# Revised Calculations for Original Research

I have reread the proper errata for The Original Research. The probability of getting a discovery using the rules as ERRATA'd is 0.3671.

My post in this thread was wrong.
https://forum.atlas-games.com/t/catnip-garlic/129/1

This requires a +3 modifier:

7+3=10 with p=0.1
8+2=10 with p=0.111
9+1=10 with p=0.1
10+0=10 with p=0.01
12-2=10 with p=0.011

8+2 is not used because an 8 is "complete failure" which must be accepted as well.

This totals to 0.332. The probability of a reroll is 0.04781 We now have an infinite series of:

0.332SUM( (20.04781)^i,i=0..infinity)
=0.332*(1/(1-2*0.04781))
=0.3671

This means the expected value of Getting a breakthrough discovery using 1 magnitude effects is =60/.244366 Which is 163 seasons, or approximately 41 years, to get the correct number of discoveries. This does not include stabilization. I personally don't like those rules and use the house rule: "Stabilization takes 1 season, with no rolls to produce a lab text."

So, inventing say, magnitude 6 spells, a PC can expect to break the Hermetic Limit in about 9-10 years (with my stabilization simplification.) This means original research (if you're willing to take the warping) can be done fairly quickly by a PC.

Hu no...
the errata takes a 8+3 example.

You count it as a 11, not a 8. It means modified effect.

After that you can decide to also use a -1 modifier in order to reach discovery.

The way i see it is:

[die] + [real modifier (decided before rolling)] = experimentation result.
If experimentation result is not a discovery, you can modifier [experimentation result] by [+ or -MT/5 round up] to get a discovry, but only a discovery.

Otherwise said, if you don't reach a discovery with the real risk modifier, you can alter it by using a second modifier (+ or - MT/5 rounded up) modifier to get it.

It basically mean that if you get 0-4 you with MT 11+, you can get a discovery if you choose an initial risk modifier.

I don't like that at all... I thought, and feel more comfortable with the old:

[die] + [modifier chosen by magus, up to +or - MT /5 round up] = experimentation result; copying it to :

• all experimentation seasons while you invent the actual effect
• the stabilizing procedure.

And, I didn't agree with some post. IMO : you only have ONE chance to stabilize a discovery. If you fail, you can't try again, and have to retry experimentation to get another discovery, which you could try again to stabilize, one time.

No wonder you find the rules easy if you allow multiple trials to get a stabilizing...

Right. I knew 8 was supposed to work for some reason and couldn't for the life of me remember why. Well then, that puts the number at 0.36701. I fixed my original post so as not to confuse further readers.

except that you only fixed the infinite series calculation, not what comes before it - now it's more confusing than ever....

Oops. Should be good now.

First, a roll of 5+3 = 8 is complete failure.

Second, the probability of a reroll is 0.169 for +3 risk modifier.
[You need 9+, you don't reroll on {0 2 3 4 5 6 7 8} and 1+{2 3 4} and 1+1+{2} = 0.831]

Third, your infinite series suppose you reroll once, not twice.
once: P = b + aP
twice: P = b + a
P^2

== Actual values for +3 risk modifier.
Complete failure = {5+3} = 0.1
Reroll = {9+3}+ = 0.169
Not a failure = 0.641

P = not a complete failure, with reroll
P = 0.641 + 0.169 * P^2
P = 0.854
Complete failure, with reroll = 0.146

== Repeat with known complete failure
Now this is a pain: if you roll 12+3, it is a discovery but you can still get complete failure on rerolls. It should reduce Discovery by 0.0016 or so but I won't consider it.
Complete failure, with reroll = 0.146
Discovery = 0.332
Reroll, without failure = 0.123
Nothing = 0.399
[Not a discovery = 0.146 + 0.399]

P = not a discovery, with reroll
P = 0.545 + 0.123 * P^2
P = 0.587
Discovery, with reroll = 0.413

Overall
+1 risk = 12 % breakthrough * 72 % stabilize = 9 % success
+2 risk = 26 % breakthrough * 72 % stabilize = 18 % success
+3 risk = 41 % breakthrough * 65 % stabilize = 27 % success

--
tl;dr

Original research with a +3 risk modifier succeeds 27% of the time.

Personally I've always Felt the luck virtue should add into research. So just how much would an varying +1, 2, or 3 bonus break the system if it always moved you to the most favorable result it could reach. And it can also vary it's self on the stabilization roll.

Actually the one time I played a serious researcher my troupe agreed to allow luck to apply but only as a plus or minus one if needed to reach a discovery. And it couldn't save me from a complete failure I think.

No, I'm assuming you roll 8 and then add 2. That is why I had to edit my initial post.

If you roll a 9, yes you have to reroll twice, but you also get a discovery because 9+1=10

Here's the errata as it looks like you're not up on it, like I once was:

I actually am not sure probability theorem you're using there. And, Yes actually, my infinite series does account for rerolls. Notice the bolded two. 0.332*SUM( (2*0.04781)^i,i=0..infinity)

The set of Rerolls that are not a 9 or 12 die roll=
1 followed by 7,8,9,0
1,1 followed by 4,5,6,7,8,9,0
1,1,1, followed by 1,2,3,4,5,6,7,8,9,0 <- means my reroll probability should be 0.479 not 4781.

I agree with your quote below that a discovery on a roll of a 9 or 12 should not result in a failure. However, this does mean that on such a roll, a discovery has been made. Only one discovery can be found per effect studied.

Failures don't matter. If at any point in the process a failure or disaster is rolled before a Discovery then the process stops.

What matters is did I get a discovery first, or, if I did not, do I have a reroll? The way I see it is:

P(Discovery) =0.332
P(reroll and not have a discovery)=0.0479
p(result I don't give a crap about)=0.6201

Are you familiar with probability trees used in decision theory? That is how I justified my methodology to find the probability of a discovery on a +3 modifier.

See in the middle of :

herkules.oulu.fi/isbn9514257855/ ... er4_2.html

And thanks for doing the +1 and +2, it verifies what I knew, the probability of breakthrough was significantly lower and simply not worthwhile.

0.332SUM( (20.0479)^i,i=0..infinity)
=0.332*(1/(1-2*0.0479))
=0.367175

Am I the only one who reads "The normal effect of experimentation" to include adding the risk factor to the roll and that adding or subtracting the bonus based on Magic Theory happens from that point? So that if you took a +3 risk factor, your roll is now 4+3=7 and then you could add an additional 3 to reach 10 if your MT was 11? The errata said subtract 1, not only add 2 in order to reach 10.

Obviously, that makes it a lot easier to reach 10.

Also, am I the only one who thinks working with magnitude 1 effects is not worth it and the thing to do is magnitude 2 effects? Sure, you have a 10% chance of a single warping point each time you stabilize a discovery. But you had at least a 1% chance of a botching every time you had a invented a spell with experimentation. You would be roughly halving the time needed for the discovery

Or make a Criamon with Self Confidence, a good score in Enigmatic Wisdom and go for broke.

I read it like that, but it was so easier, I decided that i have been wrong. Besides, IMO, confidence can't be spend on seasonal results, hence warping from lab.

Yes. I just read through it. You are right. It's pretty clear from the example. The example is 8+3=11 (so you keep 11). And for the discovery 11-1=8+3-1=10 (discovery). So with Magic Theory 11 or higher and a risk modifier of 3, you get a discovery on rolls of 2x2x2, 2x2, 2x3, 2x4, 2x5, 4, 6, 7, 8, and 9. That's a 54.1% chance of discovery any given season. Then you successfully stabilize on roughly 75% of rolls. After two seasons you will probably have stabilized any discovery on average. So one year of work should net roughly one stabilization on average.

Chris

In the example you quoted, a roll of 8 was not a Complete Failure but a Modified Effect. Therefore a roll of 5+3 = 8 is complete failure.

And yes dwightemarsh and callen, I didn't apply the Discovery modifier properly. I may have time to fix my worksheet later.

Lets take the most simple example: rolling 0 on 1d10. You are kind of saying on 2d10 the chance of rolling any 0 is 2 * 0.1 = 0.2, whereas the actual result should be 0.19. You cannot double it like that. You must do 0.9 * 0.9 = 0.81.

Even if the first roll is a Complete Failure, maybe the second roll will be a Disaster. You cannot stop rolling.

But I can agree failures shouldn't matter. All other results can add up with Discovery, why would Complete Failure be different?

This is why I think Complete Failure is different:

That can't really be a true statement if you also get something from your efforts. In the other non-botch cases you have some degree of success, but not in this one.

Chris

By that logic a complete failure should stop a disaster botch or story event too.

After all an explosion, a twilight event, or accidentally summoning a djinn are all something. (Or are they not nothing )

No, sorry, that does not necessarily follow. "Get" can be very varied in what it means. Only if you forget about many of the definitions does your statement follow logically. If, for example, here it means "succeed in attaining, achieving, or experiencing; obtain," then getting nothing would imply a lack of success or gain. In this case it would not imply you do not have a significant failure or great loss. The way I read the sentence, "get" seems to be along this idea of gain/success. You are welcome to disagree with the meaning of "get" here, but my statement clearly does not imply what you say it does.

Chris

That was my initial POV, but... is the Discovery part of the spell's extraordinary result roll?

Consider: "Consult the Extraordinary Results Chart as normal to determine the effect on your spell."
Yes, the spell is a Complete Failure.

Next: "However, you may also add or subtract all or part of your Risk Modifier in order to get a Discovery in addition to the normal effect of experimentation."
You can get a Discovery in addition to the Complete Failure.

Yes, I completely understand what you're talking about, and yes, I am using that concept, just not in the direction you're thinking about. This is why I said I'm doing probability chains from decision trees.

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

etc

Try drawing it, and then writing out the numbers this far by hand using power notation (don't compute everything) and group like terms. The algebra works out to that formula I use, otherwise I wouldn't do it.

I think the question of when to stop rerolling is up to interpretation, and basically comes down to a SG ruling. I see a two different ways of doing it:

1.) Roll all rerolls in order. If a discovery comes first, disregard failures and complete disasters. Otherwise, a failure/disaster before discovery means to stop rolling, as you now have an event happen.

2.) Roll all rerolls. Failures always override a discovery and make it somehow "false", or the notes get destroyed etc.

With my Criamon idea, the thought was that he would use the confidence to comprehend twilight and get a positive rather than negitive result. I was not considering using confidence to increase the simple die subtracted from the warping points. Or do you consider going into twilight a seasonal result? I don't know,it seems like the same sort of twilight as happens when someone botches a spell.

Of course, whether confidence can be used to comprehend twilight is debatable. When I created a thread on that question before, I thought the majority position was yes, but that hardly is definitive.