# How does the probability of success change by increasing the size of roll & keep dice pools or altering the difficulty number?

I want to better understand impact on probability of success granted by stat increases with the following dice mechanism:

$$\text{Dice Test} = \text{Roll}\ [3 + \text{STAT}]\text{d}6 : \text{Keep Highest 3}$$

• Result compared to difficulty number, success if equal or higher.
• STAT ranks from 0-5.
• AnyDice data.

At STAT 0, difficulty 11 represents a 50/50 chance of success. How much more likely is success at STAT 1-5 and difficulty 11?

Also, how is the probability of success altered by increasing/decreasing the difficulty at every STAT level? Say difficulty 5, 7, 9, 11, 13, 15, and 17 at every level of STAT 0-5.

I imagine the second question is significantly more calculation intensive than the first. I am unsure how to determine these numbers myself. If someone wanted to simply explain how the changes to probability of success could be computed for the aforementioned formulae, and not do the computation themselves, I would be much obliged.

• Say difficulty 5, 7, 9, 11, 13, 15, 17, and 19 at STAT 0-5, that's 8 different difficulties and 6 possible stat values, is this correct? What stat value goes with what difficulty? (Also, if it's always "keep highest 3", 19 is not really achievable.) Sep 22 at 18:21
• @biziclop Any dice test can be of any difficulty, and any stat level can attempt any dice test. So, the stat levels are difficulty numbers are not correlated. As to difficulty 19, you're right. I was considering extant static modifiers when I wrote that sentence. But those are irrelevant to this question. Will edit. Sep 22 at 19:20
• Ah okay, apologies then, I misread the sentence, I thought each stat value would have a set difficulty. (I don't know why I did that cos thinking about it doesn't make sense anyway.) Sep 22 at 19:31

This is a quick look at numbers to give an overview of probability changes.

I've used AnyDice program to calculate chance of success for a range of STAT and DC values, then plotted those on a line graph.

Starting with just a success chance plots:

Nothing surprising - for very high/low success chances, the value of extra rolls is diminished and, in general, each additional extra die is worth less than previous one. With DC 19 being impossible on 3d6, we can drop it from further consideration.

Second, let's look at the difference an additional die makes at different number of rolls.

So regardless of number of dice, the additional die has the highest benefit around 50% success chance. For low and mid DCs, that benefit is higher the lower the original number of dice is. But the lower the number the dice, the steeper decline in benefit after 50% chance, leading to high base number of dice benefiting more from extra dice at high DCs.

We can look at similar graf representing the value of -1 DC (so that success chance differences remain positive).

Similarly to additional die case, the effect of +/- 1 DC is most notable around 50% success chance before DC change. Other than that, I can't make any clear conclusions from this graph.

Relative versions of those 2 graphs (the increase in success chance / the base success chance) both show that relative value of additional die or -1DC grows higher with harder the check is (higher DC or lower dice number).

I've looked at some other values that felt like they could provide additional insight, like (P(DC=X)-P(DC=Y))*P(DC=X), but was unable to glean anything from them.

I was in doubt about posting this as it doesn't have a definitive answer, but having already gone through generating the graphs decided it is beneficial enough to deserve posting.