The family board game Cluedo involves distributing cards of three types as evenly as possible to up to 6 players. The numbers and types of card are: 6 people- cards, 9 room-cards, and 6 weapon-cards. Each people-card is colour coded and has a matching coloured counter. In a particular game there are five players. One card of each of the 3 types, is selected at random and hidden in an envelope for a later part of the game. Each player then chooses a coloured counter with one counter being left over in this game. The players choose their counter without knowing the contents of the envelope. All three types of card are then put together into one pile. The pack is shuffled and the cards are dealt out as evenly as possible, with the youngest player(s) getting any extra cards from the deal. What is the probability that all five players hold the people-card that matches their own counter?
2 Answers
The five players will receive $4,4,4,3,3$ cards respectively. There are thus $$ \frac{18!}{(4!)^3(3!)^2} $$ distinguishable deals.
There is from the start a $\frac{1}{6}$ chance that the right color people-card was chosen for the envelope. Assuming it was, then to have success each of the five players must be dealt the specific people card of its color. You are left with ways to distribute $13$ cards as $3,3,3,2,2$ which is $$ \frac{13!}{(3!)^3(2!)^2} $$ So the answer is $$\frac{1}{6}\frac{13!(4!)^3(3!)^2}{18!(3!)^3(2!)^2 } = \frac{4^33^2}{6\cdot 18\cdot 17 \cdot 16 \cdot 15 \cdot 14 }=\frac{1}{3\cdot 17 \cdot 15 \cdot 14} = \frac{1}{11220} $$ The previous answer forgotr the $\frac{1}{6}$.
Something like $$ {4^3* {{13} \choose {3,3,3}} * 3^2 * {4\choose {2,2}}}/{18!} $$ :
Take all permutations (18!) and the successful ones are where the first 3 players can have either one of their 4 cards their own counter. The others can be random from 13 non-counter-cards. Last two players can only have 1 out of 3 cards their own counter.
