Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on fair coin.
A fair coin is to be tossed 10 times. Let i|j, in lowest terms, be the probability that heads never occur on consecutive tosses, find i+j.
Integers
Combinatorics
Algebra
Answer: is 73.
AIME I, 1990, Question 9
Elementary Algebra by Hall and Knight
5 tails flipped, any less,
by Pigeonhole principle there will be heads that appear on consecutive tosses
(H)T(H)T(H)T(H)T(H)T(H) 5 tails occur there are 6 slots for the heads to be placed but only 5H remaining, \({6 \choose 5}\) possible combination of 6 heads there are
\(\sum_{i=6}^{11}{i \choose 11-i}\)=\({6 \choose 5} +{7 \choose 4}+{8 \choose 3}+{9 \choose 2} +{10 \choose 1} +{11 \choose 0}\)=144
there are \(2^{10}\) possible flips of 10 coins
or, probability=\(\frac{144}{1024}=\frac{9}{64}\) or, 9+64=73.
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