IMO, using confidence on twilight from adventures when you botche a spell sure.
Using confidence to mitigate the consequences of seasonal things (like aging or twilight post breakthrough), hell no.
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Twilight from one source yes but from another no? That makes no sense at all.
Sometimes people get flat tires because there is a slow leak in the tire. Sometimes they get flat tires because the tire receives damage. Once you get a flat tire, however, you change them exactly the same way.
You check for twilight when you get two or more warping points all at once. Why would the reason that you get two or more warping points matter once you have gone into twilight and are trying to comprehend it?
Try with these:
Branch 1: Roll Discovery with p=0.100
Branch 2: Roll Failure with p=0.400
Branch 3: Roll a reroll with p=0.500
Becomes:
This totals to 0.100. The probability of a reroll is 0.500 We now have an infinite series of:
0.100SUM( (20.500)^i,i=0..infinity)
=0.100*(1/(1-2*0.500))
= 1/0 oops!
Is it clear now?
If you don't get 0.200 with this test, your method is invalid.
No.... I'm using the formula: SUM( 1/p^i,i=0..infinity)=1/(1-p), this only works for p<1. For p>=1 the sum is unboundable and infinite. Your formula is SUM(1,i=0..infinity)=infinity, and you should not get 0.2. It's not a counter proof. What it's saying is that if the reroll probability is too high i.e >=0.5, you can end up with infinite rerolls.
Out of curiousity, what is your background in probability? I'm getting my M.Sc in statistics and part of the reason I enjoy RPGs is the interesting probability problems that arise.
Oh, and read the bottom of page 27 of True Lineages. It is speaking about inventing something over multiple seasons, and if you at any point roll a failure or disaster, the magus must stop their research and start over. OR something like that, I'm at work now and read it last night and forgot to include it in my post. While this does not specifically address the reroll issue, it is at least helpful.
And how can you in game, justify a failure and discovery on the same roll? It sounds like saying 1*0=1, not 0.
Mathematically, I believe Tugdual is correct. If there are three outcomes Good, Bad, and Retry (where we keep trying until obtaining either Good or Bad), with probabilities g, b, and r = 1-g-b respectively, then the probability of ending up with the Good outcome is g / (g+b), while the probability of ending up with the Bad outcome is b / (g+b). It doesn't matter how high the Retry probability is, as long as it's less than 1 - there are still well-defined probabilities for eventual Goods and Bads.
I don't know what the 2* is doing in HTH's formula. Maybe I'm not thinking about the same problem you are.
Sigh Here it goes again: The Two is because you reroll twice. If we didn't reroll twice, it wouldn't be in there. Let me demonstrate again:
Branch 1: Roll Discovery with p=0.332
Branch 2: Roll Failure with p=0.6201
Branch 3: Roll a reroll with p=0.0479
Branch 4: (from Branch 3) Roll Discovery with p=0.332
Branch 5: (from Branch 3)Roll Failure with p=0.6201
Branch 6: (from Branch 3) Roll a reroll with p=0.0479
Branch 7: (from Branch 3) Roll Discovery with p=0.332
Branch 8: (from Branch 3)Roll Failure with p=0.6201
Branch 9: (from Branch 3) Roll a reroll with p=0.0479
Branch 10: (from Branch 6) Roll Discovery with p=0.332
Branch 11: (from Branch 6)Roll Failure with p=0.6201
Branch 12: (from Branch 6) Roll a reroll with p=0.0479
Branch 13: (from Branch 6) Roll Discovery with p=0.332
Branch 14: (from Branch 6)Roll Failure with p=0.6201
Branch 15: (from Branch 6) Roll a reroll with p=0.0479
Branch 16: (from Branch 9) Roll Discovery with p=0.332
Branch 17: (from Branch 9)Roll Failure with p=0.6201
Branch 18: (from Branch 9) Roll a reroll with p=0.0479
Branch 19: (from Branch 9) Roll Discovery with p=0.332
Branch 20: (from Branch 9)Roll Failure with p=0.6201
Branch 21: (from Branch 9) Roll a reroll with p=0.0479
What are the outcomes of this? I'm going to abbreviate how to use a decision tree, using i.e. B1 as Branch 1.
P(Discovery)=B1+B3B4+B3B7+B3B6B10+B3B6B13+B3B9B16+B3B9B19+.....
=0.332+0.04790.332+0.04790.332+0.04790.04790.332+0.04790.04790.332+0.04790.04790.332+0.04790.04790.332+...
=0.332*(0.0479+0.0479+0.04790.0479+0.04790.0479+0.04790.0479+0.04790.0479+...)
=0.332*(0.04792+(0.0479^2)4+...)
=0.332(0.4792+(0.0479*2)^2+...)
QED.
I see. So your calculation yields a Discovery probability of 0.332/(1-0.04792) = 0.367 or so. But note that the same calculation would yield a Failure probability of 0.6201/(1-0.04792) = 0.686 or so. Those two probabilities add up to 1.053 or so, which can't be. Or can it? Are you allowing Discovery and Failure to occur simultaneously?
Note that in your formula for P(Discovery) you have both B3B4 and B3B6 - but that seems to double-count the event where both rerolls from B3 come up Discovery. Or do you allow two simultaneous Discoveries (in which case 0.367 is an expectation, not a probability)? Or do you stop if the first B3 reroll yields Discovery (in which case the B6 probabilities need to be multiplied by something much smaller than B3)?
In short: let's define the process exactly before trying to decide what the probabilities are.
Why is the probability of a discovery so low? I forgot about rerolls and I intentionally left out Complete Failure earlier. Even if I drop rolls of 9 and 2x5 for totals of 12 or 13, the probability is still well over 0.4 prior to rerolls. The chance of a reroll is greater than 0.1, which of course makes the overall probability of Discovery even further over 0.4. Anything that gets you a probability under 0.4 must be incorrect.
Chris
It is not double counting.B3B4 is the probality of getting a discovery on the first reroll and B3B7 is the probability of getting a discovery on the second reroll. B3*B6 is the probability of rerolling on a reroll. In fact, what you are pointing out is the fact that I'm assuming a simultaneous probability tree. In reality the tree will only evolve sequentially.Those are two different types of probabilistic processes. Damn. Now I need to rethink things a bit.
Hmm. The 0.6201 includes the events of a failure or distaster. This means that it is not the correct probability (it is too large) to divide by the chance of rerolls. The probability used should be at most 0.5101. A botch happens with at least p=0.01, and p(failure)=0.1. It should be noted again,that my numbers are glossing over the fact that if you botch/fail the reroll process stops. It's basically a best case scenario,and the difference in the results will be very small, As pointed by Tugdual.
The probability of rerolls is very well explained in my first post. If you get a total of 12 with p=0.1,you have a discovery and can't get another. The errata talks about fully using the risk modifier. It does not say you can then add subtract the risk modifier again on top of the result gotten. The flavour text does not support getting a discovery with a probability of 0.541/(1-0.0479)=0.5682. That would simply be way too fast. It would mean a player could make a breakthrough (the hardest!) discovery in about 5 years of time. Unless you want it to be that easy for your PC's to shatter the limits of Hermetic Magic multiple times, and that would make for a crazy interesting game!!
As I said:
P(Discovery)=B1+B3B4+B3B7+B3B6B10+B3B6B13+B3B9B16+B3B9B19+.....
=0.100+0.5000.100+0.5000.100+0.5000.5000.100+0.5000.5000.100+0.5000.5000.100+0.5000.5000.100+...
=0.100*(0.500+0.500+0.5000.500+0.5000.500+0.5000.500+0.5000.500+...)
=0.100*(0.5002+(0.500^2)4+...)
=0.100(0.5002+(0.500*2)^2+...)
=0.100*(1 + 1 + 1 + ...)
Reductio ad absurdum.
QED.
FYI, Physics but I develop software languages.
Did you read the example? They use a +3 risk. The roll is 8+3=11. 1 is subtracted from 11 to get a discovery. So, yes, it does say exactly that. Maybe someone made an error, but that is what it says.
Chris
He quoted it, he sure read it.
Sorry I can't help myself, but no wonder you're trying to poke the weirdest hole in my proofs. 
Anyways, I have to work new results that are immune to such silly ideas. It may take me a day or two to figure out the sequential trees... They multiply infinitely.
I'm not arguing that point. I'm saying you can't take 4+3+3=10 because that doubles the risk factor. What they are saying in the errata is: 8+3=11, and 8+2=10. Whatever gets added or subtracted to the die roll has to stay within -3 to 3.
:mrgreen:
Or you could do recursive approximations:
a) no reroll
b) reroll on 1st step only
c) reroll on step 1-2
d) reroll on step 1-3
It should converge fast enough.
ArM 05 , page 20:
No. That is not what they're saying. Read this part.
Note the use of the word "also." That means "in addition." There's the as normal part: 8+3=11. In addition to that add a number from -3 to +3. The spread is still the same, since if you want to compress it to find the domain and range you get a domain of integers [0,6] instead of [-3,3]. Either way there are 7 of them. But the chance of reaching 10 is much greater now. That may not have been what was intended, but it is what was written.
Chris
Ok, so this is how I figure it.
First, the reason we are trying to figure out the percentage chance of reaching a discovery is that we are trying to figure out on average how long it will take to reach a discovery. As long as we get to within a percentage point of the true percentage of get discovery, donāt get discovery, it is probably good enough for our purposes.
I understand the desire to handle infinite series. I donāt think the current formulations of the problem are correctly handling all the nuances. The way I see it, there are five categories of results when you make a stress die on the Extraordinary Results Table with a +3 risk factor.
Failure: if you botch or if you get an 8, you donāt have anything to stabilize. The probability of a botch is .1*(1-.9^4)=..03439 The probability of an 8 is .1 (roll a 5 and add 3). So, the probability of this result on a single die is .13439
Success unless another die is a Failure: If you get a 7, 9, 10 or 11, you succeed, unless you were forced to make more rolls and one of them was a failure. You get a 7 if you roll a 4 or if you roll a 1 then a 2, so the probability of 7 is .11 You get a 9 if you roll a 6 or if you roll a 1 then a 3 also for a probability of .11. You get a 10 if you roll an 7 for a probability of .1 Finally, you get an 11 if you roll an 8, if you roll a 1 and then a 4 or if you roll a 1, another 1 and then a 2. Thus, your probability of rolling an 11 is .111. Thus, the probability of having a success unless another die is a failure is .431
Not a success: If you get a 3, 5 or 6, you donāt succeed unless another roll was a success. You get a 3 if you roll a 0 but donāt botch, which has a probability of .06561. You get a 5 if you roll a 2 and a 6 if you roll a 3, so the probability of a not success is .26561
Success but roll more dice: If you get a 12 or 13, you will succeed unless one of your further rolls is a botch or an 8. You get a 12 when you roll a 9, which has probability of .1 You get a 13 when you roll a 1 and then a 5 which has probability of .01. Thus, the probability of getting a success but still having to roll some more and risk failure is .11
Not success, but keep rolling: The probability of the above cases is .941. So, there is a .059 chance that you have to roll twice more with no default success.
So, lets introduce the following notation for rolling two dice, that the result of the first die is F for Fail, S for success, NS for not success, SR for success but having to roll two more dice and R for to roll two more dicel.
So the following events all lead to no discovery (F,F) (F,S)(F,NS) (F,SR) (F,R) (S,F) (NS,F) (NS,NS) (SR,F) (R,F) and has probability .321268
The following events lead to a discovery: (S,S) (S,NS) (NS, S) and has probability .414717
The following events are equivalent to rolling a 12 or 13, in the first place, in that the person now has to roll two more dice but so long as they donāt botch or get a total failure, they are good. (S,R) (S, SR) (NS,SR) (SR,S) (SR,NS) (R,S) and has probability .204112
The following events have are equivalent to rolling again, and its probability can be distributed thoughout the other events. (NS, R) (R, NS) and has probability .031342
The following events mean that there is a default success, but they have to roll 4 more times without screwing it up. (SR, SR) (SR,R) (R, SR) and has probability .02508
The following events mean there is not a default success and they have to roll 4 more times (R,R) and has probability .003481
Distributing the reroll thoughtout the other events, I get the following conditional probabilities
Chance of failing without having to roll anymore: .331663
Chance of succeeding without having to roll anymore: .428135
Chance of having to roll twice more but succeeding if you donāt roll a botch or an 8: .210716
Chance of having to roll four more times: .02948512
So, multiplying all of those factors by the .059 chance of having the first result be 14+, we have an additional .0196 probability of failing, an additional .0253 probability of success, an additional .0124 chance of having to reroll but having success be the default and a vanishingly small chance .0017 that they have to roll four dice to finally get their answer
If you already have a success, it makes the chart much simpler, as you no longer have to distinguish between non successes and successes and between rerolls that count as successes and ones that donāt. Let us label this F, S and R, for failure, Success and reroll, where F has a probability of .13439, S has a probability of .69661 and R has a probability of .169.
The following events lead to no discovery: (F,F) (F,S)(F,R)(S,F)(R,F) and has probability .250719
The following events lead to a discovery without any more rolling: (S,S) and has probability .485265
The following events are equivalent to just rolling again and has probability .235454
The following event is the chance of having to roll four more times: .028561
So, once you distribute the chance of just having to roll again amongst the other options, you have
Chance of failing at this point: .327932
Chance of success without having to roll anymore: .634711
Chance of having to roll four dice: .037357
Multiplying each of those by the adjusted conditional probability of .1224, we have the following Conditional probabilities
Fail: an additional .0402
Success: an additional .0777
Roll four dice, which we only look at if we have to.: .0046
So, adding these all up, we have a probability of failing to get a discovery of at least .4+.0402+.0196=.4598 , the probability of success is at least .431+.0253+.0777=.5340 and there is that .0062 probability that someone is now required to roll 4 dice. As I said before, that seems good enough for most practical purposes.
Yes, exactly. Some may disagree on Complete Failure vs. Discovery, but they should be following the same mathematics, just getting a greater chance of Discovery. I'm not going to go through and check carefully, but at a quick glance your math looks right.
Chris