# Throw a two dices-probability.

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.

• 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. – lulu Jan 22 at 0:37