<p>Exactly ;]</p>
<p>^^ lol. 10char</p>
<p>lol, I'm stupid</p>
<p>but yeah, the Dr.Math/wiki things are very succinct and easy to understand...</p>
<p>Hell, I looked at the wiki and I still don't get it.</p>
<p>Kaznack - do the Dr. Math thing and read down to the 3 examples. That's the only thing that could really make me understand it.</p>
<p>Blackjack (21) is not a poker game.</p>
<p>Fail.</p>
<p>^ I meant like a gambing/casino game. Poker=gambling game in my book, lol. Sorry to disappoint you with my poor defining skills.</p>
<p>
</p>
<p>Assuming 365 days in a year, 184, right?</p>
<p>i think moodret's is somewhere around 30, if i remember correctly...</p>
<p>Yeah, you're right. I forgot probability like that is aggregate. I don't feel like adding up a bunch of fractions, though.</p>
<p>It's actually 23! :) </p>
<p>Here's how I computed it:
First find the probability that no one in the group has a birthday on the same day.
For two people NOT to have the same birthday, it is:
365/365 * 364/365 (which should be easy to see)
For the four people, this is simply:
365/365 * 364/365 * 363/365 * 362/365 </p>
<p>Now generalizing for n people:
(365! / (365-n)!) / 365^n </p>
<p>The complement of this is the probability that at least two people do have the same day:
1 - (365! / (365-n)!) / 365^n </p>
<p>Solving the inequality
1 - (365! / (365-n)!) / 365^n > .5
gives 23 as the lowest integer for which this is true. </p>
<p>For n=184, this is:
(1 - 2.383*10^-25)
which is almost 100%.</p>
<p>For anyone who is curious, n=15 gives 25% chance, for n=32 it jumps up to 75%!</p>
<p>Exactly, that is what I realized I had to do. Thanks for actually solving it, though. It was nice to know the answer.</p>