Post by Captain Garak on Oct 9, 2006 19:28:57 GMT -5
Nice tables, OWell! Although I have a couple of problems with some of the numbers in the "J=3" column of your Probability Table (4) , the basis of the rest of the information looks pretty solid.
So, if I had 4 Evil Blasts in my book. I would have almost a 31 percent chance of starting with one in my opening hand of 5 cards. Correct?
Lizardman: I take it you're looking at the value of 0.3080 in the "K=1" column / "N=05" row of Table (4), which would be the correct interpretation. The chance of your pulling one of your four Evil Blasts in your first hand of five cards would be:
If I had 4 Evil Blasts in a book, about what percent is the chance in opening hand of 5 cards? You should look up row of N=5 and J=1 column then, and it is 35.30%(0.3530). However, at the same time, the probability currently drawn 2 sheets exists at 4.50%(0.0450).
The formula of "Just K sheets column" is the following.
K(max)CK * (50-K(max))C(N-K) / 50CN
C is Combination.
And "1 sheets or more" is Sigma K(max)(K=1 + K=2 + ... + K=max-1 + K=max).
Description of combination. (1) Combination taken out two sheets from four cards, (4x3) / (2x1) = 6 kind.
(2) Combination taken out four sheets from six cards, (6x5x4x3) / (4x3x2x1) = 15 kind.
It is here and (1) can express 4C2, (2) can express 6C4 (C:conbination).
Now, 6 cards are 2 cards made into [O] and 4 cards are made into [@]. And it consider a combination which takes out 2 sheets form 2cads of [O] and takes out 2 sheets form 4cards of [@]. At this time, * [O]: (2x1) / (2x1) = 1 kind. * [@]: (4x3) / (2x1) = 6 kind.