Neighbor Interaction in a Random List Assuming a random arbitrarily long list where each element has a $50\%$ chance of being a $0$ or a $1$, such as:

$0001101101$

What is the chance of having a neighbor that isn't the same?  For example, in the above list the first and second item have no different neighbor but the third and fourth do.

I'm working on this for a project dealing with AB testing in a stream of items and I realized there might be interaction between the two sample sets but I'm not sure how much on average (and math class was a very long time ago so I'm rusty).  My instinct says $50\%$ but I don't trust my instincts.
Consider the following scenarios for an interior digit $d$ (i.e. not the first or last digit):


$D_1$: Exactly one neighboring digit is different from $d$
$D_2$: Exactly two neighboring digits are different from $d$
$D_{\ge1}$: At least one neighboring digit is different from $d$ (this is what you're looking for)


We can use some simple counting arguments to find that $P(D_1)=\frac{1}{2}$ and $P(D_2)=\frac{1}{4}$. From here, just note that $P(D_{\ge1})=P(D_1)+P(D_2)$ since $D_1$ or $D_2$ would qualify as a success condition, and we get our answer:

$P(D_{\ge1})=P(D_1)+P(D_2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$.

For exterior digits, $D_1$ is the only possible success condition, so we have $P(D_1)=\frac{1}{2}$.

Edit: Alternatively, for an interior digit $d$, consider the scenario $S_2$: Exactly two neighboring digits are the same as $d$. A simple counting argument shows that $P(S_2)=\frac{1}{4}$. Since $D_{\ge1}=\lnot S_2$, we have

$P(D_{\ge1})=1-P(S_2)=1-\frac{1}{4}=\frac{3}{4}$
To have a neighbor that isn't the same, assuming you pick beyond the first digit and before the last digit, is given by,

(Probability of a neighbor being different)=(probability of two digits not being the same)+(probability of one digit not being the same)=(probability of a digit not being the same)*(probability of a digit not being the same)+(probability of digit not being the same)

In math symbols,

$$P=(1-{1 \over 2})^2+(1-{1 \over 2})={3 \over 4}$$

So there is a ${3 \over 4}$ chance of you getting a neighbor that is different.

If you look at the first or last digit, the probability is indeed ${1 \over 2}$. Can you think of why?