# Bernoulli vs. uniform distribution

Here's a question that I'm struggling with:

Jack, John and Tom given 20 brownies by their mom, in a random manner. They are arguing: what are the odds that Jack will get all of them?

Jack says that there are $\binom{20+2}{2}$ different ways to give the brownies, so his chances are $\frac{1}{\binom{20+2}{2}}$

John say that they all have the same chances for each brownie, so the chances are $(\frac{1}{3})^{20}$.

Who is right? I think Jack is wrong because he didn't count similar divisions in which the brownies were given in a different order.

But I think the real issue here is - when do I choose Bernoulli with $\frac{1}{3}$ success chance, and when do I choose a uniform distribution and count all possible divisions?

Thanks!

The number of ways to distribute the brownies (assumed identical) is indeed $\dbinom{22}{2}$, by a standard "Stars and Bars" argument.
However, these $\dbinom{22}{2}$ ways are not all equally likely. So it is not correct to deduce that the probability Jack gets all of them is the reciprocal of the binomial coefficient. A distribution of type $20$-$0$-$0$, or $17$-$1$-$2$, is much less likely than a distribution of type $7$-$7$-$6$.
The argument that says the probability Jack gets all the brownies is $(1/3)^{20}$ is perfectly correct.
If every brownies are distributed randomly and independantly between each person, then you can view the number of brownies that Jacks get has a Binomial random variable $X$ with parameter $n=20$ and $p=1/3$. The probability that he gets them all is then $$\mathbb{P}(X=20)={20\choose 20}p^{20}(1-p)^{0}=\frac{1}{3^{20}}.$$
As André mentionned, the fundamental difference here is that all of the ${22\choose 2}$ ways are not equally likely. Seeing each brownie has a bernoulli R.V. simplifies things a bit.