You, your parents, your sister, go to visit grandma for her birthday. Grandma made a cake for the party. If she puts $20$ raisins in the cake at random in the cake, and she divides the cake into $5$ equal pieces, what's the probability that you get one more than than your sister?

Let X be the number of raisins you have and Y be the number of raisins she has.

$X,Y \sim Bin(20, 1/5)$

Someone told me to use the multinomial distribution but I think the hypergeometric distribution should be used and I don't understand the difference between multinomial and hypergeometric.

I think we're sampling without replacement so we should use multivariate hypergeometric.

I'd consider some cases separately and compute the probability of each case directly:

  • You get 1 raisin and your sister gets 0.
  • You get 2 raisins and your sister gets 1.
  • You get 3 raisins and your sister gets 2. $$\vdots$$
  • You get 10 raisins and your sister gets 9. (This is the last case you have to consider -- why?)

To compute any one of these, we can use a [multinomial distribution][1]. For instance, the probability that you get 3 raisins and your sister gets 2 (hence, 15 go to the unused portion of the cake) is $$\binom{20}{3, 2, 15}(1/5)^3 \cdot (1/5)^2 \cdot (3/5)^{15} > = \left( \frac{20!}{3! \cdot 2! \cdot 15!} \right) (1/5)^3 \cdot (1/5)^2 \cdot (3/5)^{15}$$ which can be simplified, of course. You then have $10$ such terms which represent disjoint events, and whose union is the event that you want, so you can add their probabilities to get your answer.

I don't understand why the multinomial distribution would solve this problem. In fact, I don't understand when to use the multinomial distribution or how that model works. I still believe this problem should be solved using the hypergeometric distribution.


1 Answer 1


There are 20 raisins and they, in effect, get distributed into three bins - one being your piece of cake, one being your sister's piece of cake, and the third being the other three pieces of cake taken together - with probabilities $1/5, 1/5, 3/5$ respectively. Each raisin's assignment to a bin is independent of every other raisin's bin assignment.

Let's write out the probability distribution, ignoring the constant of integration for the moment:

$$p(x_1, x_2, x_3) \propto 0.2^{x_1}0.2^{x_2}0.6^{20-x_1-x_2}$$

The distribution of the counts of raisins is a multinomial distribution, with size parameter $n=20$ and probabilities $0.2, 0.2, 0.6$. This is exactly the sort of process the multinomial describes.

To make the applicability of the multinomial distribution a little clearer, let's simplify the problem to one where you only want to know about how many raisins you get. Each raisin is assigned to your piece with a probability of $0.2$ and to "not your piece" with a probability of $0.8$. I'm not sure how you were taught about the binomial distribution, but coin flipping is a pretty common example; it should be clear that the raisin assignment probabilities are exactly the same, mathematically, as flipping a coin with $p(\text{heads}) = 0.2$ 20 times and counting the number of heads. The fact that there is a fixed number of raisins is irrelevant; there are a fixed number of coin flips too, but it's still a binomial. In fact, the fixed number of raisins corresponds exactly to the fixed number of coin flips, mathematically speaking.

The multinomial is just the extension of the binomial to more than two possible results, which in the problem statement is achieved by including your sister in it.

Now, if we restructured the problem such that there were, let us say, 10 yellow and 10 black raisins, and your grandmother randomly chose, say, 5 raisins from the 20 total to put on the cake, and all you cared about is the number of yellow raisins on your slice, we would be entering the realm of hypergeometric probabilities. The key is that each "draw" now has a different probability of selecting a yellow raisin, with that probability being dependent upon the results of the previous draws in the obvious way, and then the raisin is being assigned to a piece of cake. In our multinomial problem, we aren't selecting from a finite population of mixed characteristics at all, we are just repeating the same act with the same probabilities $n$ times. (You could think of the original problem as having each "draw" select a black raisin with probability $=1$, regardless of what happened on previous draws, but that is an unnecessary complication, since the probabilities are always $1$.) Since the whole first step - selecting a raisin of some color from a finite mixed bag of raisins - doesn't exist in the problem statement, the probabilities aren't hypergeometric, but multinomial.


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