Two players A and B throw a fair die alternately, A throwing first; the first player to throw a six wins the game.

A wins on throws $1,3,5,\dots$ so $P(A)=\frac16\sum_{m\ge0}\left(\frac56\right)^{2m}=\frac{1/6}{1-25/36}=\frac{6}{11}$, and $P(B)=1-\frac6{11}=\frac5{11}$. Therefore (A) is false and (B) is true. For (C): the game lasts at least four throws when the first three throws are not sixes; the fourth throw is then made by $B$, so the situation is the original game with the roles of $A$ and $B$ exchanged, and $A$ wins with probability $P(B)=\frac{5}{11}$, not $\frac6{11}$; (C) is false. For (D): the number of throws until the first six is geometric with success probability $\frac16$, so its expectation is $6$, not $\frac{11}{2}$, and (D) is false. Only (B) is correct.