If you have 10 people, what is the probability of 2 (or more) people having the same birthday?
closed as off-topic by ᴡᴏʀᴅs, Daniel W. Farlow, apnorton, Claude Leibovici, saz Feb 14 '15 at 7:08
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Well, it's very easy to see the probability that no pair shares the same birthday.
Say there are 2 people. Then one has a birthday, and the probability that the other person doesn't have that birthday is $364/365$, representing every day of the year besides the first (assuming all days are equally likely and that there is no such thing as a leap year birthday). If there is a third person, then the probability that he doesn't share a birthday with the other two is...
Continuing to 10 people, you get that...
I think of it this way. You have to pick a number between 1 and 365 (365 days in a year). The number you pick corresponds to a day (birth day).
Based on the above consideration,this how we tackle your problem
1>What is the probability that none of them have the same birthday?
2>The opposite of what you get above is the probability of 2 people or more having the same birthday.
If you have to pick a number between 1-365 ten times, the chance that none of them are same is calculated as
number of ways to pick ten different numbers divided by total number of ways to pick ten numbers
(desirable outcomes/ possible outcomes)
which is (365x364x363x...x356)/(365 raised to 10) lets call this y
y gives you the probability of all of them having different birthdays. The opposite of y is, two or more people have the same birthday.
SO 1-y is your answer
I calculated this as 0.1169 aka 11.69%