# Revised Calculations for Original Research

Hmm. I see what you're saying about what the errata says now. There really are two ways of reading it. I was reading it with the understanding that you can't use more than your risk modifier to get a full discovery. The default assumption is that you're already trying to use the whole risk modifier, so you can add part of it to get a discovery, or subtract up to the full value of your risk modifier to the die roll to make a discovery. It does sound like we're truly in the grey realm of "rules as written" vs "rules as intended" I think. This is honestly a good question for clarification.

Dwightmarsh,

Your base botch probability is 0.01, or are you assuming botching in level 3 aura? Then your 3.4% make more sense.

Anyhow. I've devised a method algebraically that makes sense no matter what numbers you use.

There are 3 probabilities of interest:
1.) P(Discovery)=d
2.) P(reroll)=r
3.) P(nothing happens {i.e, no fail or disaster})=n

What is the sequence of events that leads to a discovery when a player rolls the die?

Discovery, or reroll then discovery or reroll then nothing on the first reroll then discovery or reroll and reroll then discovery or reroll and a reroll then nothing then discovery etc.

This is d+rd+rnd+rrd+rnrd+rrnd +rnrnd+ rnrrd + rrnrd + rrrnd+rnrnrd + rnrrnd+rrnrnd+rnrnrnd+....
=d+rd+rnd+r^2d+2r^2nd+r^2n^2+r^3d + 3r^3nd+3*r^d+

This is d(Sum(r^i*(1+n)^i,i=0..infinity) = d(Sum((r+nr)^i,i=0..infinity) = d/(1-(nr+r))

So if d=0.332, r=0.0479, n=0.5101 (the highest value, doing an experiment in no aura at all,= 0.6201-0.1-0.01) then our sum is 0.3566162875.

If we use the alternate formulation d=0.541, r=0.059, n=0.26561, d/(1-(nr+r))= .5846569104

This is again the formulation saying that a discovery voids a second result saying failure/botch.

Nah, you get another botch dice for every point of the risk factor. So with a +3, you get 4 botch dice

Right, I forgot about that.