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Probability Question

Not for me, but someone was asking me and I don't know. I remember there was a thread similar to this, so here we are :


"It's rather unlikely that 2 random people have the same birthdate (month and year). The more people that you put into the group, the more likely it is that two of them have matching dates. Ignoring leap year, it's clear in a group of 366 people, at least 2 must have the same birthday. Using probability, what is the fewest number of people that would need to be a group to have better than 50% chance of having a matching birthday in the group? Explain."

Can anyone figure this out?
 
I remember doing this problem!
Too bad I forgot how I did it or the answer, but IIRC isn't it around 23??
 
The probability of 2 people having the same birthday in a group of size n is: 1 - [(365_P_n)/(365^n)]. You could graph it and see where you get a probability of .5

-Tom
 
He must have written the question wrong, then. He's asking for the same day, same month, same year, but that would be incredibly hard to calculate, wouldn't it?

He must just need the same day of the same month.
 
Originally posted by: LoKe
He must have written the question wrong, then. He's asking for the same day, same month, same year, but that would be incredibly hard to calculate, wouldn't it?

He must just need the same day of the same month.

Yes, same month and same day are usually what people mean when they say "share a birthday".
 
Do yean day and month, not month and year?

If you are looking for people that had the same exact birthdate (as in same month, same day and same year) then you cannot really answer the question without making some sort of assumptions about the distribution of ages among those present.

Also, if you are looking at matching birthdates and not just birthdays then it is quite possible even with 366 people present that you won't have a match.
 
Yeah, I believe the question was wrong. Well, I'll run back to him with the question. If he doesn't like it, well, he can't figure it out himself.

Thanks. 😛
 
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