Rules Botches MathOK - time to get rigorous:
The rules on "Botches" are on page 197 of Changeling 2nd Ed. Read the first paragraph if you need a reminder of what a Botch is per Changeling.
In the interest of space, I will only focus on rolls vs. difficulty 6, using from 1 to 5 dice.
The first thing to do is to determine the chance of rolling at least one 1 on multiple dice. Start with the chance on 1d10 - that's easy - it's 1 out of 10, or 10%.
So, how about on 2d10? Independently, there's a 10% chance on either die. But what about both at the same time? Look at it this way: the chance for both dice to roll a 1 is 1% (10% for the first die, 10% of 10% for the second die to match). That's why 2d10 are used for percentile dice. OK, sorry for the white rabbit - the chance to roll at least one 1 on either die is 19%. Hunh? you say.
You have to separate out two scenarios here: First, die #1 rolls a 1 and die #2 rolls anything - that's a 10% chance (we only care what die #1 rolls in this case). Second, die #1 rolls 2-10 and die #2 rolls a 1. Why do I say die #1 rolls 2-10, and not 1-10? Because we've already handled die #1 rolling a 1 in the First scenario. There's a 90% chance for die #1 to roll 2-10, and a 10% chance for die #2 to roll a 1, so the Second scenario yields 10% of a 90% chance, or a 9% chance. To figure the total chance, add the chances of the two (mutually exclusive) scenarios together: 10% + 9% = 19%.
For 3 dice, here are the scenarios and percentages:
1st: die #1: 1 (10%), #2: 1-10 (100%), #3: 1-10 (100%) = .10 x 1.00 x 1.00 = 10%
2nd: #1: 2-10 (90%), #2: 1 (10%), #3: 1-10 (100%) = .90 x .10 x 1.00 = 9%
3rd: #1: 2-10 (90%), #2: 2-10 (90%), #3: 1 (10%) = .90 x .90 x .10 = 8.1%
Total chance to roll at least one 1 using three dice is 10% + 9% + 8.1% = 27.1%
You also see how the 1st and 2nd scenario are repeated from our two-dice calculation? They are the same, with the same percentages. So here's the 4th and 5th scenarios:
4th: #1: 2-10 (90%), #2: 2-10 (90%), #3: 2-10 (90%), #4: 1 (10%) = 07.29%
5th: #1: 2-10 (90%), #2: 2-10 (90%), #3: 2-10 (90%), #4: 2-10 (90%), #5: 1 (10%) = 06.561%
This gives total chances using 4 dice of 34.39%; using 5 dice: 40.951% (From here on out I'm going to round figures to two decimal places.)
OK - using the same methods, here's the chances for the other ranges we'll need for the scenarios for percent chances to Botch vs difficulty 6:
1-5 (any failure): 1 die: 50%, 2: 75%, 3: 87.5%, 4: 93.75%, 5: 96.88%
2-5 (non-1 failure): 1: 40%, 2: 64%, 3: 78.4%, 4: 87.04%, 5: 92.22%
Now for the Botches - similar methods, however, the scenarios conform to the Botching rules - which are that there have to be more ones (1) than successes (6-10 vs difficulty 6). When you're analyzing the scenarios I've used, please remember that. To help with our calculations, the scenarios are mutually exclusive. Also, the chances to roll numbers are not for a single die - they are for multiple dice, being more restrictive left to right. This cuts down on the number of scenarios we have to detail, thanks to the percentages we've already calculated above.
1 die: Total chance 10%
1st scenario: rolling one 1.
1 die: 1 (10%) = 10%
2 dice: Total chance 9.5%
1st scenario: rolling one 1, and rolling a failure.
Either die: 1 (19%), other die: 1-5 (50%) = 9.5%
3 dice: Total chance 12.09% (increasing trend from here on...)
1st scenario: rolling two 1's, and rolling any number on the third die.
Any die: 1 (27.1%), one of the other two dice: 1 (19%), last die: 1-10 (100%) = 5.15%
2nd scenario: rolling only one 1, and rolling two non-1 failures.
Any die: 1 (27.1%), one of the other two dice: 2-5 (64%), last die: 2-5 (40%) = 6.94%
4 dice: Total chance 13.9% (by now you can see the more restrictive pattern for dice,
so I'll just number the sequences in the scenarios)
1st: rolling two 1's, one failure, and any number on the last die.
4: 1 (34.39%), 3: 1 (27.1%), 2: 1-5 (75%), 1: 1-10 (100%) = 7.00%
2nd: rolling one 1, and three non-1 failures.
4: 1 (34.39%), 3: 2-5 (78.4%), 2: 2-5 (64%), 1: 2-5 (40%) = 6.90%
5 dice: Total chance 17.33%
1st: rolling three 1's, and any number on the other two dice.
5: 1 (40.95%), 4: 1 (34.39%), 3: 1 (27.1%), 2: 1-10 (100%), 1: 1-10 (100%) = 3.82%
2nd: rolling two 1's, two non-1 failures, and any non-1 number on the final die.
5: 1 (40.95%), 4: 1 (34.39%), 3: 2-5 (78.4%), 2: 2-5 (64%), 1: 2-10 (90%) = 6.36%
3rd: rolling one 1, and four non-1 failures.
5: 1 (40.95%), 4: 2-5 (87.04%), 3: 2-5 (78.4%), 2: 2-5 (64%), 1: 2-5 (40%) = 7.15%
Now, how to reconcile this intuitively? Your chance of rolling a 1 on any die increases with the number of dice you roll. Your chance of succeeding on any die also increases (at a greater rate than rolling 1's), but at difficulty 6 and higher, your chance of failure on any die increases at the same (or a higher) rate! That means that having more dice doesn't help the ratio of #successes/#failures. You may get more actual successes, but you'll get even more failures. Then, to tip the scale, your chance of rolling a one (included in those failures) is increasing all the time (for more than two dice). All you need to Botch is more 1's than successes - you don't need at least half the dice to be 1's, or any other ratio test! So there goes the camel's back...
| Jhardhel'Healdan Rules Botches Math |
Site Map |