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My approach:

How many rolls of two dice are needed so that the probability of rolling a double six within this number of rolls is at least $50\%$?

Let $X$ be the number of required rolls. Therefore, we will have a binomial distribution:

So $P(X = n) = \frac{1}{36}\left(1- \frac{1}{36}\right)^{n-1} {n \choose 1}$

Hence, we solve $P(X = n) \geq 0.5$

It turns out that this equation does not have a real solution. In fact, the maximum that $P(X = n)$ goes is $\approx 0.37$. Where do you think I am messing up?

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A double-six here means at least one double-six. Your conclusion is right for exactly one double-six. The probability of that is never anywhere near $50\%$.

It will be easier to tackle the probability of the complement, that we do not roll any double-six.

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