Conditional probability - voting There is a population that consists of a mixture of "Unchangeables" and "Changeables". If you choose a person at random, then the probability that they're an Unchangeable is $p$ and the probability that they're a Changeable is $1-p$.
Every person is asked repeatedly to vote either "yes" or "no" for a proposal (which doesn't change). Unchangeables will always vote the same way on the same proposal. However, each time a Changeable votes, the probability that they will change their mind from their previous vote is $r$ and the probability that it will stay the same is $1-r$. (That is for example if a Changeable votes "yes" one time, then the probability that they vote "yes" next time is $1-r$ and "no" is $r$).
A randomly chosen person is noticed to have voted the same way on the proposal twice in succession. What is the probability that they will vote in the same way next time?

So far I've got that there are 4 possible categories of people, each combination of Unchangeable and Changeable that votes either twice the same or twice different in succession (at a particular instance), so this table should give the probabilities for getting each type of person if you pick someone randomly from the population.





twice same
twice different




Unchangeable
$p$
$0$


Changeable
$(1-r)(1-p)$
$r(1-p)$




That way, you'd expect to get $$P(\text{unchangeable } |\text{ voted same way twice}) = \frac{p}{(1-r)(1-p)+p}$$ and $$P(\text{changeable } |\text{ voted same way twice}) = \frac{(1-r)(1-p)}{(1-r)(1-p)+p}$$
and then combine these linearly to get $$P(\text{will vote the same next time}) = \frac{p}{(1-r)(1-p)+p} \times 1 + \frac{(1-r)(1-p)}{(1-r)(1-p)+p} \times (1-r). $$
Is this a correct approach? This seems to be correct to me, but I'm not completely sure that I haven't overlooked anything.
 A: I got a totally different result from yours.

Intro. 1: $$P(A \mid B) = P(A \mid BC) \cdot P(C \mid B) + P(A \mid B \overline C) \cdot P(\overline C \mid B)$$
Proof:
\begin{align}
P(A \mid B) P(B) &= P(AB) \\
&= P\left[AB(C \cup \overline C)\right] \\
&= P\left[(ABC) \cup (AB \overline C)\right] \\
&= P(ABC) + P(AB \overline C) \\
&= P(A \mid BC) \cdot P(BC) + P(A \mid B \overline C) \cdot P(B \overline C) \\
&= P(A \mid BC) \cdot P(C \mid B) \cdot P(B)  + P(A \mid B \overline C) \cdot P(\overline C \mid B) \cdot P(B) \\
&= P(B) \cdot \left[ P(A \mid BC) \cdot P(C \mid B) + P(A \mid B \overline C) \cdot P(\overline C \mid B) \right]
\end{align}
Intro. 2: $$P(A B \mid C) = P(A \mid C) \cdot P(B \mid AC)$$
Proof:
\begin{align}
P(AB \mid C) &= \frac{P(ABC)}{P(C)} \\
&= \frac{P(B \mid AC) \cdot P(AC)}{{P(AC)}/{P(A \mid C)}} \\
&= P(A \mid C) \cdot P(B \mid AC)
\end{align}

Say person A is the randomly selected one, whose proposals could be noted as $a_1, a_2, \ldots{}$ et cetera. $A_i$ stands for "yes" and $\overline A_i$ stands for "no".
If A is unchangeable, we say $X$; otherwise, $\overline X$.
If unchangeable A proposes "yes" every time, we say $Y$; otherwise, $\overline Y$. Obviously, $Y, \overline Y \subseteq X$, $Y \cup \overline Y = X$.
We can say $P(X) = p$, $P(\overline X) = 1 - p$.
Furthermore, we have
$P(a_{k+1} = a_k \mid X) = 1$,
$P(a_{k+1} \neq a_k \mid X) = 0$,
$P(a_{k+1} \neq a_k \mid \overline X) = r$,
$P(a_{k+1} = a_k \mid \overline X) = 1 - r$.
Suppose $P(A_1 \mid \overline X) = q$, $P(\overline A_1 \mid \overline X) = 1 - q$.
Suppose $P(A_1 \mid X) = s$, $P(\overline A_1 \mid X) = 1 - s$.
Now we want to calculate:
\begin{align}
P &= P(A_3 \mid A_1 A_2) + P(\overline A_3 \mid \overline A_1 \overline A_2) \\
&= P(A_3 \mid A_1 A_2 X) \cdot P(X \mid A_1 A_2) + P(A_3 \mid A_1 A_2 \overline X) \cdot P(\overline X \mid A_1 A_2) \\
&+ P(\overline A_3 \mid \overline A_1 \overline A_2 X) \cdot P(X \mid \overline A_1 \overline A_2) + P(\overline A_3 \mid \overline A_1 \overline A_2 \overline X) \cdot P(\overline X \mid \overline A_1 \overline A_2)
\end{align}

\begin{align}
P(A_1 A_2 \mid X) &= P(A_1 \mid X) \cdot P(A_2 \mid A_1 X) = P(A_1 \mid X) \cdot P(a_{k+1} = a_k \mid X) = s \\
P(A_1 A_2 \mid \overline X)
&= P(A_1 \mid \overline X) \cdot P(A_2 \mid A_1 \overline X)
= P(A_1 \mid \overline X) \cdot P(a_{k+1} = a_k \mid \overline X)
= q (1-r) \\
P(A_1 A_2) &= P(X) \cdot P(A_1 A_2 \mid X) + P(\overline X) \cdot P(A_1 A_2 \mid \overline X) = ps + (1-p)q(1-r) \\
P(X \mid A_1 A_2) &= \frac{P(X) \cdot P(A_1 A_2 \mid X)}{P(A_1 A_2)} = \frac{ps}{ps + (1-p)q(1-r)} \\
P(\overline X \mid A_1 A_2) &= \frac{P(\overline X) \cdot P(A_1 A_2 \mid \overline X)}{P(A_1 A_2)} = \frac{(1-p)q(1-r)}{ps + (1-p)q(1-r)} \\
P(\overline A_1 \overline A_2 \mid X) &= P(\overline A_1 \mid X) \cdot P(\overline A_2 \mid \overline A_1 X) = 1 - s \\
P(\overline A_1 \overline A_2 \mid \overline X) &= P(\overline A_1 \mid \overline X) \cdot P(\overline A_2 \mid \overline A_1 \overline X) = (1-q)(1-r) \\
P(\overline A_1 \overline A_2) &= P(X) \cdot P(\overline A_1 \overline A_2 \mid X) + P(\overline X) \cdot P(\overline A_1 \overline A_2 \mid \overline X) = p(1-s) + (1-p)(1-q)(1-r) \\
P(X \mid \overline A_1 \overline A_2) &= \frac{P(X) \cdot P(\overline A_1 \overline A_2 \mid X)}{P(\overline A_1 \overline A_2)} = \frac{p(1-s)}{p(1-s)+(1-p)(1-q)(1-r)} \\
P(\overline X \mid \overline A_1 \overline A_2) &= \frac{P(\overline X) \cdot P(\overline A_1 \overline A_2 \mid \overline X)}{P(\overline A_1 \overline A_2)} = \frac{(1-p)(1-q)(1-r)}{p(1-s)+(1-p)(1-q)(1-r)}
\end{align}
Therefore,
\begin{align}
P &= \frac{ps}{ps+(1-p)q(1-r)} + (1-r)\frac{(1-p)q(1-r)}{ps+(1-p)q(1-r)} + \frac{p(1-s)}{p(1-s)+(1-p)(1-q)(1-r)} + (1-r)\frac{(1-p)(1-q)(1-r)}{p(1-s)+(1-p)(1-q)(1-r)} \\
&= \frac{ps + (1-p)q(1-r)^2}{ps + (1-p)q(1-r)}+\frac{p(1-s)+(1-p)(1-q)(1-r)^2}{p(1-s)+(1-p)(1-q)(1-r)}
\end{align}
