What is the expected number of (fair) coin flips to get two consecutive heads?
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Let e be the expected number of tosses. It is clear that e is finite.
Start tossing. If we get a tail immediately (probability 12) then the expected number is e+1. If we get a head then a tail (probability 14), then the expected number is e+2. Continue …. If we get 4 heads then a tail, the expected number is e+5. Finally, if our first 5 tosses are heads, then the expected number is 5. Thuse=12(e+1)+14(e+2)+18(e+3)+116(e+4)+132(e+5)+132(5).Solve this linear equation for e. We get e=62.
Start tossing. If we get a tail immediately (probability 12) then the expected number is e+1. If we get a head then a tail (probability 14), then the expected number is e+2. Continue …. If we get 4 heads then a tail, the expected number is e+5. Finally, if our first 5 tosses are heads, then the expected number is 5. Thuse=12(e+1)+14(e+2)+18(e+3)+116(e+4)+132(e+5)+132(5).Solve this linear equation for e. We get e=62.
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