Roll a die until two consecutive equal outcomes. What’s the probability that the sum of all outcomes is an odd number?

Question

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?

Answer 1

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 $$ {\small{\frac{1}{2}}} p(0,0) + {\small{\frac{1}{2}}} p(1,1) = \frac{1}{2} {\,\cdot\,} \frac{15}{41} + \frac{1}{2} {\,\cdot\,} \frac{20}{41} = \frac{35}{82} $$