Two people, $P$ and $Q$, decide to independently roll two identical dice, each with $6$ faces, numbered $1$ to $6$. The person with the lower number wins, In case of a tie, the roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by $P$ and $Q$. Assume that all $6$ numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is ________ .
My Try:
A person with lower number wins, so sample space is $\{1, 2, 3, 4, 5\}$ and chances $= 5/6$.
The probability that one of them wins in 3rd trial
$=$ lost in first trial * lost in first trial * wins in the third trial
$= 1/6 *1/6*5/6 = 5/216 =0.023$
Could you please solve it?
Thanks in advance.
Edit1:
This examination was held yesterday, I posted this based on memory.
Edit2:
They published it. I edited the question, this is original question:
The probability that one of $P$ or $Q$ wins in the third trial is the probability that they draw (have equal dice) in the first two trials, and have different dice in the third trial.
A draw happens only when both dice show the same value.
\begin{align*} \mathbf{P}[ \text{draw} ] & = \mathbf{P}[P = Q] \\ & = \sum_{i=1}^6 \mathbf{P}[P= Q = i] \\ & = \sum_{i=1}^6 \mathbf{P}[P=i] \mathbf{P}[Q = i] \\ & = \sum_{i=1}^6 \frac16 \times \frac16\\ & = \frac16. \end{align*}
Since trials are independent
\begin{align*} \mathbf{P}[ \text{win in round $3$}] & = \mathbf{P}[ \text{draw in round $1$}] \, \mathbf{P}[ \text{draw in round $2$}] \, \mathbf{P}[ \text{not a draw in round $3$}] \\ & = \frac16 \times \frac16 \times \frac56 \\ & = \frac5{216} \end{align*}
Note that whilst our results are the same, I'm not sure based on your language (referring to loses) whether you have used the same logic. A loss suggests that one person lost, and the other won. In this calculation we must count draws: i.e. neither person won.