Expected value of a coin tossing game The game is like this:
I toss a coin and you guess whether it's a head or tail. We will play 10 rounds. If you correctly guess it, you get 1 point (wrong then 0), and, if you get the right guess consecutively, you will get 2,3,4, etc. That means if you get all 10 guesses correct, you will have 1+2+...+10 = 55 points.
For example, in the 10 rounds, your guesses are 'rrrwwrwrr' (r=right, w=wrong).
then you will get 1+2+3+0+0+1+0+1+2 = 10points
I found it is quite difficult to compute the expected value of this game. Any idea?
 A: You can think of it as getting one point for each substring of consecutive correct guesses. (For instance, $3$ consecutive correct guesses contain $6$ substrings of correct guesses and yield $6$ points.) Thus, by linearity of expectation the expected number of points is the sum over all substrings of the probability that the substring consists entirely of correct guesses. Since there are $n-k+1$ substrings of length $k$ (with $n$ the number of rounds, in your case $n=10$) and a substring of length $k$ has probability $2^{-k}$ of consisting entirely of correct guesses, the expected value of the game is
$$
\sum_{k=1}^n(n-k+1)2^{-k}=n-1+2^{-n}\;.
$$
A: With @lulu suggestion, I worked out the answer recursively. @joriki solution is smart and concise.
Consider the expected value of an $n$ round game = $E_n$. Now let's look at n+1 th round: if the player guesses wrong, then he will get 0 additional point. If the player guesses right, he gets additional points which depends on how many consecutive wins before the n+1 th round.
The number of permutation of getting n+1, n, n-1, ..., 1 consecutive wins (including the n+1 th one) will be: $1, 1, 2, 2^2,...,2^{n-1}$, so the additional points will be $1(n+1), 1(n), 2(n-1), 2^2(n-2),...,2^{n-1}$. The total permutation of the first n guesses = $2^n$.Therefore the additional expected value contributed by the n+1 th correct guess is
$$ \frac{1}{2^n}[(n+1) + n + 2(n-1) + 2^2(n-2) + ... + 2^{n-1}] = 2 - \frac{1}{2^n}$$
And because the probability of getting the n+1 th guess correct = $\frac{1}{2}$, so
$$
E_{n+1} = E_n + \frac{1}{2}(2 - \frac{1}{2^n}) = E_n + 1 - \frac{1}{2^{n+1}}
$$
With $E_1 = \frac{1}{2}$, we can arrive to 
$$
E_{n} = n-1-\frac{1}{2^n}
$$
which agrees to @joriki solution. Thank you all for the help!
