First to make the running total divisible by 7

Two players alternately throw a fair six-sided die and keep a cumulative total. The player whose throw first makes the total a multiple of $7$ wins. You may choose to throw first or second. Choosing well, what is your probability of winning?

$6/11$, by going second. The first throw can never give a multiple of $7$. After it, whatever the total, exactly one face on each throw lands on the next multiple of $7$, so every throw wins with probability $\tfrac16$. The second player wins with probability $\tfrac{1/6}{1-(5/6)^2}=\tfrac{6}{11}$.