Problem: Find The expected number of coin flips until two heads in a row or three tails in a row occurs.
Answer: $\frac{21}{5}$.
This problem is not hard to just manually solve using states and creating a system of equations—I’ve already done this myself. However, it is a little computation-heavy.
It is well-known that the expected number of coin flips until $n$ heads in a row is $2^{n + 1} - 2$. This is easy to prove using states. So, the expected flips until two heads in a row is $6$, and the expected flips until three heads in a row is $14$.
Here’s my sketchy reasoning that leads to a correct answer: If the expected is $6$ for two heads, that means that the “probability” each time is $\frac{1}{6}$. If the expected is $14$ for three tails, then the “probability” each time is $\frac{1}{14}$. It’s impossible to work toward two heads in a row or three tails in a row at the same time. Progress can only be made to one at a time. So, the probability of finishing either is $\frac{1}{6} + \frac{1}{14} = \frac{5}{21}$. Thus, the expected is $\frac{21}{5}$.
Okay, my above reasoning does give me the correct answer, but it is obvious that basically every step was extremely dubious. I’m essentially using the “rule” that $$E(A_1 \text{ or }A_2\text{ or }\cdots \text{ or }A_n) = \frac{1}{\frac{1}{E(A_1)} + \frac{1}{E(A_2)} + \cdots + \frac{1}{E(A_n)}},$$ where $E$ is expected value, but this rule isn’t true in general for all events $A_i$. However, maybe there’s a way to prove this rule works for this specific scenario I’m using it for? I’m also thinking that there might be a way to use linearity of expectations to prove that my extremely bad reasoning actually works, but I’m not sure. Help would be appreciated.
