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?
Best Answer
After any roll, provided the process is not yet complete, define the state of the process as the pair $(r,s)$ where
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 $$ {\small{\frac{1}{2}}} p(0,0) + {\small{\frac{1}{2}}} p(1,1) = \frac{1}{2} {\,\cdot\,} \frac{15}{41} + \frac{1}{2} {\,\cdot\,} \frac{20}{41} = \frac{35}{82} $$