(1) Use Stirling's formula (see textbooks or handbooks in
mathematics) to show that
the combination number of choosing 50N^2 from
100N^2 50N^2 is approximately
(\pi/2)^{-1/2} 2^{100N^2+1} (10N)^{-1}
(2)
Prove that
the probability that the game continues at 100N^2-th round
is smaller than the number obtained by multiplying
2N/2^{100N^2} to the above combination number.
(3) Prove the statement by using (1) and (2)
Hint: You can use the expectation number of coin toss game of Problem 5. Think that you get 1 cent if you set a switch to a correct poisiton, and lose 1 cent if you set to a wrong position.