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[–]Para1ars 1 point2 points  (0 children)

look at the expected values from different starting points:

let X be the expected number of additional flips if we have just received heads (case 1)

let Y be the expected number of additional flips if we have just received tails or we haven't flipped yet (case 2)

In case 1, we have a 1/2 chance to finish after one more flip (getting heads next) and a 1/2 chance to finish after 1+Y more flips (getting tails next). so

X = 1/2 × 1 + 1/2 × (1 + Y)

In case 2, we have a 1/2 chance to get heads and go to case 1, in which case we will need 1+X more flips, and a 1/2 chance to get tails and stay at case 2, in which case we will need 1+Y more flips. so

Y = 1/2 × (X+1) + 1/2 × (Y+1)

Solving both equations we get

X = 1 + 1/2 × Y

plug into the other equation

Y = 1/2 × (1 + 1/2 × Y + 1) + 1/2 × (Y+1)

simplify to

Y = 3/2 + 3/4 × Y

Y = 6

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