I have found a neat way to reduce a certain number of die rolls for my game.
The 5/6th per roll, is used a lot. (Rolling 6 means a fail, and the sequence stops immidiately)
And it happens that the number of rolls can be a lot too.
This time, I don't want to alter the "exponential" effect into a "lineair" effect.
I did some calculations and thought of the following:
1 rolls of 5/6th equals 1 roll of 5/6th.
2 rolls of 5/6th equals 1 roll of 4/6th.
4 rolls of 5/6th equals 1 roll of 3/6th.
6 rolls of 5/6th equals 1 roll of 2/6th.
10 rolls of 5/6th equals 1 roll of 1/6th.
To compare the expected chances with the new chances:
0,833333333
0,694444444 is lowered with 4% to 4/6th.
0,482253086 is increased with 3,68% to 3/6th.
0,334897977 is lowered with 0,5% to 2/6th.
0,161505583 is increased with 3,2% to 1/6th.
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I can allow "attacking" players to determine, which kind of rolls they use. But the bottomline is that 10 dice rolls will become 1 if they want to.
I bet that all players will combine the positiver outcomes with normal rolls. And that they will not use the negative rolls.
Chances grow exponential small, if players know that the target will make use of these rolls. So the high number of rolls, will be rare.
The only valid options now are 4 and 10 rolls. Since they have a slightly better outcome.
4 is an average situation.
And 10 is often the upper limit, caused by the new "slower projectile". Any chance on a higher number of rolls would mean that attacking players would not even take chances.
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I can't imagine that no one has thought of this before: Is there a similar case with this?
Are there better ways to do this?
How do you guys feel about this approach?
I don't see how CRT helps me with this.
I am already using it too. I didn't know it was called CRT though. I was calling it, "having 2 prints on each die side". (I actually got that idea from MtG, when I thought of rolling creatures)
Black numbers:1,2,3,4,5,6
Red numbers:0,1,1,2,3,3
Black is used for most, "roll this or less, then...", situations.
Red is used for damage out put.
So 2 numbers on each side. But there is room for a third. But in what way? Since the black numbers got it covered.