We throw two cubical dice. How to show that we only need $300$ throws in order to have a $95\%$ chance that in at least $100$ throws, we’ll get a smaller number on the first cube? Is $250$ throws enough?
I thought about a Bernoulli trial, where $k=300$, $n=100$, and the probability of success is equal to $\frac{15}{36}$. But these are not calculations for a human.
According to Wolfram alpha, there's a probability of $0.2755$ (approximately) that in $250$ throws the first die will show the smaller number fewer than $100$ times. That is, the probability that it shows the smaller number at least $100$ times is only $0.7245$, so $250$ throws are not enough to get a $95\%$ chance of that occuring.
With $271$ throws there's only a probability of about $0.0484$ that the first die will show the smaller number fewer than $100$ times. Thus $271$ throws are enough to have chance of at least $95\%$ that the first die will show the smaller number at least $100$ times.
On the other hand, with $270$ throws, the probability that the first die shows the lower number fewer than $100$ times is $0.0536$, so $270$ throws aren't enough to get a $95\%$ chance of the first die's showing the smaller number at least $100$ times.