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I'm going to assume you mean rolling the same number twice in a row in 25 rolls. The first won't cause a double, so you just need to consider the odds of rolling the same number as the last for the last 24 rolls. The easier approach is to realize that the probability of rolling at least one double is 1 minus the probability of rolling no doubles. One roll has this probability of not rolling the same as the last:

P(different number from last) = 5/6

Since they are independent events:

P(no doubles in 25 rolls) = (5/6)24

Now the final probability, of at least one double, is 1 - (5/6)24

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βˆ™ 14y ago
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Q: What are the odds of rolling doubles with one die in 25 rolls?
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