I would like to ask if I were to "design" a game where the less likely of the recognized dice combinations is the better, which would be the "best" of all?
The theme is a railyard. Building a ready-to-depart train is the goal. Six standard dice are used. Wagons are the numbers 1, 2, 3, 4, 5 and number 6 is the engine. Without an engine, only a loose consist can be scored. Regardless of probability, a loose consist will be inferior to any train.
Loose consist: 123, 234, 345, 1234, 2345, 12345.
Single-headed trains: 3456, 23456, 123456.
Double-headed trains: 34566, 234566.
Triple-headed-trains do not count as there should be more wagons than engines in a train.
An intermodal is(?) the (planned) highest scoring combination, which can be single or double-headed, with four (and no less than four) wagons. In ascending order, these are 665555, 655555, 664444, 644444, 663333, 633333, 662222, 622222, 661111, 611111.
My question is, whether an intermodal is more or less likely than a single or double-headed train? Which should be the "ultimate train" in the game? An intermodal (engine(s) + identical wagons) or a train (engine(s) + sequence)?
Kindly help me out.
Hint:
The probability of a certain combination of a given length $n$ is proportional to number of ways to permute it:
$$ \frac{n!}{n_1!n_2!\cdots n_6!}, $$ where $n_i$ is the number of dice showing $i$ in the sequence.
Therefore the probability of a doubly headed intermodal is proportional to: $$ 5\frac{6!}{2!4!}=75, $$ where the factor $5$ counts the ways to choose the wagon number, whereas the probability of the double headed train is proportional to $$ \frac{6!}{2!}=360. $$
One observes that the probability of an intermodal is 4.8 times less than that of the double-headed train.