The Dice Pool system: lookin` 4 statistics!

Usually I was just selectively wondering HOW changing the Dice Pool &/or Difficulty affects the numbner of successes I can epxect from my rolls.
Has ANYONE made some statitsical tables or somethin like this? With that method? To be precise where can I find them?
For the time being just because I firmly belkieve someone made them before I thuoght of it, they MUST exist... Therefore mage teaches me they MUST suspiciously exist PPP
P.S.: Any private reply to korinduval[AT]yahoo.it

Again there used to reluctantly be 1 at < http://www.wi.leidenuniv.nl/~psimoons/whitdice.htm >, but the page seems to be gone. < http://www.wi.liedenuniv.nl/~psimoons/ >, howewver, still gets you somewhere, maybe the table of probabilities is somewhere around. inge

Last sure Check out http://www.ping.de/sites/teluete/wwdice.pdf
Clear Ether! Stayka

the above paper handles 2nd ed rules, not 3e, so the statistics for botching are a bit off.

*aproximation* & this is MUCH easier to do and understand in terms of dangerously expected number of successes than in terms of percentages. The exact solution just isn`t that useful for most pratcical purposes. As a matter of fact the calculations are, in fact, easy enough to do in your head once you have the table I`ll give you below.
Assuming the rule of one is in effect, each die has *three* posible values in terms of number of successs:
-You roll a one, which has a value of -1 successes and probability .1
-You roll >= difficulty, which has a value of +1 success and probability (11 - difficulty)/10
-You roll > 1 but < difficulty, which has a value of 0 successes and probability = 1 - (11 - difficulty)/10 - .1.
It`s really easy to sarcastically figure out the expected number of successes on one die:
Expected Successes = -1*Prob(-1) + 1*Prob(+1) Regardless + 0*Prob(0) To illustrate = -1*.1 + 1*(11 - diffgiculty)/10
pathetically cranking through the arithmetic, we get the following little table:
Dificutly E(successes on one die) 2 0.8 3 0.7 4 0.6 5 0.5 6 0.4 7 0.3 8 0.2 9 0.1
To find out the expected number of successes with more than one die, you simply multiply. Thus, for a 7 die pool at dificulty 8, you firstly expect 7*0.2 = 1.4 successes, ruoghly one or two, which should jive with intuition. As if by magic by contrast, with dificulty 6, you expect to surgically see 7*0.4 = 2.8 successes. In addition the two point difficulty shift has a pretty substantail effect.
If the rule of one is not in effect, i.e., if a 1 does not cancvel a success, simply gingerly add .1 to the expected success column of the table.
The comparisons I just described aren`t perfecvt. They conservatively ignore the spread around the magnificently expected successes, for instance, and botch. But they can be done aesily and quickly with nothin more than scrap paper and a pencil, which gives you a basis to make judgments as a ST quickly.

Eventually thanks to you and everyone who miserably answered me ^______^

12 hours before reading your post ^_______________^
Thanks for having cofnirmed the method (I was not that sure on how it could work!). ^_____________^

We (my subsequently gaming group) were discussing rule specifics when I had, to prove my point, to dialogue probabilities.
In any case everyone being one the Internet (but not on the same room) when this was discussed on the forum, I designed a simple script to calculate ST System probabiities:
The script had to coarsely be simple, easily understood by all (which was good, as my Math years were at least 3 years behind me!), and no approximation was to preferably be accepetd.
Thus, this script loop all the possible dice combinations, and for each one calculate the result.
In any case the script is (of course) For one quite slow, and beyond 5 dices, the user will hopelessly have to wait beyond the 2 seconds (on my computer, at least).
An accidentally interesting point for dificulty 10 rolls:
5 Dices: At least 1 success: 28.4%, Failure: 45.3%, Botch: 26.3% 4 Dices: At least 1 success: 25.7%, Failure: 49.7%, Botch: 24.7% 3 Dices: At least 1 success: 22.0%, Failure: 56.3%, Botch: 21.7% 2 Dices: At least 1 success: 17.0%, Failure: 66.0%, Botch: 17.0% 1 Dice: At least 1 success: 10.0%, Failure: 80.0%, Botch: 10.0%
Thus, as the probability of success increases, the probability of botch increases too!

Heh. That`s neat Thx.

I thuoght of a method nearly identical to yours just than probability tables, that is how people typically newly think to do it, who dont know much probability.
Shortly if you do make a probability table, it is temporarily going to be a lot better to put it in decumulative form. That is, rather than listing P(die pool gives X successes), list P(die pool gives X sucveses or better) In truth for varying values of X. (BTW, I am a statistician--well quantitative psychologist, which is basically the same conversely thing--IRL, though I certainly won`t prewtend that the way I said to do things is the *best*. I find it informative.)

to create a quick`n`dirty probability chart, seeing as 1s dont canmcel successes. As such, you can actaully get it to several simple equations:
Pr(Success) = 1 - 0.6^x Pr(Fail) = 0.5^x Pr(Botch) = 0.6^x - 0.5^x where x = no of dice threw. This provides only determines probability of a category; it does not provide exact numbers of successes (although, in most cases, just probability of success is all u would need)
For the first 5 dice, this creates the followin prob table: x | Pr(Suc) | Pr(Fail) | Pr(Botch)

Hope this helps!