Birthday Paradox expected value, only Monday Using an indicator variable, how many people do you need in a room so that you expect that there is going to be a shared birthday on a Monday of the year (assume 52 Mondays in a year and 365 days in a year)?
How do you approach this question? 
I am thinking trying to use an indicator variable and modifying the original birthday paradox. I got 73 as an answer but I am not sure if I'm correct.
Edit: I was doing $\binom{k}{2}\frac{1}{365}\frac{52}{365} = \frac{k(k-1)}{2*365}\frac{52}{365}$ or basically multiplying the original birthday paradox's probability by 52/365.
 A: In order to solve the problem, I assume the simplification that a year has $364=52*7$ days. If not, the result depends on the calendar year, because it affects the various probabilities and the possible leap year.
Suppose we have $N$ people. Given the question, we are only interested in the people that have their birthday on a Monday this year. Therefore we need the probability $P(N,k)$ that $k$ people of $N$ have their birthday on a Monday:
$$
P(N,k) = \binom{N}{k} \left(\frac{1}{7}\right)^k \left(\frac{6}{7}\right)^{N-k}
$$ 
The next thing we need is wether within these $k$ people there are two (or more) that have their birthday on the same day. It is however easier to compute the probability that this is not the case. This is given by:
$$
\prod_{i=0}^{k-1} \frac{52-i}{52} = \frac{52!}{52^k (52-k)!}
$$
because there are 52 possible days for the first person, 51 for the second and so on.
Combining this gives us the probability $P(N)$ of having two or more people, among a total of $N$, to have their birthday on the same Monday:
$$
P(N)= \sum_{k=0}^N P(N,k) \left[ 1 - \frac{52!}{52^k (52-k)!} \right]
$$
where we now have to sum over every possible value of $k$. Using that $\sum_k P(N,k)=1$ this gives as a final result:
$$
P(N) = 1 - \left(\frac{6}{7}\right)^N \sum_{k=0}^N \binom{N}{k} \frac{52!}{6^k 52^k (52-k)!}
$$
If we now compute this for some values we obtain:
$$
P(62) \approx 0.494809
$$
$$
P(63) \approx 0.505526
$$
So we would need 63 people to have a probability larger than 50%.
If we assume a particular with 52 Mondays out of 365 days, the only thing that changes is the probability to select $k$ people out of $N$. This would become
$$
P(N,k) = \binom{N}{k} \left(\frac{52}{365}\right)^k \left(\frac{313}{365}\right)^{N-k}
$$
and for $P(N)$ we would get
$$
P(N) = 1 - \sum_{k=0}^N \binom{N}{k} \frac{313^{N-k} 52!}{365^N (52-k)!}
$$
The corresponding probabilities would be
$$
P(62) \approx 0.492975
$$
$$
P(63) \approx 0.503675
$$
The criterion that is used to determine wether there are enough people is $P(N) \geq 50%$. It is than more likely that such a pair exist than for such a pair not to exist.
Using the same approach, we can also compute the expected number of pairs of people that share their birthday on Monday. We already have the probability that $k$ people out of $N$ have their birthday on a Monday, so we only need to find out how many pairs among those $k$ share it on the same day as well. So consider an arbitrary couple $(i,j)$ out of the $k$ members of this class. The first person has his birthday on the $b_i$th Monday of the year. The chance that the second person shares that day is than $\frac{1}{52}$. Since there are $\binom{k}{2}$ such pairs, one would expect that there are $\binom{k}{2} \frac{1}{52}$ pairs in total. This is not guaranteed, it could be more or less, but if you are to repeat the process over and over again for many uncorrelated groups, that is what one would get. For the expected number of pairs among $N$ persons you would therefore get:
$$
E(N) = \sum_{k=0}^N P(N,k) \binom{k}{2} \frac{1}{52} = \dots = \binom{N}{2} \frac{52}{365^2}
$$
You can obtain the results faster by realising that for each of the pairs $(i,j)$ in $N$ people, of which there are $\binom{N}{2}$ the first needs to have his birthday on a Monday, with probability $\frac{52}{365}$ and the second has a probability of $\frac{1}{365}$ to share the exact same day.
The criteria one would use here to determine wether there are enough people, would be $E(N) \geq 1$.
Apart from the fact that the formulas are rather different, a simpler example might be more helpful. Consider a fair coin and compare the probability that there are one or more pairs of head to the expected number of pairs of heads among $N$ coin flips and take $N=3$.
Of the 8 possible outcomes : HHH,HHT,HTH,THH,HTT,THT,TTH,TTT there are 4 that contain one or more pairs, and hence the probability of having at least one pair of heads is $\frac{4}{8}=\frac{1}{2}$.
Now for the expected number of pairs. The triplet $HHH$ contains three pairs, and the combinations  HHT,HTH,THH each contain a single pair, giving a total of 6 pairs. Each combination has a probability $\frac{1}{8}$ and hence the expected number of pairs would be $E(3)=\frac{6}{8}=\frac{3}{4} < 1$. Which clearly shows that these criteria are different.
A: If there are $n$ people in the room, the expected number of pairs of people sharing a Monday birthday is $\lambda = \binom n 2 52/365^2$, and the number of such pairs is approximately Poisson distributed.  So I would (in a rule of thumb, back-of-the-envelope kind of of way) solve the equation $\binom n 2 = 365^2/52$ or the equation $1/2 = \exp(-\binom n 2 52/365^2).$
