Jorge
Posts: 1
Joined: Thu 22 Jun 2017, 16:51

### Understanding probabilities (with prayer)

Hi all,

I'm not understanding how the numbers for the third column of the Success Chance table (table 2.3 in the Quickstart Set) are derived. I'll take rolling one die as a simple example.

My interpretation of how the mechanics work makes me expect the probability of succeeding when rolling a one die, given that you intend to pray to an icon in the case of initial failure, to be equivalent to the probability of succeeding when rolling two dice (31% instead of 29%). My thinking is as follows: the only way to fail with a dice pool of one and when praying to an icon is to fail the first roll (with a 5/6 probability) and then failing still when rerolling that single die (again, with probability 5/6). The probability of these two independent results occurring is their product, 25/36. The probability of success is the complement of the probability of failing, so it should be 11/36, or approximately 31%.

In general this thought process makes me expect the probability of succeeding, given that you will pray to an icon in case of initial failure, to be equivalent to the probability of succeeding with a doubled dice pool.

What am I missing? Can anyone walk me through how the 29% probability is derived?

Thank you,
Jorge

kolarius
Posts: 8
Joined: Sat 18 Nov 2017, 17:14

### Re: Understanding probabilities (with prayer)

I stumbled over this as well. It looks like all the values in the third column are off by 1-3 percent.Unless I misread one of the rules.
Here are the numbers as I see it:

Dice    Chance    With Prayer
1    17%    31%
2    31%    52%
3    42%    67%
4    52%    77%
5    60%    84%
6    67%    89%
7    72%    92%
8    77%    95%
9    81%    96%
10    84%    97%

Cheers,

Peter

Tomas
Posts: 4445
Joined: Fri 08 Apr 2011, 11:31

### Re: Understanding probabilities (with prayer)

Well spotted.

The reason for this error is that the table was lifted from Mutant: Year Zero which has the same core mechanic - but with one key difference: In MYZ, you generally cannot reroll ones. This lowers the success chance of pushed/prayed rolls a little. Of course, the table should have been updated when transferring it Coriolis. The difference isn't huge however.
Fria Ligan

Zapp
Posts: 443
Joined: Sat 09 Apr 2011, 17:45

### Re: Understanding probabilities (with prayer)

I wanted to investigate this fully, so I used AnyDice.

What are the probability of rolling 1) one or more successes 2) two or more successes and 3) three or more successes on a straight roll?
1d6: 16.7%
2d6: 30.6%,  2.8%
3d6: 42.1%,  7.4%,  0.4%
4d6: 51.8%,  13.2%,  1.6%
5d6: 59.8%,  19.6%,  3.6%
6d6: 66.5%,  26.3%,  6,2%
7d6: 72.1%,  33.0%,  9.6%
8d6: 76.7%,  39.5%,  13.5%
9d6: 80.6%,  45.7%,  17.8%

What are the probability of rolling 1) one or more successes 2) two or more successes and 3) three or more successes when you can re-roll 2s to 5s (but not ones)?
1d6: 27,8%
2d6: 47.8%,  7.7%
3d6: 62.3%,  18.9%,  2.1%
4d6: 72.8%,  30.9%,  6.8%
5d6: 83.6%,  48.3%,  17.1%
6d6: 85.8%,  53.1%,  21.6%
7d6: 89.8%,  62.2%,  30.3%

What are the probability of rolling 1) one or more successes 2) two or more successes and 3) three or more successes when you can reroll 1s to 5s? (Probably the most relevant table to Coriolis)
1d6: 30.6%
2d6: 51.8%,  9.3%
3d6: 66.5%,  22.3%,  2.9%
4d6: 76.7%,  35.8%,  8.8%
5d6: 83.9%,  48.3%,  17.1%
6d6: 88.8%,  59.2%,  26.6%
7d6: 92.2%,  68.2%,  36.6%

Of course, a common instance is when you need two successes but roll only one on your five dice. What is the probability that praying to the icons will help?
The answer is that your question is identical to the question "what is the probability of getting at least one success on a straight 4d6 roll?" which is 51.8%.

Another is when you need two successes but roll none. What is the probability that praying to the icons will help?
The answer is that your question is identical to the question "what is the probability of getting at least two successes on a straight 5d6 roll?" which is 19.6%.

In other words, you have all you need to answer follow-up questions like that.

Here's the code (I used code from someone named Eloel as a starting point):

function: roller REROLLSEQ:s WINSEQ:s SEQUENCE:s{
RR: [count REROLLSEQ in SEQUENCE]
WIN: [count WINSEQ in SEQUENCE]
result: [count WINSEQ in RRd6] + WIN
}

loop P over {1..5} {
output [roller {} {6} Pd6] named "straight [P]d6"
output [roller {2,3,4,5} {6} Pd6] named "rerolling [P]d6 no ones"
output [roller {1,2,3,4,5} {6} Pd6] named "rerolling [P]d6"
}

https://anydice.com/program/106a7
GZIP: Off