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, don’t get discovery, it is probably good enough for our purposes.
I understand the desire to handle infinite series. I don’t 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 don’t 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 don’t succeed unless another roll was a success. You get a 3 if you roll a 0 but don’t 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 don’t 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 don’t 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 don’t. 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.