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The original problem is: Flip a fair coin until the third head appears, and then stop right after that flip. What is the probability that it took you "i" flips to accomplish this?

Why is it: (combination of 2 out of (i-1))/2^i and not: (combination of 3 out of i)/2^i ?

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    The $i$-th flip is always the third heads. – 2017-01-12
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    HHHTT would not fit with i=5, but your second expression DOES count this sequence – 2017-01-12
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    Thanks Ritz, I've got the trick : the i-th flip is fixed, so you'll have to choose, from the remaining flips, 2 combinations out of i-1. – 2017-01-12

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The probability for head on any single attempt is $\frac12$.

You need to get:

  • $\color\red{\text{$2$ heads out of $i-1$ attempts}}$
  • $\color\green{\text{$1$ head at the $i$th attempt}}$

The probability for that is:

$$\color\red{\binom{i-1}{2}\cdot\left(\frac12\right)^{2}\cdot\left(1-\frac12\right)^{i-1-2}}\cdot\color\green{\left(\frac12\right)}$$

Which is equal to:

$$\binom{i-1}{2}\cdot\left(\frac12\right)^{i}$$

Which is equal to:

$$\frac{(i-1)(i-2)}{2}\cdot\frac{1}{2^{i}}$$

Which is equal to:

$$\frac{(i-1)(i-2)}{2^{i+1}}$$

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    Thanks Barak, I've got the trick: the ith attempt is fixed, so you'll have to choose, from the remaining flips, 3-1=2 combinations out of i-1. – 2017-01-12
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    @Hicham: You're welcome. – 2017-01-12