I often play with dice pools. Where each die represents a projectile or an event.
I hardly ever use 2d6 or 3d6 or #d10 or whatever is possible.
No! All I need is a d6. And all the tricks you can do with it to get to a certain chance.
Limits
Before you apply tricks to a d6 to get to a certain chance of succes. You need to understand the limits of a d6. The d6 offers the following chances for succes:
16.7% = 1/6
33.3% = 2/6 or 1/3
50.0% = 3/6 or 1/2
66.7% = 4/6 or 2/3
83.3% = 5/6
100.% = 6/6
What you see here are chances with a base number. We got the 6. But also the 2 and 3.
This means, we can build up smaller portions of chances with more rolls to come.
2x2 = 4, 25% is possible with 2 rolls of 3/6.
3x3 = 9, 11% is possible with 2 rolls of 2/6.
Of course, with a tripple roll, we also have the bases 8, 12, 18 and 27.
Additions?
Are not really possible. So we use 1 re-roll. Which can be different. This re-roll is only applied to the failures. With this, we can add to a certain roll.
If we want a chance of 4/9 for example. We roll 3/9 instead.
What is remaining is 6/9 being a fail. This 6 is a fail on itself. But can be divided by 6. Thus adding 1 more.
Let's look at the base 9 series.
1/9 = 2/6 x 2/6
2/9 = 2/6 x 4/6
3/9 = 2/6
4/9 = 2/6 + R1/6
We aren't there yet.
The next one in the series is 5/9. Which is the opposite of 4/9. So instead of rolling for succes. You roll for fail.
5/9 = F(2/6 + R1/6)
6/9 = 4/6
7/9 = F(2/6 x 4/6)
8/9 = F(2/6 x 2/6)
Before I continue. I should share the list of base 8.
1/8 = 3/6 x 3/6 x 3/6
2/8 = 3/6 x 3/6
3/8 = 3/6 x 3/6 + R1/6
4/8 = 3/6
5/8 = F(3/6 x 3/6 + R1/6)
6/8 = F(3/6 x 3/6)
7/8 = F(3/6 x 3/6 x 3/6)
How to interpret the 3/8?
First the player rolls each die 2 times, 3/6. If both times are a succes, it counts. But if even one of the 2 fails, it is a fail. In that case, the player may roll once more, 1/6. If the roll is a succes. The die counts as a succes.
If the player has 8 dice like this. The first roll would put 4 of them in the fail pool. The second roll would put 2 more in the fail pool, totalling to 6. Then the fail pool is rolled and 1 gets put back in the succes pool. We have 3 in the succes pool and 5 in the fail pool.
If the player needed the 5/8 instead. The whole roll can be done as described above. But at the and, the fail and succes pool switch sides.
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Cumulative chain rolls
When we roll for a 1/9 chance. We roll 2/6 x 2/6.
Both rolls need to be a succes for the total roll to be a succes.
But we can also count each succes.
The first roll on itself can be one. And both rolls can be 1.
2/6 is a 1/3 chance.
If this counts towards the total chance and we roll twice again, we get:
1/3 + 1/9 = 4/9
Is this the correct calculation?
While the answer is correct, the calculation should have done differently. Since the second roll depends on the first roll. And a chance on having 2 succesfull rolls should count twice. But the chance on having only 1 succesfull roll is 1/3 x 2/3, since a portion would become 2 successes.
1/3 x 2/3 + 2x1/9 = 4/9
I put that correct calculation in there because confusion often happens.
Explosive dice
Once the second roll is a succes. Whatever comes after. We just doubled the number of dice. Hence I call these explosive dice. And we can set a limit to how many rolls are allowed. Which doubles as a maximum possible number of successes.
_7/36 = L2E1/6
16/36 = L2E2/6 this is the 4/9.
27/36 = L2E3/6 which is another 75%.
40/36 = L2E4/6
55/36 = L2E5/6
72/36 = L2E6/6 which is indeed 2 succes.
_43/216 = L3E1/6
104/216 = L3E2/6
189/216 = L3E3/6 which is 7/8.
304/216 = L3E4/6
455/216 = L3E5/6
648/216 = L3E6/6 which is indeed 3 succes.
By combining one of these with another roll. You can get many other chances. But keep in mind that the result is cumulative. And the chance is an average result instead.
No limits?
While I don't like infinite rolls. Like for example discarding each 6. Just to get to that base 5. Which is regressive.
I do like the progressive way, where the number of succes only climb up. It is just that after a dozen successes, you pause that roll and see how they fair later on.
And we will be meeting a very interesting number here.
1/5 = L∞E1/6 Base 5?
1/2 = L∞E2/6
1 = L∞E3/6
2 = L∞E4/6
5 = L∞E5/6
∞ = L∞E6/6
While this roll with base 5 is an option. There are several things to keep an eye on:
1. The 1/5 is an average result. Meaning that it is not a true base 5.
2. Explosive rolls cannot be imbedded into the other 2 rolls. We cannot have a FL#E or RL#E. But, we do have L#EF and L#E(+ R).
With L#EF, we roll and the failures count as succes. Then this number of dice are rolled again with the same rules. Adding more and more to the pool of succes. While you won't see L∞EF1/6, it can be used when the base is for example 8 and the chance for each die has to be for example, 7/8th. We can only get 7/8th by having a fail roll imbedded. Example:
105/64 = L2EF(3/6 x 3/6 x 3/6)
L#E(+ R), is simpler to understand. We are still rolling for seeing how much succes the current roll has. The failures get another chance here.
33/64 = L2E(3/6 x 3/6 + R1/6)
Fake base 5:
1/5 = L∞E1/6
2/5 = L∞E2/7 Imposible...
3/5 = L∞E(3/6 x 3/6 + R1/6)
4/5 = L∞E(2/6 + R1/6)
Since the explosive rolls can multiply the number of dice. You might as well use multiple dice to begin with for that fake base 5. Thus 2/5 could be 2x L∞E1/6.