Anil and Bela play a game with one fair die…

Step 1. Anil. Anil wins on throw $1$ with probability $\dfrac{1}{6}$. He wins on throw $3$ if the first two throws are not sixes and the third is: $\left(\dfrac{5}{6}\right)^2 \times \dfrac{1}{6}$. In general he wins on throw $2k + 1$ with probability $\left(\dfrac{5}{6}\right)^{2k} \times \dfrac{1}{6}$.

Step 2. The sum. The terms form a geometric series with first term $\dfrac{1}{6}$ and common ratio $\dfrac{25}{36}$. The sum is $\dfrac{1/6}{1 - 25/36} = \dfrac{1/6}{11/36} = \dfrac{6}{11} \approx 0.5455$.

Step 3. Bela. Bela's terms are Anil's terms multiplied by $\dfrac{5}{6}$, so $P(\text{Bela}) = \dfrac{5}{6} \times \dfrac{6}{11} = \dfrac{5}{11} \approx 0.4545$. The two add up to $\dfrac{11}{11} = 1$.

Step 4. Within three throws. The game is still going after three throws only if the first three throws are not sixes: $\left(\dfrac{5}{6}\right)^3 = \dfrac{125}{216}$. So the game ends within three throws with probability $1 - \dfrac{125}{216} = \dfrac{91}{216} \approx 0.4213$.

Another way. Let $x$ be the probability that the player who throws first wins. That player wins at once with probability $\dfrac{1}{6}$. Otherwise the roles swap, and the first player wins only if the other fails: $x = \dfrac{1}{6} + \dfrac{5}{6}(1 - x)$, so $x = \dfrac{6}{11}$.