151 reputation
4
bio website
location
age
visits member for 3 years
seen Jul 29 at 14:40

Aug
29
comment How many different 4-player UNO starting games are possible?
@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.
Jan
31
comment How do I think about solving this sort of problem without having to count intersections?
although this is an answer to my example problem (is it correct? do you get 300,704), I was looking at a more general way at looking at this type of problem. When I thought about the question again in the terms of a tree, and how close I was to achieving my objective, it reminded me of a markov chain. My other die question (small straight 6d6) linked above didn't use the same method to achieve the solution. Brute force was the easiest solution for many of the people who provided answers.
Jan
31
comment How do I think about solving this sort of problem without having to count intersections?
@ArturoMagidin, yes. That isn't how I would word it, but you do have it correct. At least three dice are not "6 or more" (there exists at most two dice that have values of 6 or more), and of the three that are not 6 or more, at most one of them may have a value of 5. Is there some easier way of stating this? Each time I rewrite it, it seems less intuitive.
Jan
31
comment How do I think about solving this sort of problem without having to count intersections?
I don't think you got my meaning in the question. This might because you don't have a role playing background. I will update the question so that you can better understand it. 5d20 means "roll five 20-sided polyhedral dice". What I was interested was how to think about the problem in a general sense so that I can apply this to other dice questions (instead of brute forcing every example).
Jan
31
comment How do I think about solving this sort of problem without having to count intersections?
@ArturoMagidin, it must have been a typo. What I meant was at least one die with 5 or less, and at least two other/different dice with 4 or less. This is of course different than at least one '5', and two '4's. At least one '5' & two '4's doesn't count a result of '3', '2', '2', 'X', 'X' which would be a success in my example.