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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?

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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 :)

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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.

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