# Problem 589: Poohsticks Marathon
Christopher Robin and Pooh Bear love the game of Poohsticks so much that
they invented a new version which allows them to play for longer before
one of them wins and they have to go home for tea. The game starts as
normal with both dropping a stick simultaneously on the upstream side of
a bridge. But rather than the game ending when one of the sticks emerges
on the downstream side, instead they fish their sticks out of the water,
and drop them back in again on the upstream side. The game only ends
when one of the sticks emerges from under the bridge ahead of the other
one having also 'lapped' the other stick - that is, having made one
additional journey under the bridge compared to the other stick. On a
particular day when playing this game, the time taken for a stick to
travel under the bridge varies between a minimum of 30 seconds, and a
maximum of 60 seconds. The time taken to fish a stick out of the water
and drop it back in again on the other side is 5 seconds. The current
under the bridge has the unusual property that the sticks' journey time
is always an integral number of seconds, and it is equally likely to
emerge at any of the possible times between 30 and 60 seconds
(inclusive). It turns out that under these circumstances, the expected
time for playing a single game is 1036.15 seconds (rounded to 2 decimal
places). This time is measured from the point of dropping the sticks for
the first time, to the point where the winning stick emerges from under
the bridge having lapped the other. The stream flows at different rates
each day, but maintains the property that the journey time in seconds is
equally distributed amongst the integers from a minimum, \$n\$, to a
maximum, \$m\$, inclusive. Let the expected time of play in seconds be
\$E(m,n)\$. Hence \$E(60,30)=1036.15...\$ Let
\$S(k)=\\sum\_{m=2}\^k\\sum\_{n=1}\^{m-1}E(m,n)\$. For example
\$S(5)=7722.82\$ rounded to 2 decimal places. Find \$S(100)\$ and give
your answer rounded to 2 decimal places.