# Combinatorics Probability for a Board Game

There's a board game I recently played called "Welcome to the Dungeon." In it, there is a deck of 13 cards that consists of 8 types of monsters. Each player may draw a new card each turn. Each player can end the game at any point it is their turn.

There is a mechanism in the game where you win if at the end of the game each monster that has been drawn is unique (i.e. no duplicates).

The 13 card deck consists of the following:

• Monster A - 2 cards
• Monster B - 2 cards
• Monster C - 2 cards
• Monster D - 2 cards
• Monster E - 2 cards
• Monster F - 1 card
• Monster G - 1 card
• Monster H - 1 card

If I were to end the game after 5 cards were drawn, what is the probability that each card is unique? (And would you please help me understand how to calculate it?)

I understand it's a combinatorics problem, however I'm not sure how to calculate it since not every type of monster has an equal number of cards associated with it.

Thank you!

• maybe you will need to treat separately hands with 0,1,2 & 3 of F, G, H in them – Cato Dec 22 '16 at 15:48

## 1 Answer

Break the possible drawings up by how many of $F, G, H$ we draw.

If we draw 0 of $F, G, H$, we're drawing 5 of $A, B, C, D, E$, and 2 ways to do each, for a total of $\binom{3}{0}\binom{5}{5}2^5$. If we draw 1 of $F, G, H$, we similarly get $\binom{3}{1}\binom{5}{4}2^4$. If we draw 2 of $F, G, H$, we get $\binom{3}{2}\binom{5}{3}2^3$. If we draw all 3 of $F, G, H$, we get $\binom{3}{3}\binom{5}{2}2^2$.

Thus, the total number of ways to draw 5 unique cards is $\binom{3}{0}\binom{5}{5}2^5 + \binom{3}{1}\binom{5}{4}2^4 + \binom{3}{2}\binom{5}{3}2^3 + \binom{3}{3}\binom{5}{2}2^2 = 632$.

We're drawing 5 of 13 cards, so there are $\binom{13}{5} = 1287$ total ways to draw them.

Therefore, the probability is $\frac{632}{1287}$.