Probability that a group of people all shares a 'free' day in their schedule? My old stats teacher shared this 'challenge' problem with me, and it has me pretty stuck:
"Suppose 5 people are trying to find a date they can all meet together. There are 22 possible dates in question, and each person has 5 random dates that will NOT work for them. What is the probability that they can find a date that works for everyone?"
Each person's dates are independent from each other, so multiple people could have the same date(s) be unavailable.
I figure this could be generally framed as P(at least 1 date works for all), which should be equal to 1-P(no date works for anyone).
So far, I think I've worked out that, for any one day, the probability that it doesn't work for at least 1 person is 0.2755 [1-(P(Doesn't work for 1 person))^5 = 1-(17/25)^5 = 0.2755].
From there, though, I'm not really sure where to go. How do I apply this to find the probability that EVERY date doesn't work for at least 1 person. I don't imagine it's as simple as just taking 0.2755^22, is it?
 A: We can solve this by inclusion-exclusion. First, add up the probability that a particular date works over all $22$ dates. Then subtract off the overlaps: the probability that two particular dates both worked, summed over all $\binom{22}{2}$ pairs of dates. Then add back in the triple overlaps, and so forth.
If we choose a particular set of $k$ dates, the probability for someone to have all $k$ of them free is $\frac{\binom{22-k}{5}}{\binom{22}{5}}$, so the probability that all $5$ people have all $k$ of those dates free is $\frac{\binom{22-k}{5}^5}{\binom{22}{5}^5}$. Finally, there are $\binom{22}{k}$ ways to pick $k$ dates to apply this calculation to. Therefore, in the $k^{\text{th}}$ level of the inclusion-exclusion calculation, we will be adding or subtracting $\binom{22}{k} \cdot \frac{\binom{22-k}{5}^5}{\binom{22}{5}^5}$.
For our final answer, we take the sum
$$
   \sum_{k=1}^{17} (-1)^{k+1}\binom{22}{k} \frac{\binom{22-k}{5}^5}{\binom{22}{5}^5}.
$$
The factor of $(-1)^{k+1}$ means we add the $k=1$ term, subtract the $k=2$ term, and so forth, just as the inclusion-exclusion principle tells us. We go up to $k=17$ because that's the highest value of $k$ for which an overlap is possible: we cannot have $18$ dates that all work for everyone, because each person only has $17$ dates that work for them.
Mathematica tells me that this sum evaluates to $\frac{458964963421}{458967886146}$ or about $0.999994$. (Intuitively, we should expect a probability close to $1$: the $5$ people only veto $25$ dates total, counted with multiplicity, so they'd have to try pretty hard to rule out all $22$ dates in this way.) Mathematica has not found a general formula for this sum if we replace the parameters ($22$, $5$, and $5$) by variables.
A: Before continuing, it is worth checking if this is a trick question.  Is it both possible that they can find a date that works for everyone and is it possible they can not find a date that works for everyone.  Questions like this which are trick questions are prime candidates to be solved by pigeonhole principle.
Sadly, both are possible... so the probability will be nonzero and will be less than $1$, as evidenced by the two extreme examples where all five people's bad days are the days 1-5, as well as the example where the 1'st person's bad days were 1-5, second person's bad days were 6-10, third person's were 11-15, fourth were 16-20, and fifth were 21,22, and three others.

 A proof that it would always be possible might have been phrased something like "let there be a box for every date and let each person receive $17$ balls.  Let them place a ball in each box corresponding to a date they are available.  The average number of balls in the boxes is $\frac{17\cdot 5}{22}\approx 3.86$.  Since there are always a whole number of balls in each box, that means there must necessarily be some boxes with $4$ or more balls in them and others with $3$ or less."  (It is at this point that if the average number of balls was strictly greater than four, that we could have concluded that there must have been a box with five balls and been done... saying that there must always be a date good for all people involved.  Sadly, that failed here).


As to your attempt.  You are correct that in problems like this it might be worth looking at the complement.
The probability that at least one date works for everyone is indeed going to be $1$ minus the probability that there are no dates that happen to work for everyone.
You are correct that the first day is disliked by at least someone with probability $1-\left(\frac{17}{22}\right)^5\approx 0.7245$.

 In fact, by linearity of expectation we can multiply this by $22$ to get the average number of days who are disliked by at least one person to be $\approx 15.939$ which could have been useful to say that this implies the existence of an arrangement of preferences such that there were $15$ or fewer days which had at least one person disliking it, thus proving that the probability there is a day liked by all must be nonzero (in case if we were too tired to come up with our examples at the start, or if they were too hard to spot).  See more techniques and problems solved by techniques like this in a course titled 'Probabilistic Methods'.

The really difficult part comes next.  We want to try to use this to come up with a probability that all days are disliked by at least someone.  IF one day being disliked by at least someone is a truly independent event of another day being disliked by at least someone... then it would have been as simple as taking this probability and raising it to the power of $22$... however, that is not going to be the case here.  These are not independent events.  Still... this isn't the most horrible guesstimate if we wanted to be incredibly quick and dirty about this.
Instead, perhaps let us try looking at "The first day works for everyone or the second day works for everyone or the third day works for everyone etc..." and breaking this apart by the inclusion-exclusion principle.
The first day works for everyone with probability $\left(\frac{17}{22}\right)^5$.  To explain this, we could see it a couple of different ways... One might be to imagine a bag of $22$ balls, each corresponding to a different date.  One of those balls is colored red and corresponds to the first day.  The first person draws $17$ balls.  The red was among the balls drawn with probability $\dfrac{17}{22}$.  They put it back and the next person has a go, and so on... multiplying these together.  If that is unsatisfactory, then consider the more tedious explanation of that there are $\binom{22}{17}^5$ different possible equally likely scenarios we could be in, and of those $\binom{21}{16}^5$ correspond to every person having the first date and 16 other random dates selected.  The arithmetic works out that $\dfrac{\binom{21}{16}^5}{\binom{22}{17}^5}=\left(\dfrac{17}{22}\right)^5$
The first two days work for everyone with probability $\left(\frac{17\cdot 16}{22\cdot 21}\right)^5$.  This can be seen by a similar argument as before.

At this point, before continuing I will introduce a new notation for falling factorials.  I will write $x\frac{n}{~}$ to represent $\overbrace{x(x-1)(x-2)\cdots (x-n+1)}^{n~\text{factors}}$, what some authors might have written as $~_xP_n$ or $P(x,n)$ or as $\frac{x!}{(x-n)!}$.  It will help simplify notation here.

We continue by noting that the first three days are all liked by everyone with probability $\left(\dfrac{17\frac{3}{~}}{22\frac{3}{~}}\right)^5$
In general, the first $k$ days are all liked by everyone with probability $\left(\dfrac{17\frac{k}{~}}{22\frac{k}{~}}\right)^5$.  Note that when $k>17$ the falling factorial on the numerator will include a factor of zero, implying it is impossible that more than $17$ days are liked by everyone.  This should be clear since each person has five days they dislike.
We can now put all of this together into a big formula.  Inclusion exclusion would have had us split apart a long string of "or"s as the sum of the probabilities if each event were taken one at a time... minus the sum of the probabilities if each event were taken two at a time, plus if taken three, minus if taken four, plus if taken five, etc...  The number of ways of taking these events $k$ at a time would be $\binom{22}{k}$
We have then:
$$22\cdot \left(\frac{17}{22}\right)^5 - \binom{22}{2}\left(\frac{17\cdot 16}{22\cdot 21}\right)^5 + \binom{22}{3}\left(\frac{17\frac{3}{~}}{22\frac{3}{~}}\right)^5-\dots \pm\binom{22}{k}\left(\frac{17\frac{k}{~}}{22\frac{k}{~}}\right)^5\pm\dots$$
$$=\sum\limits_{k=1}^{17}(-1)^{k+1}\binom{22}{k}\left(\frac{17\frac{k}{~}}{22\frac{k}{~}}\right)^5$$
At this point, we plug it into a calculator and get a final answer of $$\approx 0.99999363196\dots$$
