</div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Power PMV @ Jan 5 2004, 01:00 PM)</td></tr><tr><td id='QUOTE'> </div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (DB_Hunter &#064; Jan 5 2004, 09:44 AM)</td></tr><tr><td id='QUOTE'> </div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (samsonlonghair &#064; Jan 3 2004, 04:28 PM)</td></tr><tr><td id='QUOTE'> </div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Anbu Girl &#064; Dec 30 2003, 02:54 PM)</td></tr><tr><td id='QUOTE'> May. 15th. 1981

Wow same as u &quot;samsonlonghair&quot; </td></tr></table><div class='postcolor'>
That&#39;s pretty cool.

</div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (DB_Hunter)</td></tr><tr><td id='QUOTE'>Apparantly the probability of finding someone who shares your birthday is 1 in 14 i.e. in a room of 14 people you ar elikely to find one other person who has the same birthday as you.
</td></tr></table><div class='postcolor'>

How do you get that? I don&#39;t know what kind of math that is, but I would think that the odds are 1 in 365. </td></tr></table><div class='postcolor'>
I think the guy was a Maths Lecturer and I was at the London School of Economics (LSE) when I learnt this. </td></tr></table><div class='postcolor'>
well what you can prove is that in a room with more than 14 (or something like that) people the probability of 2 of them having the same birthday is greater that a half so it&#39;s a highly probable event. It&#39;s not hard to prove and if someone is interested I can do it....





....and yeah I&#39;m a math major. </td></tr></table><div class='postcolor'>
Say:

Red Lorry Yellow Lorry

As fast as you can without screwing up .