Anaya and Bhavik take turns rolling a fair die, Anaya rolling first, and whoever rolls the first six wins. What is the probability that Anaya wins?

Anaya wins on her 1st, 2nd, 3rd, … roll, and each extra round needs both players to have missed. So $P=1/6+(5/6{)}^2(1/6)+(5/6{)}^4(1/6)$ + …, a geometric series with first term 1/6 and common ratio $(5/6{)}^2=25/36$. Its sum is $(1/6)÷(1-25/36)=(1/6)÷(11/36)=6/11$. A neater route: let $p$ be Anaya's chance. She wins at once with probability 1/6; otherwise, with probability 25/36 both miss and the position repeats exactly, so $p=1/6+(25/36)p$, which gives $p=6/11$. The value 5/11 is Bhavik's chance, and the two correctly add to 1. The value 1/2 assumes the game is even-handed, but rolling first is a real advantage — Anaya may win before Bhavik has thrown at all. The value 1/6 counts only Anaya's opening roll.