Is it a bad question, or a difficult one?
This is my third time typing this grrrr. Let's hope my login holds this time.
I have written a dice game for the computer and now I am trying to calculate the probabilities necessary to make a paytable so that it can be played against the computer like one would a casino game.
Quick description:
The game is called 24. It involves 6 dice tossed up to 6 times with at least one die retained each toss. Once retained a die can not be reintroduced. The game is scored when there is a 2 and a 4 retained, the remaining dice are summed. For example 2,4,6,6,6,6 is a perfect score of 30.
If this is tossed and retained in one throw I have calculated there to be 30 ways (6!/4!) and there are 6^6 possibilities. Obviously the odds on this are phenomenal, .06% but of course there are 163 ways to get this perfect score, such as tossing and retaining 66662 then toss the last one again to get 4. There are 5 ways to get 66662 and 1 way to get 4. And so on all the way until you throw all the dice and retain 1 each time like 6+6+6+6+4+2 of which there are 30 ways as well.
If I add them all up I get 900.
The questions come now:
In doing this, do I have to change the 6^6? Or is the perfect score of 24 (900 out of 6^6) something like 1.9%?
I drew all of these combs/perms out but I assume there would be a good formula to calculate them instead? I mean the 900. I kind of like (6!/4!)^2 but can't come up with a justification for it.
Obviously, I will have to calculate them for 23 and 22 and so on. But if you can give me a good push in the right direction I think I can do it.
-- Abs
Last edited by Absinthe; 05-26-2012 at 08:30 PM.
Is it a bad question, or a difficult one?
When I first read it it wasn't sufficiently clear to me what you're actually asking.
I don't have emotions and sometimes that makes me very sad.
My bad. Six dice one toss yields 6!/4! Different ways to get perfect score. (four 6 and one each 2 and 4).
You can toss up to six times keeping at least 1 die each time. Figuring the permutations by hand I get 900 ways to get a perfect score.
Intuitively that is the square of the number of ways to get perfect score in one toss. However intuition does not prove the calc enough to apply it to the next score ...
Does that makeclearer what I am looking for?
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