Roll a die until two consecutive equal outcomes. What’s the probability that the sum of all outcomes is an odd number? Keep rolling a die until two consecutive outcomes are equal. What’s the probability that the sum of all outcomes(including the last two) is an odd number?
This is not a homework. It’s a combination (just came to my mind) of two easy problems:
What’s the probability took N rolls to see two consecutive equal outcomes?
The probability that the sum of N rolls of a die is odd?(which means there are odd number of odd outcomes, then it’s not hard to find the result is 1/2.)
My initial thought about the solution of this new problem was combining the results of the two simple questions. But then I realize these two events are not independent, applying the stopping rule changes probability of the second event, which is no longer 1/2.
I am wondering if it can be solved by calculating the probability of the second event (sum is odd) if we treated N as a fix number and then times the probability that the game is stopped at N rolls, then the solution is a form of summing up all weighted probabilities as an infinite series?
 A: After any roll, provided the process is not yet complete, define the state of the process as the pair $(r,s)$ where

*

*$r=0$ or $r=1$ according as the previous roll is even or odd.$\\[4pt]$

*$s=0$ or $s=1$ according as the sum of all previous rolls is even or odd.

From each state $(r,s)$, let $p(r,s)$ be the probability that, when the process completes, the sum of all rolls is odd.

Considering how the value of $p(r,s)$ is affected by the parity of the next roll, we get
$$
\left\{
\begin{align*}
\;p(0,0)&=\,{\small{\frac{1}{3}}}p(0,0)+{\small{\frac{1}{2}}}p(1,1)\\[4pt]
\;p(0,1)&=\,{\small{\frac{1}{6}}}+{\small{\frac{1}{3}}}p(0,1)+{\small{\frac{1}{2}}}p(1,0)\\[4pt]
\;p(1,0)&=\,{\small{\frac{1}{6}}}+{\small{\frac{1}{3}}}p(1,1)+{\small{\frac{1}{2}}}p(0,0)\\[4pt]
\;p(1,1)&=\,{\small{\frac{1}{3}}}p(1,0)+{\small{\frac{1}{2}}}p(0,1)\\[4pt]
\end{align*}
\right.
$$
which can be regarded as a system of $4$ linear equations in $4$ unknowns.

Solving the system yields
$$
p(0,0)=\frac{15}{41},
\;\;\;\;\;
p(0,1)=\frac{26}{41},
\;\;\;\;\;
p(1,0)=\frac{21}{41},
\;\;\;\;\;
p(1,1)=\frac{20}{41}
$$
hence, since the state after the first roll is either $(0,0)$ or $(1,1)$, according as the first roll is even or odd, it follows that the probability of an odd total sum when the process completes is
$$
{\small{\frac{1}{2}}}
p(0,0)
+
{\small{\frac{1}{2}}}
p(1,1)
=
\frac{1}{2}
{\,\cdot\,}
\frac{15}{41}
+
\frac{1}{2}
{\,\cdot\,}
\frac{20}{41}
=
\frac{35}{82}
$$
