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4 Answers

Best answer
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53 votes

Answer is (C)

Let the expected number of coin flips be $X$. The case analysis goes as follows:

a. If the first flip is a tails, then we have wasted one flip. The probability of this event is $\frac{1}{2}$ and the total number of flips required is $X+1.$
b. If the first flip is a heads and second flip is a tails, then we have wasted two flips. The probability of this event is $\frac{1}{4}$ and the total number of flips required is $X+2.$ as the same scenario as beginning is there even after 2 tosses.
c. If the first flip is a heads and second flip is also heads, then we are done. The probability of this event is $\frac{1}{4}$ and the total number of flips required is $2$. 

Adding, the equation that we get is -
$X = \frac{1}{2}(X+1) + \frac{1}{4} (X+2) + \frac{1}{4}2$

Solving, we get $X = 6$.

Thus, the expected number of coin flips for getting two consecutive heads is 6.

edited by
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9 votes

Check the first answer from William Chen,

on -> https://www.quora.com/What-is-the-expected-number-of-coin-flips-until-you-get-two-heads-in-a-row

link found by @Anurag Pandey !

Nice answer !

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