It would not take too many Foci (10+ total force) to force you to make an addiction test every week. At first, I thought big deal! So what if I am addicted to my Foci. I love them, would not leave home without them, etc. But the section talks about how the addiction can turn from mild, to severe, to burnout, to permanent stat loss, to coma. It all depends on how many cumulative times you fail your weekly check (threshold 2).

So, if you have 10 dice (5 body, 5 will, 5 logic) of physical and mental addiction resistance, how long will it be before you are addicted, and need to spend karma and abstinence to cure yourself? I guess we'd have to ignore edge. I am not sure you can spend edge on something that is continuous. (or can you?)

This is where I need a math expert. After how many weeks will there be a 50% that you have been addicted. You are rolling 10 dice, and for you to fail, 9-10 of those dice must be 1-4.

If you show you work, you get extra credit

Basically, you'l fail 1 roll in 10. Which, if you're running 10+ in total Force, means you're progressing to be a grade worse every 10 weeks if the addiction is only physiio- or psychological, or if the addiction counts as both every 5.

OK, I gave a try myself, but my math is shaky and would appreciate someone checking it out:
Prob of 1 dice 1-4 = 2/3
Prob of 9 dice 1-4 = (2/3)^9 = .026
Prob of 10 dice 1-4 = (2/3)^10 = .01734

Prob of 9 OR 10 dice 1-4 = 1-(1-.026)(1-.01734) = 1-(.974)(.98266) = .04289

Call this the prob of failing each week. PF=.04289

Prob of succ = (1-PF)^N where N is the number of weeks

Week Prob of NOT being addicted
1 0.95711
2 0.91606
3 0.87677
4 0.839165
5 0.803173
6 0.768725
7 0.735755
8 0.704198
9 0.673995
10 0.645087
11 0.61742
12 0.590938
13 0.565593
14 0.541335
15 0.518117
16 0.495895

Conclusion: Your addiction to Foci will increase by one step every 4 months. Within 1.5 years you should be in a coma.

Friends don't let friends have Foci. Same holds for skillwires...

I'm away from all books, but here's some help on the math.

I'm not clear what the expression is you're trying to evaluate. If you want to know the probability of "rolling 4 or less at least 9 times with 10d6" then that's ~10.40%. Another way to state that (if I understand your condition correctly), is that the probability of "rolling 5 or more at least 2 times with 10d6" is ~89.60%. I think that's what you're looking for, which is basically what RHat said.

What you're doing is independently evaluating the probability of rolling 9d6 and getting no successes (~2.60%), and then rolling 10d6 and getting no successes (~1.73%); that's actually a much rarer event and not the same thing. So that part of your math is wacky.

Your cumulative probabilities are basically right if what you're doing is assuming they succeed until the odds reach roughly 50/50, at which point they fail. Your initial numbers are off (as explained above), but you're figuring out the cumulative probabilities correctly.

You mentioned edge. Again, I don't know if edge is allowed or not, but it's not that hard to calculate. Again, roughly speaking here, the inclusion of a single point of Edge into the roll jumps the probability from ~90% to ~93%. After that, each point of Edge increases the probability by ~1%.

I believe using Edge to Try Again has been calculated to about a 5/9 success rate per die on the other boards; based on that calculation, odds of success when able to use Edge to try again are about 97.5% - but I know I wouldn't be inclined to allow Edge to be used for this test.

That's a good point and I should clarify: I wasn't making a distinction between pre-rolling or post-rolling Edge, or rerolling successes (partially because that's a pain, and partially because I wasn't entirely sure what conditions to evaluate). The numbers I mentioned were only in reference to the effect of adding dice to the dicepool. When you take into consideration what Edge does beyond giving you extra dice, then the odds continue to bend in your favor.

The differences between pre-rolling and post-rolling in this case aren't that large to begin with (~1.2%), and they converge as the edge pool grows larger (at Edge 6 we're talking about ~0.2%). With respect to re-rolling, my back-of-the-napkin results give me ~98.8%, which is probably well within the margin of error for the fractions you're using.

If Edge isn't allowed, then that's a moot point...but still good to know.

And_Knowing_Is_Half_The_Battle.jpg

Well, thanks for the reply. I tried to find derive your answer and could not. I googled a number of pages that tried to explain it, but their math went over my head. Thanks anyway

So if I take your 10%:
1 0.896
2 0.802816
3 0.719323136
4 0.64451353
5 0.577484123
6 0.517425774
7 0.463613493
8 0.41539769
9 0.37219633
10 0.333487912

You will get addicted in 7 weeks, and therefore go into a coma in half a year. I bet the writers did not take into account how unusable that makes Foci and skillwires.

That seems to be the case.

I believe there was some discussion about that in one of the larger errata threads that sprang up shortly after the SR5 release last July. There's also the issue that foci addiction as described in the Magic chapter doesn't necessarily line up with foci addiction in the...you know I forget the name of it, but it describes addiction rules in general...chapter.

Some of the relevant threads can be found here.

As far as Foci go, I believe what the writers took into account is wanting there to be a cost to keeping as many active at once as you could. You don't risk addiction AT ALL if you don't go crazy with it (that is, you don't have more than Force=Magic foci active at once), and your risk is very much modulated by the extent to which you go past the "safe" point. As for Skillwires, at Addiction Rating 5 it's pretty easy to not be using them for a long enough period of time to set the Threshold to 0, and thus be safe.

I'd quibble with that. It's possible that in your games skillwire-users can take two weeks off (or more) between runs, but that's hugely problematic for 1) games where the pace of shadowrunning is more desperate and / or frenetic, and 2) the canonical legions of skillwire-implanted wage-slaves who are supposedly the primary users of this technology.

I wrote a probability calculating program for SR5 a while back and, in the process, computed the probability of success on edge rerolls. It's exactly 5/9. Good job other board.

(The other super fun fact to come from that is the threshold for deciding whether to add edge before a roll or reroll. In the absend of limit considerations, you should add before the roll when edge is 2.5 times bigger than the pool. Good rule of thumb)

Turning that ponderous machinery towards this problem I found:

P(0 hits) = 0.017 (which comes from 2/3^10)
P(1 hits) = 0.086 (which comes from combination(10,1)*(2/3)^9*(1/3)^1, i.e.: there are 10 different ways to pick 9 failures (which show up w/ prob (2/3)^9) and 1 success (which shows up w/ prob 1/3)

So P(0 hits or 1 hit) = P(0 hits) + P(1 hit) = 0.103, which JesterZero already relayed to you. Glad we're in agreement.

So the expected value of the number of tests until you fail a roll is ~10. That comes from a pretty complex derivation (that I don't entirely remember) that you can find here. http://mikespivey.wordpress.com/2012/02/09/consecutiveheads/

But if you want grungier probabilities:
P(I fail a test after burnout addiction on roll N) = P(I have failed 5 addiction tests on roll N) =

N P(N)
1 0
2 0
3 0
4 0
5 1.22e-05
6 6.67e-05
7 2.13e-04
8 5.19e-04
9 1.07e-03
10 1.95e-03
11 3.28e-03
12 5.14e-03
13 7.64e-03
14 0.011
15 0.015
16 0.020
17 0.026
18 0.033
19 0.041
20 0.050

So the answer to your original question is 20 weeks until 5% chance of burnout.

Yeah, I've been meaning to get around to writing something to crunch complex probabilities for SR5 for a while now... I'll get around to it sometime soon, I'm sure.

spending edge after the fact seems far more beneficial. (edit: apparently it even removes your limit from the initial roll, as i understand it... though i can't double check, as i'm afb. /edit)

oh, and i seem to recall focus addiction is based on total force of focuses used, not focuses used at one time. so if you have 3 different focuses, each force 3, and use each of them once in the time period, it's the same as if you used them all at once.

on the flip side, addictive stuff is really addictive, which is probably how it should be. i mean, if it was easy to avoid addiction, it wouldn't be an addiction, right?

There was some debate on the subject, but I'd submit this interpretation is absolutely not RAI.

My combinatorics made it 10.405% for less than 2 hits, but that is probably rounding that gave you 10.3%

But, as far as I know you can use Edge to greatly slow down the process

Using edge only on failed addiction tests (to conservate edge, the chances to succeed are slightly higher if you always use edge, however the benefit of using edge beforehand is extremely small for this test. Basically, using edge before rolling only has an effect if you get exactly one 6 on the original pool and that reroll gives a hit. A complete waste.

Here is the table for risk of failed addiction test if you use edge after your roll to "fix" when you roll 0 or 1 hit

Edge Using edge
1 7,5%
2 5,4%
3 3,9%
4 2,7%
5 1,9%
6 1,4%
7 1,0%


Already at edge 2 you have halved your risk (doubled your expected life as an addict)

At edge 5 you have decent chances of survival.

The downside is that addiction test every week is lot of bookkeeping and is quite boring for your other players that have to sit through your addiction. Try to make it so that you and GM takes care of this between game sessions or have a good quick way to pick your dice and roll them without breaking up gameplay too much.

I expect, however, many a GM would say no way in hell you're Edging that.

Ehm.. given the power of foci ... yeah, I would not allow it either now that you mention it... I tend to say that all edge spent during downtime (for finding gear etc) is "spent" the next adventure starts.

The edge regained every night rule only applies during runs for me to prevent players from using edge every day to do something stupid and start with full edge every run anyway.



Anyway, the full table for risk of addiction based on a pool is

Pool Risk of addiction (threshold 2)
1 100,0%
2 88,9%
3 74,1%
4 59,3%
5 46,1%
6 35,1%
7 26,3%
8 19,5%
9 14,3%
10 10,4%
11 7,5%
12 5,4%
13 3,9%
14 2,7%
15 1,9%
16 1,4%
17 1,0%
18 0,7%
19 0,5%
20 0,3%


And, for reference

risk of addiction Threshold 3 (or risk of failure to get 3+ hits with any test for that matter)

Pool Threshold 3
1 100,0%
2 100,0%
3 96,3%
4 88,9%
5 79,0%
6 68,0%
7 57,1%
8 46,8%
9 37,7%
10 29,9%
11 23,4%
12 18,1%
13 13,9%
14 10,5%
15 7,9%
16 5,9%
17 4,4%
18 3,3%
19 2,4%
20 1,8%

Just for future reference: anydice.com and "output [count {5, 6} in 10d6]"

What is the probability of rolling 10 dice and getting either 9 1-4 or 10 1-4?

For the second case, that's pretty easy. You have a .4 chance of getting a 1, 2, 3, or 4 on a single ten-sided die. So the probability is .4^10.

For the first case, it's a little tougher. The easiest way is to invert it, so what is the chance of getting EXACTLY 1 success (5-10) and 9 failures (1-4) on 10 ten-sided dice So your probably of a single success is .6, times 9 failures is .4^9. There are 10 permutations (the first die can be a success, the second one, the third one, etc.), so we multiply that answer by ten.
= 10 (number of permutations) * (.6^1 (your one success) * .4^9 (all those failed dice))
= .001572864
Plus the answer from the first part you get .001677722. So each roll, you have a 0.17% chance of getting one or fewer successes when throwing that many dice.

I have seen this quote before (you don't have a problem if FociForce <= Magic), but I have not seen it in the rules. Can anyone point me to the page#?

Anyway, if this is the "real" rule, then the actual RAI (No more than Magic number of individual Foci, with their sum Force no greater than Magic*5) is rendered completely moot by the addiction mechanics. After all, thanks to addiction the real limit is 5X less than the practical limit!

Oh, and thanks to those who were helping me with math. Pragma came closest to getting some of those neurons sparking. I seem to remember something with parentheses and factorials above and below the fraction: N!/(N-r)!r! then the spark went out, and I lost it...

Glad to be of service. I went back and finished the answer, by the way, so check out the final result. It turns out that you have a 5% chance of attribute loss on week 20.

And I'll post my code (to save you the trouble, RHat) when my server comes back up. Dyndns just went to paid hosting, and I'm a cheapskate.

Um.

You playing World of Darkness over there?

Sorry, I was on the phone and yes, thinking of another game

Changing it to 1-4 on a six-sided die turns it to .67^10 + 10 (.33^1 * .67^9) = 10.8% chance of failure.

Not to be that guy, but...if you evaluate the actual fractions* instead of just using rounded decimals, you'll get 10.4%, which is more correct.

* (4/6)^10+10*((2/6)^1*(4/6)^9) = 0.1040 <-- (I didn't reduce the fractions to make it more obvious where the sides / dice are).

Hopefully that also helps TimTurry see how to evaluate other expressions as well. Although if you don't feel like breaking out calculators, Excel, and the BINOM.DIST and / or COMBIN functions, something like SmallRoller can be a very helpful tool.

Reminds me of TRoll.
http://topps.diku.dk/torbenm/troll.msp

The 'T' stands for 'Turing Complete'

I had totally forgotten about TRoll. Which is strange, because I remember preferring it to AnyDice, so much so that I wrote a number of the contributed expressions for it. Thanks!

I got the author to include a specific way of allocating and counting dice so we could look at the distribution curves.

Basically the idea of distributing 24 dice between the 6 attributes (min 3, max 6) then rolling and taking the best 3. But the important part we needed was not "what does a distribution of 5d6 look like?" but more "how do the different allocation possibilities compare?"

So it was important to know that there was 4 dice in STR but 6 in CON.