Expected Value with random variables for coin toss scenario

Question

Question:

I flip a fair coin, independently, 6 times, resulting in a sequence of heads (H) and tails (T). For each (consecutive) HTH in this sequence, you win $5.

Define the random variable X to be the "amount of dollars that you win".

For example, if the sequence is THTHTH; then X=10. What is the expected value of X?

Answer: $\frac{5}{2}$

Attempt:

I am very confused on how to approach this. Like do I have to count the amount HTH possible and find the probability of it for all $64$ sequences and mulitply by 5? Writing out all the sequences will take me forever to do...

Answer 1

For any three consecutive tosses, there is a $1/8$ chance of hitting HTH, so on tosses 1-3, you expect to win $5(1/8)$ dollars. Similarly for tosses 2-4, 3-5, and 4-6. By Linearity of Expectation, the expected winnings for the whole game is $4(5/8)=5/2$ dollars.

Answer 2

It is not that hard. Let $x$ be the random variable for the number of “HTH” in the sequence. Let $y$ be the corresponding income. Then, y=5x, so $\mathbb{E}(y)=5\mathbb{E}(x)$. Now, all you need to do is finding $p(x)$, which is essentially a combinatorics problem.