Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that for poker, there are $54!$ ways to arrange the cards in a poker deck which only contains unique cards. An UNO deck on the other hand has 108 cards. bit contains duplicates. It contains 25 cards in each color, a single 0, two copies of the numbered cards 1-9, two each of Skip/Draw-2/Reverse. The deck also contains four copies of Wild and Wild-Draw-4.

So, I believe there are $\dfrac{108!}{(2^9*2^3)^4 *4!*4!}$ ways of arranging an UNO deck.

This is all possible ways 108! (not considered duplicates), dividing out the duplicates in each color (9 number cards and 3 action cards, two copies each) to the 4th power for each color. Then divide out the Wilds which each have 12 possible arrangements of the duplicate cards that are being over counted.

What I am not sure of is how to remove the starting hands from each of the 4 players. Do I just divide by another $(7!)^4$?

share|cite|improve this question
Sounds right , now transform it into a number – Willemien Aug 29 '13 at 9:46
@Willemien, I am not so sure. Dominion starts with a 10 card deck for each player composed of 7 Cooper and 3 Estates. Starting Hand size is 5 cards. Using this formula, you get one possible starting hand,but I know there are more possible hands (all 3 estates & 2 coins, 3 Coin& 2Estates, 4Coin&1Estate,5Coin). This formula solves for the remaining ways of arranging the deck, but what I am interested in is Starting Game states. – user1873 Aug 29 '13 at 13:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.