What are the chances that 5 people are all born on the same day#? Assuming 30-day months, given 10 people in a room.  What are the chances that 5 or more people are all born on the same day#?  (i.e., 5 born on the 28th, or 5 born on the 6th, etc)
(EDIT: changed from chances of 5 to chances of 5 or more)
I have tried two answers so far.
In the first, you pick any person, and see what the chances are of the other 9, then 8, etc to match the first.  This seems to be 10 * 9/30 * 8/30 * 7/30 * 6/30.
In the second, I suppose you could calculate the chances of 5 of the 10 having a birthday on day 1 + the chances of 5 having a birthday on day 2, etc.
These answers seem quite different.  What do you all think?
 A: You use the Binomial distribution formula here
p = 1/30, q = 1-p = 29/30
Probability 5 born on the same day
$$P(X=5) = {10 \choose 5}p^5q^5=8.75*10^{-6}$$
A: Suppose the max number of the ten who share one birthday is $n$.  Then $n=5,6,7,8,9,10$ constitute a partition of the event of interest.  To count the number of ways for fixed $n$, we first choose $n$ of the ten people to have one birthday, you can do that in ${10}\choose{n}$ ways.  Those $n$ can be assigned one birthday in $30$ ways.  The remaining $10-n$ can be assigned to other birthdays in $(29)^{10-n}$ possible ways (remember we're counting the number of ways to have exactly $n$ different days among the $10$ people).  But we have to avoid, when $n=5$, the case that we assign the other $10-n=5$ people to one birthday (otherwise we will double count it).  So for $n=5$ we have to correct by subtracting $29$ possibilities and then adding back all ways to have five for one day and the other five for another day, which happens in ${10\choose5}{30\choose2}$ ways.
Thus there are
$${10\choose5}(30)((29)^{5}-29)+{10\choose5}{30\choose2}+{10\choose6}(30)(29)^{4}+{10\choose7}(30)(29)^{3}+{10\choose8}(30)(29)^{2}+{10\choose9}(30)(29)+{10\choose10}(30)$$
ways to assign ten people to birthdays so that five share one day.  So just divide this by $30^{10}$ to get the probability.
This gives $0.0002702992287761012$ which agrees perfectly with Brian Tung's answer.  
A: Let us first calculate the chance that exactly five people share a birthday.  There are ${10 \choose 5}=252$ ways to choose the people who will share a birthday, $30$ ways to choose the birthday, and $29^5$ ways to choose the birthdays of the other people, giving $\frac{252\cdot 30 \cdot 29^5}{30^{10}}\approx 0.000262$.  This is not quite right because we have double counted the cases where there are two groups of five in the room, but that correction will be very small.  Since the chance of five is small the chance of six is smaller yet, so I would just ignore it.
A: it's simpler than you can imagine
it's the probability that 5 were born on the same day (1/30 ^ 5) plus the probability that 6 were born on the same day (1/30 ^ 6) ... until you reach 10
p(5 <= x <= 10) = ((1/30)^5) + ((1/30)^6) + ((1/30)^7) + ((1/30)^8) + ((1/30)^9) + ((1/30)^10)
= 4.25713069e-8

