# Simple probability question (similar to birthday paradox)

If $x$ objects are randomly distributed to $n$ groups, what is the formula for working out how big $x$ needs to be for the probability that at least one of the groups gets an amount $y$ (or larger) to exceed $50\%$?

Specifically, I am interested in knowing how big $x$ needs to be if there are $n=7$ groups and I need there to be a $50\%$ probability at least one of the groups gets $y=30$ objects.

• Do you mean "uniformly randomly distributed"? – EngrStudent Jul 27 '17 at 14:45
• @Engr That's a fair assumption that universally is made when the groups are not otherwise distinguished from each other. Incidentally, $x=163$ objects are needed; $162$ won't quite do it. – whuber Jul 27 '17 at 14:46

You have $x$ objects and throw them randomly (with equal probabilities) into one of $n$ boxes. Another formulation is that you have a $n$-sided regular dice and throws it $x$ times. (I will use your notation even if it is somewhat unconventional for this case). The number of times the dice comes up $i$ is $X_i$, and then $X_1, X_2, \dotsc, X_n$ has a multinomial distribution. Define $$M = \max_{i=1,\dotsc,n} X_i$$ and to solve your question we need to find the probability distribution of $M$. But that question is answered completely in Die 100 rolls no face appearing more than 20 times