Happy Birthday
I expect everyone knows this one, but it's better than nothing.
In a random group of 23 people, there is a slightly better than a 50% chance that at least two of them share the same birthday. How is that possible?
How big would the group need to be to get the chance of a shared birthday up to at least 95%?
Discard all leap year complications. Assume each birthday is equally likely. Only consider the day and month, not the year of birth. It is not a trick problem.
In a random group of 23 people, there is a slightly better than a 50% chance that at least two of them share the same birthday. How is that possible?
How big would the group need to be to get the chance of a shared birthday up to at least 95%?
Discard all leap year complications. Assume each birthday is equally likely. Only consider the day and month, not the year of birth. It is not a trick problem.
Labels: logic, mathschallenge, SharedPuzzle
posted by Chris at
6:08 PM
5 Comments:
47
You are right. Explanation is as follows. Total number of variants to choose birthdays of n persons among 365 days is 365^n. They will not coincide in the 365*364*...*(365-n+1) cases. In the rest cases at least two will coincide.
A more detailed explanation:
Let's call P the probability of at least 2 persons out of n have birthday at the same day.
Then, P'=1-P is the probability of no person shares their aniversary day, and this is easy to calculate.
Suppose n=2. The first person's aniversary can be any day, but the second can be only in 364 out of 365. For n=2 P' is then 364/365.
Now, if n=3, the third person has only 363 days to be born, out of 365: P'=364*363/(365*365) for n=3
P'(n) can be easy determinated as P'(n)=365!/(365^n*(365-n)!), and P=1-P'
P(23)=50.7297234%
P(46)=94,8252843%
P(47)=95,4774403%
and thus, 47 is the correct answer, although the result is nothing but intuitive.
Note on calculating this in EXCEL.
Although EXCEL has the FACT function, this is not suitable to perform this calculation, as the numbers are very big. The workarround is to use the COMBIN(n,k) = n/(k!(n-k)!), and the formula in excel is then (assuming that your n is in cell C3)
=1-COMBIN(365,C3)*FACT(C3)/POWER(365,C3).
(sry for my english, I hope it is clear enough)
As I feared, too well known or too easy for some.
Obviously I posted it because the uninitiated will not believe the probabilities.
Well done quantense and a very well done Miguel. Thank you.
The following link pretty well covers the whole topic:
http://en.wikipedia.org/wiki/Birthday_problem
NB The Windows freebie calculator can calculate 365!
PS Miguel, I have posted the non-sneaky find the Oak instructions.
the anwser is 47
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