1. ## Football

I was at a football (soccer) match the other day, it was a local derby and there was a great drink up/party afterwards. It was a very special party so only the referee and the two sides were invited. So only 23 people attended.

One of the players shouted "It's my birthday today, the drinks are on me" but then another player shouted " It's mine too, we can share the bill"

What are the chances that of 23 people, two share the same birthday?

2. ## Re: Football

<span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Almost even odds:
about 51% at least 2 will share, 49% all different</span hide>

Or are you asking the odds that exactly 2 will share?

Steve

3. ## Re: Football

Hi Steve

No that will do. I was rounding up and down so would have been happy with 50% <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>

4. ## Re: Football

I recall hearing some stat that proved that there is close to a 100% chance that in a group of 100 people there will be at least 2 with a birthday on the same day! I dunno how accurate that is..but I remember reading it somewhere!

5. ## Re: Football

Weeeeellll Thank you Rudi for that <img src=/S/bubbles.gif border=0 alt=bubbles width=31 height=17>

6. ## Re: Football

Are you sure that it isn't the story about 365 people in a bar, none of whom has a birthday in common with any other person in the bar?
The next person who comes in has a 100% chance of having a birthday in common with one of the persons already in the bar*!

John

(That's probability for you!)

* except in a leap year, of course...!

7. ## Re: Football

The probability that at least two persons in a group of n people have the same birthday = 1 - the probability that they all have different birthdays (1 = 100%).
Ignoring leap years and assuming that birthdays are distributed evenly over the year, the probability of different birthdays is

(365/365) * (364/365) * (363/365) * ...

with n terms in the multiplication. It's easy to create an Excel worksheet to calculate these probabilities. It turns out that the probability of all different birthdays is:

less than 50% from n=23
less than 10% from n=41
less than 1% from n=57
less than 0,1% from n=70
and of course exactly 0% from n=365

8. ## Re: Football

My calc assumed 366 possible birthdays, and I did not take into the prob of only 1 leap year in about 4 years. So my answer was based on a "366" denominator, so will be a little different than the answer derived from this...

Steve

9. ## Re: Football

Phew...thank goodness I didn't say its my story! <img src=/S/laugh.gif border=0 alt=laugh width=15 height=15>

Thanx for the stats guys! In future I will not believe EVERYTHING I hear! <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>

10. ## Re: Football

My high school maths teacher use to win bets with this question. The only time he got close to losing was when he had to use his own birthday to match with one of the kids.

11. ## Re: Football

There was a guy at work who had heard about this and he liked to test it. The problem was he would test it on the same group of people (the people from work who got together). My wife and and other of the chemist's wife had the same birthday, and it screwed up the probability. There could be just him and his wife, my wife and I, the other chemist and his wife (6 people) and people had the same birthday all the time...

12. ## Re: Football

duh! Maybe he should test it with his twin!

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