Break-even loss in a double-six versus single-six game

Two fair dice are thrown. Double sixes win you $100. Exactly one six (with the other die not a six) loses you \text{\char36}$x$. Any other result means both dice are thrown again, repeating until one of the two deciding outcomes occurs. What is the largest $x$ for which the game's expected value is still non-negative?

$x=10$. Double six has probability $\tfrac1{36}$ and exactly one six has $\tfrac{10}{36}$; ignoring rerolls, the game ends in a win with probability $\tfrac1{11}$ and a loss with $\tfrac{10}{11}$. Requiring $\tfrac{100}{11}-\tfrac{10x}{11}\ge0$ gives $x\le10$.