The birthday problem and the lottery

[Frank]

Heres a conundrum. Imagine a room full of people who chat and ask each other what is the date of their birthday. What is the minimum number of people there would need to be in the room such that the probability of two people having the same birthday was 50% ?

It a well known conundrum which leads on to lotteries and how many draws need to be out before a result repeats.

I won't tell you the answer yet and if you don't know it some of you might like to think about it . Some of you might google it anyway. (Better if you don't as its good food for the brain).


Anyway I'll be back after a while and we'll talk about how this affects lotteries.

[Frank]

Its a pity theres no spoiler facility in this forum, I would have used it here.

You might be tempted to think linerarly and think that about half of 365 people , say 160 need to be in the room. You'd be wong. The answer is 23 !

How can so few people when comparing birthday dates have a 50% chance of 2 people having the same birthday ? Doesn't seem possible does it ?
It all comes down to comparisons, and how many comparisons are being made.

Lest consider the first six people to enter the room.

person A compares birthday with persons B,C,D, E and F thats 5 comparisons.
person B compares birthday with persons C,D , E and F thats 4 comparisons.
person C compares birthday with persons D , E and F thats 3 comparisons.
person D compares birthday with persons E and F thats 2 comparisons.
person E compares birthday with person F thats 1 comparison.

So thats a total of 5+4+3+2+1 =15 comparisons of birthdays for 6 in the room.

I think you can see a pattern forming here.... So when we get to 20 people in the room, you can work out that there are (add 1 to 19) =190 comparisons. Already thats more than half the 365 possible birthday dates. In fact a quick approximation to the answer can be obtained by taking the square root of the possibilities (365) and adding 17% to that result to give 22.
A good explanation is here:- http://betterexplained.com/articles/unde...y-paradox/

For a more complicated but exact method of arriving at the answer of 23 by plotting on a graph you can read it here..http://mathforum.org/dr.math/faq/faq.birthdayprob.html

The good news is that you can use a spreadsheet to do the calculations. Use each row of the sheet to work out the probability of no 'n' people sharing the same birthday. Have a different value of n on each row. This means that the probability of n people sharing the same birthday is 1- that value.




You can read off when the probability of NOT sharing a birthday is 49.27% which means that the probability of at least 2 people sharing a birthday is 100-49.27%=50.73%.

Okay, I'll leave you with a thought. How many lotto results need to be out such that the probability of a result repeating is 50% ?

[Frank]

In response to the millions of people enthusing about this thread, not letting me get a word in edgeways, reponding with guesses and suggestions about this revelation, I have no further comment.
Here >https://spreadsheets.google.com/ccc?key=...l=en#gid=1 Work it out for yourselves.