We throw a two symmetric dices. How to show, that we only need to $300$ throws to have $95\%$ chance, that in at least $100$ throws we will get a smaller number on the first cube. Is $250$ throws enough?

I thought about Bernoulli trial, where $k=300,n=100$ and probability of success is equal to $\frac{15}{36}$. But these are not calculations for human.

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    $\begingroup$ the normal approximation to the bernoulli distribution should be good enough. Note: the mean is the number of throws times $\frac {15}{36}$. Thus, for $250$ throws the mean is just over $104$ and $\sigma \approx 7.8$ so there is a very high probability of the first one winning fewer than $100$ times. $\endgroup$ – lulu Jan 22 at 0:37

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