Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on fair coin.

Fair Coin Problem – AIME I, 1990

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.

  • is 107
  • is 73
  • is 840
  • cannot be determined from the given information

Key Concepts




Check the Answer

But try the problem first…

Answer: is 73.

Suggested Reading

AIME I, 1990, Question 9

Elementary Algebra by Hall and Knight

Try with Hints

First hint

5 tails flipped, any less,

by Pigeonhole principle there will be heads that appear on consecutive tosses

Second Hint

(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

Final Step

\(\sum_{i=6}^{11}{i \choose 11-i}\)=\({6 \choose 5} +{7 \choose 4}+{8 \choose 3}+{9 \choose 2} +{10 \choose 1} +{11 \choose 0}\)=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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