<p>ok, someone please help me with this math problem. If there are 40 kids in my class, what is the probability that 2 have the same birthday? (please show me how you did this)</p>
<p>I remember this problem from a theory of probability class I took, but I forgot the math behind it. Apparently it's about a 50% chance in a group of 23, so I'd assume its around 90-100% in a group of 40. Sorry I couldn't help more.</p>
<p>Math has never been my fort</p>
<p>^
Sorry, but that's not correct (any two of the kids could have the matching birthday, not just the 1st and the 40th like in your calculation)</p>
<p>I didn't do an exact calculation, but the probability in 40 people is about 90%</p>
<p>Check these websites for details:
<a href="http://www.mste.uiuc.edu/reese/birthday/%5B/url%5D">http://www.mste.uiuc.edu/reese/birthday/</a>
<a href="http://mathworld.wolfram.com/BirthdayProblem.html%5B/url%5D">http://mathworld.wolfram.com/BirthdayProblem.html</a></p>
<p>I don't have a calculator with me except the one on the computer, and I took Stats last year. Bear with me, and check my work!</p>
<p>Start by calculating the probability that all forty kids have different birthdays. It's easier. We're assuming that the probability of each birthday date is the same (except for leap years, which I'll get into).</p>
<p>The first kid can be born on any day: 365/365.
The second kid can be born on all but one of those days: 364/365
The third kid can be born on all but two of those days (the birthdays of kid 1 and kid 2): 363/365
Et cetera. </p>
<p>So the total probability of having forty different birthdays is:
(364<em>363</em>362<em>361</em> [...] *326)/(365^39) = p</p>
<p>So the probability that this is not the case, and at least two kids have the same birthday is: 1 - p = q</p>
<p>To get this more exact, you need to factor in leap year. If your teacher doesn't need this, just leave this part out. Leap years happen every four years except on the turn of the century excluding millenia (so there wasn't one in 1900 but there was in 2000 and was in 2004). </p>
<p>If you want to do this the easy way (ignoring the century stuff) the probability is simply 1/((4*365)+1) = 1/1461.</p>
<p>If you want to be more accurate and factor in the century and millenia stuff, then the probability is 97/((400*365)+97) = 97/146097. This is for 400 years worth of leap years, which is 97. This is just the number of years I sort of recall using in another problem once. </p>
<p>We will call either of these probabilities (1/1461 or 97/146097, whichever you use): r.
None of the forty kids has a birthday on this date with probability = (1-r)^40
The probability that two or more kids were born on leap year would then be 1-(1-r)^40.</p>
<p>Multiply this number by the number found in the first step, and you should get the right answer. I hope.</p>
<p>Didn't see MrMan's post before I posted. Sorry!</p>
<p>But MathWorld is right about everything, so do whatever they say.</p>
<p>Didn't Grammbla ask specifically for two kids to have the same birthday? I read through MathWorld, and the calculations shown there are for two or more kids.</p>
<p>Accounting for the issue of combinatorics, would it suffice to just add 40C2 to my calculations above?</p>
<p>thanks guys soo much!</p>
<p>
[quote]
Didn't Grammbla ask specifically for two kids to have the same birthday?
[/quote]
Since the number of students wasn't specified as being exact, it can be commonly assumed to mean that number or more. </p>
<p>I don't know if you used my method (or if it works!), but no problem, Grammbla.</p>