First head on the fourth toss

Question

You flip a fair coin repeatedly, stopping at the first head. What is the probability that the first head appears exactly on the fourth toss?

Accepted Answer

For the first head to land exactly on toss four, three specific things must happen and then one more: tosses one, two, and three must all be tails, and toss four must be heads. The flips are independent, so multiply their probabilities: P(	ext{TTTH}) = \frac12 	imes \frac12 	imes \frac12 	imes \frac12 = \left(\frac12ight)^4 = \frac{1}{16} = 0.0625. In general, the chance the first success happens exactly on try k is "fail, fail, ..., fail, then succeed," which is (1-p)^{k-1}\,p. Here that is (	frac12)^3 \cdot 	frac12 = 	frac{1}{16}. Takeaway. The probability the first success falls o…

First head on the fourth toss

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