Three coins are set on a table all showing Heads. Each turn one flips one of them independently with a probability of $\frac{1}{2}$ for the 1st coin, $\frac{1}{3}$ for the 2nd coin and $\frac{1}{6}$ for the 3rd coin. Let $X_{n}$ be the number of Heads showing after $n$ turns. What is the expected value of $X_{3}?$
My attempt:
Given these, I realized the only values $X_{3}$ can get are $0$ and $2$, so I tried to calculate $\mathbb{P}\left(X_{3}=k\right)$ for $k\in\left\{ 0,2\right\} $:
- $\mathbb{P}\left(X_{3}=0\right)=\mathbb{P}\left(\text{All coins got flipped}\right)=\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}=\frac{1}{36}$
- $\mathbb{P}\left(X_{3}=2\right)=\mathbb{P}\left(\text{One of the coins got flipped twice}\right)=\mathbb{P}\left(\biguplus_{i=1}^{3}\text{Coin } i \text{ got flipped twice}\right)=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{6}\right)^{2}=\frac{7}{18}$
Therefore:
$$ \mathbb{E}X_{3}=\sum_{x\in\text{supp}\mu_{X_{3}}}x\cdot\mathbb{P}\left(X_{3}=x\right)=2\cdot\frac{7}{18}=\frac{7}{9} $$ Where $\mu_{X_{3}}$ is the distribution function of $X_{3}$.
Now one thing that bothers me here is that $$ \sum_{x\in\text{supp}\mu_{X_{3}}}\mu_{X_{3}}(x)\neq1 $$
So i'm thinking maybe I did something wrong with the calculations*
Any ideas?
The probability that all coins are flipped is: $$3!\times\frac12\times\frac13\times\frac16=\frac1{6}$$
Observe that $\frac12\times\frac13\times\frac16$ can e.g. be recognized as the probability that at first the first coin is flipped, then the second and at last the third. But there are $3!$ equiprobable orders leading to the same result.
So $P(X_3=0)=\frac16$.
Further it is clear $\{X_3=2\}$ is exactly the event that not all coins are flipped (if not all coins are flipped then it is inevitable that we end up with exactly two heads).
So $P(X_3=2)=1-\frac16=\frac56$.
That leads to: $$\mathbb EX_3=\frac16\times 0+\frac56\times 2=\frac53$$
edit:
Finding $P(X_3=2)$ by brute force gives:
$P(X_3=2)=3\times[\frac12\frac12(\frac13+\frac16)+\frac13\frac13(\frac12+\frac16)+\frac16\frac16(\frac12+\frac13)]+\frac12\frac12\frac12+\frac13\frac13\frac13+\frac16\frac16\frac16$
I haven't checked yet whether that indeed equals $\frac56$ .
Feel welcome to do and ring the alarm bells if it does not.