10) + 30
____c(12) = c(11) + 31 = 365.25
For 2 persons the probability without matching
birth-month is
365*(365-c(1)) / (365 ^2) = 1- c(1)/365
For 3 persons
[365*(365-c(1)*(365-c(2)] / 365^3 =
= (1- c(1)/365) * (1-c(2)/365)
The probability of no-matching birth-months among m
persons is
_p(m) =
___ = (1-c(1)/365)*
___ * (1-c(2)/365)*…*(1-c(m-1))/365)
where m < = 12.
To the Co-readers of this News:
I apologize for the different errors made along the
formula´s derivation.
All comments will be welcomed.
_____licas (Luis A, Afonso)
Luis, let's look at your formula for m=2 persons. It depends only on the number of days in January and the total number of days in a year.
Suppose that January had 31 days, December had 334 days, while the other months had zero days. Then the real probability that the two people would not have birthdays in the same month would be
1 - (31/365)^2 - (334/365)^2 = 0.1554.
Your formula gives 1 - 31/365 = 0.9151.
Jack
.
- References:
- The matching birth-month formula by Afonso
- From: \"Luis A. Afonso\"
- The matching birth-month formula by Afonso
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