On each round of a game, 20 marbles are distributed at random among five children: A, B, C, D and E.
- A: 4, B: 4, C: 5, D: 4 and E: 3
- A: 4, B: 4, C: 4, D: 4 and E: 4
In many rounds of the game, will there be more results of type I or type II?
Can we say that the probability of both scenarios will be same as $\frac{1}{(4 from 20)}$?
No, the probability will not be the same, at least not under the scenario that one would usually call 'distributed at random'.
Just like there is a slightly higher probability to see 50 heads out of 100 than to see 49, there is a slightly higher probability of seeing the second configuration (assuming, per asdf's comment, that for the first, we only count exactly the configuration $(4,4,5,4,3)$ and don't also include symmetry-related configurations like $(5,4,4,4,3)$).
The probability of the different configurations here follows a multinomial distribution, but we can just count them up manually. We have five buckets to choose from for each of the twenty objects, so there are $5^{20}$ total possible allocations. For the first configuration, we choose $4$ for bucket A, $4$ of the remaining for bucket B, five of the remaining for bucket C and so on. So that's $$ {20 \choose 4}{16 \choose 4}{12 \choose 5}{7 \choose 4}{3 \choose 3}$$ different allocations that have the configuration $(4,4,5,4,3).$ Dividing this by $5^{20}$ gives approximately $0.00256$ or $0.256\%.$
On the other hand the total number for configuration $(4,4,4,4,4)$ is$$ {20 \choose 4}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}$$ which, divided by $5^{20}$ gives approximately $0.00320,$ or $0.320\%,$ which is slightly higher, in accord with our intuition from the coin flip example.