As a space fight, it puts me in mind of lights on a display panel or HUD. The colouring could be reconsidered in that light.
The colours don't communicate any information, they're just what I had handy for the playtest. One die type has dots, the other has numbers, which is the important difference. But no, the colours can be anything, so they won't necessarily be those colours in the final thing.As a space fight, it puts me in mind of lights on a display panel or HUD. The colouring could be reconsidered in that light.
You've probably already considered this, but are there likely to be any sight-related readability issues on the red-yellow scheme? There's a type of colour-blindness for which red looks more green and less bright... which seems quite close to saying it looks yellow.
those % are too high, there is no way one die has a 2% chance of exploding 5 timesI'm not sure I'm reading it right. If I'm reading it right, the table then looks like this:
The % in anydice is 0.78. Less than 1%. I calculate it as lower still.those % are too high, there is no way one die has a 2% chance of exploding 5 times
it certainly is lower than that: (1/6)^5 / 2, so 0.0000643 (to meet TN 6, exploding 5 times is twice that)The % in anydice is 0.78. Less than 1%. I calculate it as lower still.
Though with 1 die you can't succeed vs TN 3 (0%) but with 2 dice you can (25%).This makes a too-small-by-one pool twice as likely to succeed at a task than an just-enough-to-succeed pool.
TN 2. 1 die = 50% chance, 2 dice = 25%
TN 3. 2 dice = 25%, 3 dice = 12.5%
...
Basically, you are replacing a 50% success die with a guarenteed 100% success.
In other words, it would often make mechanical sense to have a less skilled character try something to improve the chance of success, if you know the TN going in.
I am having the darnedest time finding the free basic rules. Do you have a link?Again, please have a look at Slayers. The basic rules are free.
I rounded up from 1.56%. This is what AnyDice shows me. I think I must be reading it wrong.those % are too high, there is no way one die has a 2% chance of exploding 5 times
Forgive me if I've totally gotten this wrong, but if I understand, when using a d6, the definition of exploding means you roll a 6, right?I rounded up from 1.56%. This is what AnyDice shows me. I think I must be reading it wrong.
View attachment 283646
yes, hence the chance for one die to meet a target of 6 is (1/6)^5 / 2 (the /2 because there is a 50% chance the last roll scores one point), or 0.00643% chance of this.Forgive me if I've totally gotten this wrong, but if I understand, when using a d6, the definition of exploding means you roll a 6, right?
So in order to get 5 exploding dice, you need to roll 5 sixes in a row, is that correct? And that would be defined as 6*6*6*6*6, or 1 in 7,776 chance.
Apologies, I wasn't precise - the data is spread over a few freely available downloads.I am having the darnedest time finding the free basic rules. Do you have a link?
TARGET NUMBER -> | 1 | 2 | 3 | 4 | 5 | 6 |
1 die | 50% | 8.33% | 1.39% | 0.23% | 0.04% | 0.01% |
2 dice | 75% | 33.33% | 9.03% | 2.08% | 0.44% | 0.09% |
3 dice | 87.5% | 56.25% | 25% | 8.22% | 2.29% | 0.57% |
4 dice | 93.75% | 72.92% | 43.4% | 19.68% | 7.16% | 2.24% |
5 dice | 96.88% | 83.85% | 59.98% | 34.30% | 15.86% | 6.15% |
6 dice | 98.44% | 90.63% | 73.05% | 49.32% | 27.58% | 12.96% |
# of explosions needed | is 1 in.... | as a % |
1 (TN 2) | 6 | 16.66666667 |
2 (TN 3) | 36 | 2.777777778 |
3 (TN 4) | 216 | 0.462962963 |
4 (TN 5) | 1296 | 0.07716049383 |
5 (TN 6) | 7776 | 0.0128600823 |
1 in | 1 in | 1 in | % | ||
TN | base chance | # explosions needed | chance of those explosions | and then 50% final roll | as a % |
1 | 6 | 0 | 1 | 2 | 50.000 |
2 | 0 | 1 | 6 | 12 | 8.333 |
3 | 0 | 2 | 36 | 72 | 1.389 |
4 | 0 | 3 | 216 | 432 | 0.231 |
5 | 0 | 4 | 1296 | 2592 | 0.039 |
6 | 0 | 5 | 7776 | 15552 | 0.006 |
This calculates 6s adding an additional die.Prince Valiant has a rule that if all the dice are successes, an additional success.
The Burning Wheel family of games have various rules which make 6s "open ended" ie each 6 adds another die to the pool, which is also open-ended.
Calculating the odds for open-ended dice is a bit tricky for the reasons @Thomas Shey gives, but some rough noodling around is possible.
Eg with 3D, the probability of no sixes is 5/6 cubed, so the probability of at least 1 six, ie actually counting as a pool of 4D+, is 91/216, or not much short of half. So it is a meaningful change to the possible range of outcomes.