Flipping a coin - expected value

Question

We have seven coins, two of them have tails on both sides. We choose (randomly) a coin and we flip a coin untill two tails appear (not necessarly in row). Calculate expected number of throws.

My attempt:

$\mathbf{P}\left( X=0\right) =\mathbf{P}\left( X=1\right) =0 \\ \mathbf{P}\left( X=2\right) =\frac{5}{7}\cdot\frac{1}{2^2} +\frac{2}{7} \\ \mathbf{P}\left( X=3\right) =\frac{5}{7}\cdot\frac{1}{2^3}\cdot {3\choose 2} \\ \mathbf{P}\left( X=n\right) =\frac{5}{7}\cdot\frac{1}{2^n}\cdot {n\choose 2}$

So the expected value is given by this formula:

$\mathbf{E}X=\lim_{n\to\infty}2\cdot\left(\frac{5}{7}\cdot\frac{1}{2^2}+\frac{2}{7}\right)+\sum_{k=3}^n \frac{5}{7}\cdot\frac{1}{2^k}\cdot{k \choose 2}\cdot k=\frac{13}{14}+\frac{5}{14}\sum_{k=3}^\infty\frac{k^2\cdot(k-1)}{2^k}$

Wolfram shows that this sum is equal about $4$, which seems to be fine, but I highly doubt that professor would give us such a sum on an exam.

Is there any other way to solve this?

Answer 1

You have $\frac 27$ chance of being guaranteed two tails in two throws. You have $\frac 57$ chance of expecting to take four throws, so the expected number of throws is $2 \cdot \frac 27+ 4\cdot \frac 57=\frac {24}7$ It should definitely be under $4$ as it would be four without the two two-tailed coins. Those will lower the expectation.