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Spoil odds.

Postby Woltato on Sun Sep 05, 2010 5:59 pm

has anyone calculated the odds of hitting set/s?

I've managed to figure out for myself.

3 spoils - set = 1/3
4 spoils - set = 7/9 or 2/3 (if previously no set with 3)

But what are the odds of 2/3 sets after a ko?

6 spoils - 2 sets ?
7 spoils - 2 sets ?
9 spoils - 3 sets ?
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Re: Spoil odds.

Postby Zemljanin on Sun Sep 12, 2010 9:20 am

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 5:46 am, edited 2 times in total.
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Re: Spoil odds.

Postby BaldAdonis on Sun Sep 12, 2010 4:44 pm

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 7:22 pm, edited 1 time in total.
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Re: Spoil odds.

Postby Zemljanin on Mon Sep 13, 2010 5:25 am

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:

6 cards - 2 sets = (90 + 90 + 60 + 3)/729 = 243/729 = 1/3
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Re: Spoil odds.

Postby Woltato on Tue Sep 14, 2010 6:40 pm

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.
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Re: Spoil odds.

Postby Woltato on Tue Sep 14, 2010 6:44 pm

thanks both for your help.
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