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in order to avoid losing my backgammon match, I had to avoid rolling a one on both of two successive rolls with two dice on each roll. If I rolled a one once or double ones I was ok, I just couldnt roll 1-x then 1-x on two successive rolls. What were my odds of avoiding rolling a one twice in a row?

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    $\begingroup$ I assume there are only two rolls total going on (i.e. that you're not asking about it happening twice in a row but anywhere in a succession of rolls). What's the chances of getting 1-x in one roll? Assuming rolls are independent, what's the chances of having that happen in two rolls? $\endgroup$
    – Glen_b
    May 18, 2017 at 1:57

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There are 10 bad rolls for you: 1-2, 1-3, 1-4, 1-5, 1-6, and the five reverse sequences of these. There are 36 possible outcomes for a pair of dice. Since dice rolls are independent, the chance that you roll 1-x twice in a row and lose the match is given by

$$\left( \frac{10}{36} \right) \left( \frac{10}{36} \right) =\left( \frac{100}{1296} \right) \approx0.0772 $$

That means your chance of avoiding the loss is $1-0.0772=0.9228, $ about $92\%.$

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