I was doing a little math to see how much better or worse a stress die is than a simple die.
My reasoning (I am not a mathmetician):
Max on a simple die = 10. Average (mean) roll on a simple die (SUM 1...10)/10 = 5.5
A stress die gives > 10, 5% of the time (roll of 1 followed by >=6)
If you ignore the potential infinite roll on a stress die and limit doubles to x2 the average (mean) roll is (SUM 0,2...9,[(SUM 2,4,6,8, 10,12,14,16,18,20)/10])/10 = 5.5
Of course the median is 10, but I do not think that is the most useful here (I am not a statistician either).
So the average value of either roll over time is the same 5.5.
Stress dice will give a result > 10, 5% of the time (5/100), but only give a botch (on 2 dice) 1.9% of the time (19/1000).
The variance is much greater on stress dice.
You are 2.63 times more likely to beat 10 on a stress die than you are to botch!
I THINK the chances of botching match the chances of beating 10 when you have 5 botch dice (4.99% v 5%) or more.
Assuming that beating 10 is AS GOOD as botching IS BAD (the cancel each other out), then, if you have the choice, stress dice are the way to go as long as you have 4 botch dice or fewer.
For the mathematicians out there, does this scan?
[color=red]Warning- Mathematics involved!
Could be boring or even offensive to some!!!
(You have been warned...)
Your errour is in substituting the value of "2" for a roll of "1" after the first doubling and stopping there. That is where the value of a stress die begins to explode logarithmically (and I may have badly undervalued it below by stopping at 2 doubles.)
No practicing mathematician either, but I remember the principles, so let's fake it via the brute force method...
The "average" roll of a stress douse, "only" considering the possibility of two doublings, is:
The average of the rolls
0, #, 2, 3, 4, 5, 6, 7, 8, & 9, where value of # is the average of:
!, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, where the value of ! is the average of:
@, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 (and etc for @... but enough for now*.)
So, ! = 220/10 (+ #/10, but we aren't going there...), or ~22(+).
That value, fed back up, makes * = 132/10, or ~13(+).
So, our average of a stress douse (limiting double 1's to a max of two!) is, in fact, closer to something over 5.7. Probably not quite 6.0, but possibly close, or possibly asymptotic toward that or somewhere near it (ie, "approaching 6 but never reaching it".)
(And yes, you are indeed no statistician- the median, the "middle roll", with an equal number above and below it, is not 10, but somewhere theoretically between 5 and 6. The median here is not useful, except to determine what the odds are of whether one will roll "below" or "above" (possibly WAY above) average. I'm not sure where you're getting "10" from.
The casual "of course" worries me too... I am reminded of faking my way thru proofs in high school that way, or trying to.) 
All that aside, your analysis of High Roll vs. Botch is in the ballpark.
But it would take someone who remembers their asymptotic formulae & etc to dis/prove us both. Any takers? Bueller? Bueller?...
8)
(Actually, I know I've seen complete proofs/analyses of this somewhere. I'll bet someone has saved the link...)
* If we go one more doubling, to approximate a value for "@", we get:
?, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80 (or 440/10 = 44+?/10)
That 440/10 44 adds 44/10 to !, which adds 4.4/10 to *, which is our roll of "1" in our initial roll. That additional +.44, averaged with the others, adds another .044 to the final value, making that closer to 5.75, and tells me we are closing in on our "asymptote".
Tho' each "doubling" gets bigger, it also gets divided by another power of 10, so is getting smaller and smaller in a reverse logarithmic dive, and so never spirals out of control, just adds less and even less each time.The 10 is from the median value of an exploded die with one doubling 2,4,6, etc, although, I guess you can't get a 2, oh well, so I messed that up. I meant to give the median between 20 (10 x2) and 2 (1 x 2), but how do you get a one on a stress die?
BTW, I stopped with one doubling.
I did a brute force simulation in MS Excel by creating hundreds of cells that would simulate dozens of stress dice and I did get an average value somewhere around 5.75.
You did proofs in high school? I could hardly find a teacher that could do arithmetic in high school. I learned basic algebra in college (I learned to factor polynomials) and got A's, but that is a far as I went. I swear, I must have gone to the worst high school (with the least skilled and knowledegble teachers) in the country based on what people have told me about their high school experiences. My World and American History teacher never said a word to us he just sat in the corner reading Sports Illustrated and would ignore anyone who asked a question.
The average roll for a non-stress die is 5.5.
The average roll for a stress die is 5.75.
The initial roll (X) breaks down as: 90% (2 to 0) unmodified, 10% (on a 1) double value of an "extended" roll (Y).
"Extended" rolls (Y) break down as: 90% (2 to 10) unmodified, 10% (on a 1) double value of a further "extended" roll (Y).
The recursion is easily accounted for in the equations by simply reintroducing the suitably-weighed expected value on the right-hand side:
E(Y) = (1/10)(2+3+4+5+6+7+8+9+10) + (1/10)2E(Y)
E(Y) = 5.4 + (1/5)E(Y)
(4/5)*E(Y) = 5.4
E(Y) = 6.75
That would cover the old "quality die", which doubled on a 1, but always counted the initial zero as a 10. Filling back into the first roll:
E(X) = (1/10)*(2+3+4+5+6+7+8+9+0) + (1/10)2E(Y)
E(X) = 4.4 + (1/5)*6.75 = 4.4 + 1.35 = 5.75
It's like watching two Bonisagus argue magic theory.
Fascinating, but somewhat incomprehensible.
I'm enjoying this, though.....
Eric
Um, three by my count, not two. You are indeed math challenged. 8)
gaaaah. 
Worse, THAT is like watching three of them argue, falling asleep and have one of them ask your oppinion. 
Hi,
I might have missed something here, but:
(Edit: I did miss something! 
Fixing.)
A d10 has 10 sides. Thus, the chance that any particular outcome of a single roll occurs is 0.1, or 10%.
On a stress roll, an outcome of 0 is worth 0, the outcome of any number other than one is worth the number rolled, and a 1 is doubled.
Therefore, the value of a die equals the sum of the integers 2 through 9 times 0.1 plus 0.1 times double the value of a die where the 0 is worth 10, or
stress_die = 4.4 + 0.1*2*subsequent_die
subsequent_die = (5.4 * 0.9) + 0.12subsequent_die
Solving yields
stress_die = 4.4 + 5.4* (0.1*2**1 + 0.01*2**2 + 0.001*2**3 + 0.0001*2**4 + ....)
= 4.4 * sigma(n=1, n-->inf, 0.2**n)
= 4.4 + 5.4* 0.25
= 5.75
Thus, a stress die has a slightly better value than a simple die. On the one hand, a stress die can achieve much higher results. On the other hand, the simple die can roll a natural ten or natural one.
The probability of rolling an arbitrary number of ones is accounted for. The one thing not accounted for here is the real value of a botch.
(The first time around, I forgot to account for the fact that 0s become 10s after a 1 is rolled.)
Anyway,
Ken
So basicly, what has been determined is that with stress one is more likely to achive great successes (as opposed to minor ones).
What has not been determined, is if stress or simple gives a better chance of success...
And the math so far is also lacking any accounting for botching (which in many cases act as a negative value...)
You are clearly wise in the ways of mathemagics. 8)
And with a much clearer memory of the actual formula involved, thanks! (Altho' it's good to know that the old brute force method gave me the same answer!)
HKD's stab at the botches was in the right direction, formal math or no. For "success", one would use the "median", or the "middle" roll- where the same number of rolls fall above and below it. And again, it is slightly better with a stress douse than a simple one.
[Edited to make another sarcastic comment]
That is the point where my math skills curl up in a ball and cry.
Anyway, I did make the assumption that a single botch has the same subjective absolute value as a roll of >10. On 2 botch dice the chances of a double botch are pretty slim (1/10 x 1/100 = 1/1000).
Would y'all say that, subjectively, a roll of >10 is as good as a single botch is bad?
Actually, no, but that's a diff discussion.
In my experience, while nice, a roll of around 16 may not get a tough job done (esp in combat), but a botch is almost sure to make a simple challenge tougher.
Think of it this way- if you entered into a significant situation knowing, guaranteed, that you would roll >10 and then a botch, alternating, one then the other, every other turn- would you be excited?
A 1 is better than the chance of a botch, but does not balance the botch itself, imo.
(As an aside, as SG I have, at times, houseruled that any "1" will yield at least a value of "10", regardless of the second roll. Those "4's" and "6's" are sometimes just too disappointing to watch.)
Definitly NO! I.e. in combat a single botch often results in a value of a negative value >10 (reducing the total to 0), and in spellcasting a single botch results in warping, total failure of spelland gods know what else...
In many cases you succeed as long as you don't botch (many simple spells work that way...)
So I'd say each botch would be closer to -10 in effective/subjective value...
Since they reduce your total to zero, the bigger your bonuses, the more you'll be affected by botches. It has the interesting consequence that, if you're outmatched in combat, you should seek conditions that add lots of botch die to even the chances.
Anyway, since someone asked, here are the chances you have to beat a given difficulty on a stress or simple roll, assuming you don't have any bonuses.roll stress simple
3: 80.0% 80%
6: 49.0% 50%
9: 16.9% 20%
12: 5.9% 0%
15: 3.8%
18: 2.7%
21: 0.6%
24: 0.6%
That cant be right
You have a 16.9% chance to beat a 9 on a stress die.
(10% (roll a 9) + 6% (1 folowed by > 5) + 0.9% (1 followed by 1 followed by anything but a 2)
?? Looks like you are agreeing...
And I do agree with Fruny, I worked this out with another player a little while ago (cut and paste from e-mail):
(edit: Note "target" means that number or greater)
Anyway, great mechanic. Remove stress when you can. Take a risk if you want to do something truly brilliant.
Hope this helps.
-K!
It does now, I could have swworn that it said something different when I posted, yet if it had then my quote would not have shown the same numbers.
Guess I was stressed and rolled a single botch.
Math can do that to ya.
It means 0.21 + 0.22 + 0.23 + 0.24 + ... 0.2**infinity.
Anyway,
Ken