It's of course possible that I erred somewhere, but this is my (partial) result:
6 cards - 2 sets = (90 + 90 + 60 + 3)/729 ... ................................ ... >< 153/729 = 17/81 ~ 21% (I am spreading it, so it's easier to find a possible error)
EDIT: An error has found. See next few posts...
7 cards - 2 sets = (2187 - 126))/2187 = 2061/2187 = 229/243 ~ 94.24% (I counted 5-2 as only case that doesn't deliver 2 sets. If there is an additional case that I currently don't see, then my result is correspondingly wrong)
9 cards - 3 sets - Sorry, this case is complicated and I don't have much time&energy at the moment. If nobody solves it for 2 days, I'll be back...
Last edited by Zemljanin on Mon Sep 13, 2010 6:46 am, edited 2 times in total.
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With 6 cards, you always have one set, and the only way to have 2 sets is if the remaining cards make a set. So the odds of having two is the same as the odds of having one with 3 cards. 9 cards, 3 sets is exactly the same.
The math on 2 sets with 7 cards above is correct (7 choose 2 for the cards, times 6 for the colours).
Last edited by BaldAdonis on Mon Sep 13, 2010 8:22 pm, edited 1 time in total.
BaldAdonis wrote:With 6 cards, you always have one set, and the only way to have 2 sets is if the remaining cards make a set. So the odds of having two is the same as the odds of having one with 3 cards.
You're right, thanks. I had a feeling I was missing something, but have been too hasty... I counted 90 cases of 2-2-2, 60 cases of 3-3-0 and 3 cases of 6-0-0. However, I overlooked further 90 cases (4-1-1). So the correct math is:
BaldAdonis wrote:With 6 cards, you always have one set, and the only way to have 2 sets is if the remaining cards make a set. So the odds of having two is the same as the odds of having one with 3 cards. 9 cards, 3 sets is exactly the same.
nice one, I guess that's pretty obvious when you think about it.
