# The expected outcome of a random distribution of marbles among several people

The problem is:

On each round of a game, 20 marbles are distributed at random among five children: Alan, Ben, Carl, Dan, and Ed. Consider the following distributions:

• I) Alan: 4, Ben: 4, Carl: 5, Dan: 4, Ed: 3
• II) Alan: 4, Ben: 4, Carl:4, Dan: 4, Ed: 4

In many rounds of the game, will there be more results of type I or type II?

I guess number II is more probable because they are around $$\mu=4$$, but I'm not sure that the problem is that easy! Am I correct?

Your intuition is correct, but it's better to approach from a more calculus perspective:

Distributing marbles according to Type I is equivalent to counting the number of permutations of the following string and dividing by $$5^{20}$$, i.e. total number of probabilities: $$AAAABBBBCCCCDDDDEEE$$, which is $$\left(\frac{20!}{(4!)^3 5!3!}\right)/5^{20}$$.

For type II, this is $$\left(\frac{20!}{(4!)^5}\right)/5^{20}$$ since there are $$4$$ of each letter. The denominator of the second one is larger, since $$5!3!>(4!)^2$$, and therefore probability of obtaining type II is larger.

Seems I'm a little bit late :)

Your intuition is correct, and can be verified by explicitly calculating the probability. This is a multinomial distribution so that the probably of getting $$n_1,n_2,n_3,n_4,n_5$$ marbles for Alan,Ben,Carl,Dan,Ed respective is:

$$\frac{20!}{n_1!n_2!n_3!n_4!n_5!}.$$

For type I, the denominator is $$(4!)^33!5!$$.

For type 2, the denominator is $$(4!)^5$$.

Then $$(4!)^2 = 3!\cdot 4 \cdot 4!\leq 3!5!.$$

You'll notice the denominator for type I is bigger than type II, hence type II is more probable.