Understanding the "Birthday Problem" I found on this website http://www.cut-the-knot.org/do_you_know/coincidence.shtml proof that the probability of two people in a room having the same birthday equates to 50% when when there are 23 people in the room
The only thing is, the explanation is too complicated for the average Middle School student to comprehend. So, my question is, is there any simplifies explanation of how this works? I tried Wikipedia but the explanations were even more complicated.
Thanks!
 A: If you accept that having more than $2$ people with the same birthday is (for small numbers of people) much less likely than a single collision and can be ignored to within the accuracy used for this explanation/calculation, then the probability of a collision and the expected number of collisions are the same.  The latter is (number of unordered pairs of people)/365 = $\frac{n(n-1)}{2} \cdot \frac{1}{365}$ for $n$ people.  This will exceed $1/2$ when $n$ is about $20$.
The point is that the answer is approximately the square root of the number of days in the year, which much less than the intuitive estimate of half the number of days.  
A: I revised this Reddit comment that contains typos and wasn't formatted with MathJax.
Look at 2 dots. How many straight lines can be drawn between them? Only one.
Now add another dot to give you 3. How many straight lines can you draw between all the dots? Three
Four dots? Six.
Five dots? Ten
Six dots? Fifteen.
See the pattern? 1, 3, 6, 10, 15. . . You have a Triangular number. If you have 23 dots, there are 253 straight lines that can be drawn between all the dots, since 253 is the $23^{rd}$ number in the triangular sequence.
Now how does this relate to the birthday paradox?
Take two people. You're simply seeing if A has the same birthday as B. There's one relationship you need to look at.
Now with three people, you need to see if A has the same b-day as B, if A has the same b-day as C and if B has the same b-day as C. You're now looking at three relationships to see if there's a match.
With four people you look at AB, AC, AD, BC, BD & CD or six relationships for a possible match.
Every person you add increases the number of relationships (by the triangular number sequence) you need to examine to see if the people involved have the same birthday.
To figure out the combined probability of multiple people having the same birthday, it's easier to look at the odds of NOT having the same birthday and subtract that by 100%.
So with two people the odds of them not sharing a birthday are 364/365 = 0.9973 or a 99.73% chance. Subtract this from 100% and there is a 0.0027 or 0.27% chance they do share a birthday.
For three people, the odds that AB don't share a birthday is still 364/365, the odds for AC is also 364/365, and the odds for AC is also 364/365. To figure out the odds of $\color{red}{\text{NONE OF THEM SHARING THE SAME BIRTHDAY}}$, simply multiply them together to give you $\dfrac{364 \times 364 \times 364}{365 \times 365 \times 365}$,  which is the same as $(364/365)^3 = 0.9918$. So the odds of NOT having the same birthday is 99.18%. That means the odds of having the same birthday is $100 - 99.18 =$ 0.82%
Add another person and the odds of NOT having the same birthday is $(364/365)^6 = 0.9837$. Making the odds of anyone sharing a birthday 1.63%
Five people? $(364/365)^{10} = 0.9729$ or 97.29%, which means the chances of anyone sharing a birthday is 2.71%.
Six people? $(364/365)^{15} = 95.97%$, which means probability of anyone sharing a birthday is 4.03%.
By the time you have 23 people and are examining 253 relationships to see if any of them match, you have $(364/365)^{253} =$ 49.95% of nobody sharing the same birthday. That means that there is a 50.05% chance at least two people in that group have the same birthday.
