Would you bet even money any two kids in a group of 25 had the same birthday?

charles_92688

Put it this way, imagine an elementary school or any classroom really with 25 kids in it. That's probably a typical number of kids. Think there's at least a 50/50 chance that there are two kids in that class that have the same birthday?

scudzilla

charles_92688 wrote: »

Put it this way, imagine an elementary school or any classroom really with 25 kids in it. That's probably a typical number of kids. Think there's at least a 50/50 chance that there are two kids in that class that have the same birthday?

Are you having a laugh?? I'm no mathematical genius but there's 365 days in a year, so the chances of any 2 of 25 kids having the same birthday has gotta be somewhere near the 15/1 mark

charles_92688

scudzilla wrote: »

Are you having a laugh?? I'm no mathematical genius but there's 365 days in a year, so the chances of any 2 of 25 kids having the same birthday has gotta be somewhere near the 15/1 mark

Do you know any school teachers? Ask one of them. See what they tell you. I'll bet they'll tell you most certainly they'd take the bet.

scudzilla

charles_92688 wrote: »

Do you know any school teachers? Ask one of them. See what they tell you. I'll bet they'll tell you most certainly they'd take the bet.

But there's 365 different days in a year, and only 25 kids, no way will this be a 50/50 shot on 2 kids having the same birthday

charles_92688

scudzilla wrote: »

But there's 365 different days in a year, and only 25 kids, no way will this be a 50/50 shot on 2 kids having the same birthday

Yes way. In fact, it's about 50/50 with 23 (actually 22.4) kids. With 25 kids it's better than 50/50

Look at at it this way. Any two kids have a 1/365 chance of the same birthday, or 364/365 of not the same birthday. How many tests are there? It would be

((N^2) - N))/2 tests for N kids in class. Imagine an N by N matrix. Each kid asks every other kid his birthday. Each kid can't ask him self and billy asking johnny is the same as johnny asking billy. That's how we get the equation above.

Now, what's the probability no one has the same birthday?

it would be (364/365)^(number of tests)

So the probability of at least two kids having the same birthday is 1 - the probability no kids have the same birthday.

(364/365)^(((N^2) - N))/2) ~ 0.5 for N = 23

Illkillya

The odds of being born on a particular day of the year are not completely random - e.g., David McWilliam's book "The Pope's Children" says that there was a huge spike in births 9 months from the day of the Pope's visit to Ireland in 1979. There are also some other factors (e.g., having a set of twins in the class).

Having said that, even money is definitely not right, should be much longer odds than that.

charles_92688

Illkillya wrote: »

Having said that, even money is definitely not right, should be much longer odds than that.

It's better than a 50/50 chance with 25 random people.

scudzilla

charles_92688 wrote: »

Yes way. In fact, it's about 50/50 with 23 (actually 22.4) kids. With 25 kids it's better than 50/50

Look at at it this way. Any two kids have a 1/365 chance of the same birthday, or 364/365 of not the same birthday. How many tests are there? It would be

((N^2) - N))/2 tests for N kids in class. Imagine an N by N matrix. Each kid asks every other kid his birthday. Each kid can't ask him self and billy asking johnny is the same as johnny asking billy. That's how we get the equation above.

Now, what's the probability no one has the same birthday?

it would be (364/365)^(number of tests)

So the probability of at least two kids having the same birthday is 1 - the probability no kids have the same birthday.

(364/365)^(((N^2) - N))/2) ~ 0.5 for N = 23

Can somebody please tear this theory a new hole

charles_92688

scudzilla wrote: »

Can somebody please tear this theory a new hole

Do it yourself. Run a Monte Carlo or ask a local school teacher.

event

yeah, first year in college our maths lecturer asked us this in a class of 30. we all said no way.
there were 3 of us with the same birthday day.

go figure.

Jesus Wept

It's called The Birthday Problem. I read about it 10 years ago in a book by a well known road gambler who claimed to have hornswaggled money out of people with the bet.

If you picked a specific date, then the odds wouldn't be good that 2 people in 25/30/etc would share it.

But if the question is, will any two people share the same birthday in the group, it becomes quite likely...

scudzilla wrote: »

Are you having a laugh?? I'm no mathematical genius but there's 365 days in a year, so the chances of any 2 of 25 kids having the same birthday has gotta be somewhere near the 15/1 mark

scudzilla wrote: »

But there's 365 different days in a year, and only 25 kids, no way will this be a 50/50 shot on 2 kids having the same birthday

scudzilla wrote: »

Can somebody please tear this theory a new hole

There aren't just 25 combinations, each persons date goes up against everyone elses.

Anyhow, the wiki explains it well.

red_fox

charles_92688 wrote: »

Now, what's the probability no one has the same birthday?

it would be (364/365)^(number of tests)

I don't have time to think about this now but surely that's not right, with 367 kids (allowing for a leap year) then the pigeon hole principal will guarantee that at least two birthdays fall on the same day - but the number above is always above 0...

The tests are not independent, say you have Ann, Beth and Cathy, then A=B, and B!=C gives A!=C, and you can't have more than 366 people with different birthdays.

Edit: I'm lazy, wikipedia, the probability of having n distinct birthdays:


Edit 2: I would rather that first 1 be expressed as 1-(0/365) for the sake of clarity, then the expression is Product[ 1 - (i/365)] for i=0..n-1. (and this zero for n>365)

Yakuza

charles_92688 wrote: »

Put it this way, imagine an elementary school or any classroom really with 25 kids in it. That's probably a typical number of kids. Think there's at least a 50/50 chance that there are two kids in that class that have the same birthday?

If you were in a room with someone whose birthday is September 26, would the chances of their sharing a birthday with you be 100%?

I'm not actually looking for an answer, but it never ceases to amaze me that in this day and age, people use usernames with what could be construed as a date of birth in them, giving a potential hacker some an extra vector of attack. Just sayin', is all.

Welcome to Boards:)

alan4cult

The reason most people find this problem hard to grasp is because they ask the question:

"What is the probability somebody in this room shares their birthday with me?"

when the question really is:

"What is the probability that any two people in this room share the same birthday?"

In a group of 23 people there are 253 different pairs of people you can make however only 22 of them contain you...!

finlma

As a teacher I can verify that for the last 5 years there have been at least 1 set of "birthday twins" each year. This is true.

Turtwig

Instead of scoffing about it why not just experiment? If it's true then any game in euro 2012 will have two players sharing the same birthday. 99% guaranteed! This has been true in every game so far.

equivariant

If we chose n (n < 366) people at random, the probability that no two of them share the same birthday is

[latex] \frac{365!}{365^{n-1}(366-n)!}[/latex]

For larger values of n, we approximate this number by [latex]e^{-\frac{n^2}{730}[/latex]

These formula ignore the (small) effects of leap years and the (slightly larger) effects of a non uniform distribution of birthdays.

charles_92688

Yakuza wrote: »

If you were in a room with someone whose birthday is September 26, would the chances of their sharing a birthday with you be 100%?

I'm not actually looking for an answer, but it never ceases to amaze me that in this day and age, people use usernames with what could be construed as a date of birth in them, giving a potential hacker some an extra vector of attack. Just sayin', is all.

Welcome to Boards:)

92688 is my zip code in Las Flores, Orange County, California.

charles_92688

Jernal wrote: »

Instead of scoffing about it why not just experiment? If it's true then any game in euro 2012 will have two players sharing the same birthday. 99% guaranteed! This has been true in every game so far.

How many people on a soccer team? 11? (I forgot) so that is N = 22. And for my calculations, 50/50 was at N ~22.4 so there is slightly less than a 50/50 chance of any two soccer players on the "pitch" having the same birthday.

Turtwig

charles_92688 wrote: »

How many people on a soccer team? 11? (I forgot) so that is N = 22. And for my calculations, 50/50 was at N ~22.4 so there is slightly less than a 50/50 chance of any two soccer players on the "pitch" having the same birthday.

Sorry I rushed that comment I meant that you'd include the entire squads of both teams playing so you'd have just under fifty people. (According to wiki 46 players) Either way it's been true so far.

doc_17

23 people? Take person no 1. They can pair themselves up with 22 other people. Person no 2 can pair up with 21 others, making 43 different unique pairs. person no 3 can pair off with 20 thers etc.

I love using this every year with a different class. Class size is roughly 20-25 so y success rate is roughly 50%.

Isn't there soething else about only needing 7 people to have a very good chance of 2 of them being born with a week. That one works every time for me.

2rkehij30qtza5

A mathematics lecturer did this with us when we were in first year studying maths in then UCG!!! He asked the lecture hall that question and then proceeded to ask the front row of the lecture hall to say their birthdays!!! I would Definately take the bet!!!!