I read somewhere once that in a class of 30 children the chances of 2 of the children sharing a birthday are about 50/50.
Is this correct and if it is, why is it?
I find this hard to credit. I have never met anyone who has the same birthday at me, even when I was at University with a much larger class size than 30.
From Wikipedia, “The birthday paradox states that if there are 23 people in a room then there is a slightly more than 50/50 chance that at least two of them will have the same birthday. For 60 or more people, the probability is greater than 99%. This is not a paradox in the sense of it leading to a logical contradiction; it is a paradox in the sense that it is a mathematical truth that contradicts common intuition. Most people estimate that the chance is much lower.”
The article goes into the maths involved.
I thought it was a much smaller number - in a group of 17 people at least two will have the same birthday more than 50% of the time. I think this was discussed here 2-3 years ago.
FYI, it means the same month and day - not the same year. The chances of that happening are much, much lower.
In a class of 23 people, there’s a 50% chance that two people have the same birthday. This is a pretty standard demonstration of the fact that probability can be counterintuitive.
However, if you’re in a class of 23 people, the probability that somebody shares your birthday is significantly lower than the probability that two people share the same birthday.
23 people sounds more right - I think I misremembered the 17.
You seemed to have misunderestimated.
If it’s a standard, elementary-school type classroom, the chances that it’s the same year are pretty good, I’d think.
A common confusion. In a class of more than 30, there’s a good chance that some pair of people have the same birthday, but you won’t necessarily be one of that pair.
If you are only looking at the chances that someone else has the same birthday as you, you need to have a group of 253 other people for there to be more than a 50% chance that someone else shares your birthday.
(364/365)^252 = 0.500895161
(364/365)^253 = 0.499522846
That’s assuming that there are only 365 days in a year – to take into consideration leap year would be a more cmplex calculation.
Actually, the calculation wouldn’t be that much more complex. Instead of calculating using a base year, you’d calculate using a base period of four years +1 day (365*4 + 1 =1461) That’s the denominator of your fraction. Since four of those days match your birthday, the numerator is 1461-4 = 1457. Since this is a small adjustment to your prior calculation, you can start using your prior result:
(1457/1461)^253 = .4997605448
(1457/1461)^252 = .5011325710
As you can see, the practical answer [integer # of days] doesn’t happen to change if you consider leap years. The bracketing values remain the same. I did the calculation on a pocket calculator, which probably uses logarithms internatlly, and might be off on the lest significant figures. However, a little back-of the envelope calculation convinces me that it is certainly accurate to the first four decimal plaes, which is enough to verify the answer given.
For th OP, the first step is to calculate the chances that N people all have different birthdays. Ignoring leap years (which may not affect the answer anyway), we can do this by calculating p(x) - the chance that adding an xth person will not cause a match. Then we multiply all the p(x) from x=1 to x=N to get P(N). [For each value of x, multiplying the lower results gives the chances that you will reach the xth person without a match, and p(x) is the chance that the xth person won’t be a match – which gives you P(x) the chances that there is not match in x people (note the distinction between p(x) and P(x) - upper/lower case)
p(x) = (“allowable [nonmatching] days” remaining) / (days in the year)
= [365-(x-1)] / 365
P(x) = p(1) * p(2) * p(3) … * p(x)
= ( [365-0] * [365-1] * [365-2]… 365-[x-1]x) / 365^x
= (365!)/[(365-x)!] / 365^x
= 365! / [(365-x)! * (365^x)]
You an do the same calculation, factoring in leap years, by changing the fraction p(x) to relect the number of nonmatching days in a 4-year period
Actually, the probability of birth is not evenly distributed throughout the year. Some month consistently have more births than others. This concentrates a disproportionate number of births in a “target rich environment” where collisions are more likely. I don’t know, offhand if this effect would be enough to lower the integer solution to 22 in real life. It might depend on the country, since birth cycles through the year depend on cultural and climate conditions [among other things]
Real Life Anecdote: In my piano class of 15 people, I discovered that another girl shared my exact same birthday – December 6th, 1984. What are the odds of this? 15 is quite a smaller number than 30 or 50.
100%, apparently.
Seriously, though, this is just confirmation bias, a case where you’d only notice a positive effect. Consider the tens of thousands of 15-people groups you’ve been in during your life, the vast majority of which did NOT have a person with you exact birthday in them…and you never noticed.
I don’t care a whit about confirmation bias, I was asking what the odds were for 2 people in any random grouping of 15 to share the exact same birthdate.
This puts it the probability at about .25.
The first time I met someone who shared my birthday was at age 24, an Australian who was travelling Europe and had settled down to temp in the UK.
My Dad teaches math at a state college, and at the beginning of every class he bets the class a case of cola that two people in the class share the same birthday. Needless to say, they always take the bet, and he nearly always wins.
It’s the old 17/23 correlation. The Master spoke of it here. 
22 people work at our maintenance department. Three of us have january 26 as our birthday. Three different years, though.
When he wrote “100%” the poster meant that since it’s already an established fact, it’s probability is 100%.