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?



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.