Did the teams cheat at this charity golf tournament? I was at a charity golf tournament yesterday, run by a very well known Fortune 100 company. There were 4 golf courses at the club. On each golf course there were 36 teams playing, so a total of 144 teams. At 12 holes on each of the golf courses, a charity organisation was stationed with a tent at the tees, so a total of 48 charities across the entire 4 courses. In order to encourage the teams to speak to the charities and find out more about each organisation, they put on a poker competition with prizes. Each of the 4 courses would have a prize for the best poker hand submitted at the end of the day. A deck of cards was given to each charity, and each team was to speak to the charities on the course and take a single card from each deck, as they passed each charity throughout the day.
Each of the 36 teams started with 2 teams to a hole on each of the 18 holes on that course. That way everyone moved from hole to hole around the course in the most efficient and timely manner. The 12 charities were spread across the 18 holes on each course, in no particular order. So each team ended up with 12 cards at the end of the day.
At the end of the day, the winning hand on each of the four courses was a royal flush. On announcement of the winners, the CEO remarked about needing to audit the process next year, due to the remarkable results - he naturally suspected there was some cheating involved. 
So, I'm curious - what is the probability that a royal flush hand would appear as the winning hand on each of the 4 different courses? 
Ask questions if you need clarity on anything. I'm curious to know the math used - I have basic understanding of combinatorics and probability, but this one is too complicated for me.
 A: Calculating the exact probability would be somewhat complicated, but we can get an upper bound that shows that this was exceedingly improbable.
The probability that there is at least one royal flush on a course is bounded above by the expected number of royal flushes on a course (see first moment method), and this in turn is the sum of the probabilities of forming a royal flush of all $5$-tuples of cards. There are $36$ teams per course and $\binom{12}5$ $5$-tuples per team, and the probability for any given $5$-tuple to form a royal flush is $4\cdot5!/52^5$ ($4$ for the $4$ suits and $5!$ for the possible orderings), so the expected number of royal flushes on a course is
$$
\frac{36\cdot\binom{12}5\cdot4\cdot5!}{52^5}=\frac{13365}{371293}\approx0.036\;.
$$
This is an upper bound for the probability of a course having a royal flush, and the four courses are independent, so an upper bound for the probability of all four courses having a royal flush is
$$
\left(\frac{13365}{371293}\right)^4\approx1.7\cdot10^{-6}\;,
$$
i.e. a bit more than one in a million.
