Assuming there really is anything to be found, that is.
Of course you can always throw a die for each individual searcher (I like to use a different coloured d6 for each character so I know who has and hasn't succeeded, just in case they want to keep the news to themselves or something, the sneaky bastards).
However, if you want to skip over all that and get the show on the road, you can always convert all those individual rolls into one combined d100 roll.
The formula for finding the probability from a whole bunch of d6 (or any other die type) is this:
1 - [(chance of NOT showing the desired pips) to the power of (the total number of dice)]
Assuming that you might just want to know the numbers and not have to do a whole lot of tedious maths to find out, here are some I prepared earlier:
Number of d6 | Chance of at least one 6 showing on a handful of d6: 1 - [(5/6)^N] |
Chance of at least one 1 OR 2 showing on a handful of d6: 1 - [(2/3)^N] |
---|---|---|
1d6 | 16.7% | 33.3% |
2d6 | 30.6% | 55.6% |
3d6 | 42.1% | 70.4% |
4d6 | 51.8% | 80.2% |
5d6 | 59.8% | 86.8% |
6d6 | 66.5% | 91.2% |
7d6 | 72.1% | 94.1% |
8d6 | 76.7% | 96.0% |
9d6 | 80.6% | 97.4% |
10d6 | 83.9% | 98.2% |
11d6 | 86.5% | 98.8% |
12d6 | 88.7% | 99.2% |
13d6 | 90.6% | 99.4% |
14d6 | 92.2% | 99.6% |
15d6 | 93.5% | 99.7% |
16d6 | 94.6% | 99.8% |
17d6 | 95.4% | 99.89% |
18d6 | 96.2% | 99.93% |
19d6 | 96.9% | 99.95% |
20d6 | 97.4% | 99.96% |
I would rather only let the best (or worst... depending whats going on) character roll for a given action.
ReplyDeleteIf Conan cant bash the door open then there is no reason to give Dumbledore a chance to succeed. This also works the other way around: The thief can climb up the wall without a roll but the fighter with plate armour has to succeed a difficult climb check.
Neat! I wonder if this is how the FASERIP chart was born.
ReplyDeleteAnd thus we see that 9 elves are worth twenty humans.
ReplyDelete