At least how many persons are required in a group so that the probability of two persons were born on the same day of the week is 0.5? 
At least how many persons are required in a group so that the probability of two persons were born on the same day of the week is $0.5$?

I simplified it so there's only 365 days. Next find the probability that no one is born on the same day of the week then subtract by $1$ to get the solution. 
But the answer is wrong, so I'm kind of out of ideas.
 A: Given a group of $n$ people, the sample space of days of the week in which they were born is $\Omega = \left \{ 1, ..., 7 \right \}^n$, which give $\left | \Omega \right |=7^n$.
Let $A$ be the event that at least two of them were born on the same day. Then $P(A)=1-P(A^c)\ $. Now $A^c$ is the set that none of them share a birthday, so you need to pick $n$ different days out of $7$. This can be done in ${7 \choose n}$ ways. Assuming the people are ordered, given a set of $n$ days we have to consider all permutations, hence:
$$P(A^c)={n!{7 \choose n} \over 7^n}$$
Notice that if $n≥8$ then obviously at least two share a birthday, which agrees with our formula since $P(A^c)=0$ for $n>7$. Now use the formula above to compute the probabilities for $2≤n≤7$.
A: This is a variant on a very classic question known as the "Birthday Paradox."  Instead of trying to calculate the probability that two people have the same birthday, consider the probability that two people have different birthdays for a given size group.  If you can get this probability under .5, then you know the probability of two people having the same birthday is greater than .5
