Ok, did an empirical 100,000 interation test of 6 Openning Dice vs 3 Opposing Dice, no TN modifiers. 1 or more successes came in at 39.85%, only +/-.01% off of our calculations.

However 2 or more successes came in at only 12.82% (compare to our sheet's prediction of 18.80%), and 3 or more successe came in at 2.40% (compare to our sheet's prediction of 6.90%). At 100,000 interations i would expect the variation to easily be <1%. Notice how close the first number is.

Everything on that grid above 1 Success needs the formula redone.

OK. I'll re-do those cells with a more exact (rather than relative) formula and see how things go. It'll have to wait for a bit though as I'm revising for my exam (on tuesday) right now.

The "at least 1 success" calculation in our spreadsheet is dead on, i stand by it. You are merely showing how flawed it is to calculate using the average TN. The underlying reason, as far as i can tell, that the simple average doesn't work is because the lower TNs are MUCH more probably for the lower number of dice opposing to make, the average TN is quite tough, and the higher only marginally tougher than the average. So when weighted by that the overall probability raises past what the probability is for the simple average TN.

P.S. Lilt, i've looked into it and figured why the "2 or more success" columns and up don't work with how you are progressively calculating. It is because the "1 or more success" contains the "more" part. If we were using a "1 and only 1 success" calculation we could iterate to get the multiple successes properly. I also figured out how roughly the algorithm required on paper, but haven't translated it into computerese yet. I should be able to do that tomorrow night.

I ran a simpler test than I described in my previous post, as it was simpler to write and was the basis for the more complicated test. I have updated http://www.anasazisystems.com/~gknoy/sotsw...enTestGraph.xls (1:45 PST) to include my new test data (on the second sheet).

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Note that using an open-style roll for the guard (4 dice) is the SAME as saying "what's the highest number at which he can make one success". Which is what we care about in thiscase, as 1 success is enough.
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I've run a test of 1 million open rolls with 12 dice.

50% of them were an 11 or better.
25% were 14 or better.
10% were 17 or better.
3.7% were 21 or better.

I then ran the same test with only 4 skill dice.

20% were 11 or better.
8% were 14 or better.
3.7% were 17 or better
1.2% were 21 or better.

What does this tell us?

12 dice are 2.4 times more likely to roll a 11 or better than 4 dice.
12 dice are 2.7 times more likely to roll a 14 or better than 4 dice.
12 dice are 2.9 times more likely to roll a 17 or better than 4 dice.
12 dice are 3.0 times more likely to roll a 21 or better than 4 dice.

If we line up by percents >= $roll, we can see that:
12 dice are as likely to roll a 21 as 4 dice are likely to roll a 17.
12 dice are as likely to roll a 15 as 4 dice are likely to roll an 11.

It appears that the highest roll from 12 dice is likely to be about 3 or 4 higher than the highest roll from 4 dice. The "middle" cases, where half are that roll or better, and half are worse, come at 11 and 7 (for 12 and 4 dice, respectively).

If we divide the "% N or better" for 4 dice vs that for 12 dice, we get about 25-30%. Which I think (could be wrong ) might mean that the 4-dice guard has about a 25-30% chance of beating the adept ... but I might be quite wrong there. In fact, I suspect that that interpretation (one massaged number vs another massaged number?) could be in the category of "damn lies and statistics", and might NOT be the right interpretation. I wouldnt' stake anything of value on that. (what DOES it mean? I don't know.)

I think it's pretty safe to say (without doing a REALLY exhaustive test ) that your perception 4 guard is not very likely to beat your adept. However, if my guess above is correct, he has a better chance than I previously thought.