Is 3d8 better than 2d10 when all you need is one die to be high?


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Pathfinder Rulebook Subscriber

So the context is that I'm noodling on an idea where skill in a game is represented by the die you roll, as well as how many you roll. But instead of adding the resulting rolls together, you're just hoping for ONE of the dice to roll high.

So clearly a d6 is better than a d4 because 1/3 of the time you will roll higher than the d4 can roll in the first place. And a d8 is better than the d6, etc.

And clearly if you can roll 2d6 and take the highest that is (mathematically) better than 1d6, as now you have two chances for a preferable number.

All of that makes easy sense and my brain has no trouble coping with it.

Here's where it breaks down for me and I'm struggling with how to figure things:

Say a character rolls 3d8 and takes the highest result. How do I compare the relative strength of that roll to some other combination? Is 3d8 better than 2d10, on mathematical average? Is 3d8 better than 5d6?

How should I set up the equation to figure out the probability of getting any particular number as the high result in a given pool of dice?


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General rule of dice: more dice equals more consistency.

3d8 v 5d6...you will more consistently be in the 4-6 range on the 5d6 option than the 3d8 option, but you have 0 chance for a 7-8, eliminating your ability to make a skill roll requiring that high of a roll.

Basically you need to balance your need for a consistently good chance of getting at least one die in the upper half of the range versus a peak number.

Without knowing anything about the system beyond your description above, I'd probably take the 3d8 over the 5d6, as 3 dice is still a good chance of getting both upper range numbers and higher peak numbers.

I also would take the 3d8 over 2d10, simply because it's easier to roll a high number with 3 dice than 2.

This seems contradictory, but the trick is looking for balance in consistency versus peak. More dice gives consistency, bigger dice equals peak.

Example, with some limited math: If you are consistently successful with a result of 4, then 5d6 is going to be better than 3d8. A d6 rolls a 4 or better 50% of the time, and you have 5 chances at that. That's 5*.5 = 2.5. the d8 rolls at least a 4 on 62.5% odds, but you only have 3 chances. 3*.625 = 1.875. 2.5 is bigger than 1.875, so the 5d6 is better. Now if the desired results is a 6, the answer changes quickly. A 6 on a d6 is only a 16.7% shot. 5*.167 = .833. on a d8 you still have a 37.5% chance per die. 3*.375 = 1.125, which is bigger, and therefore better.

Does that give you a better idea of what you would be looking for?


jdripley wrote:

So the context is that I'm noodling on an idea where skill in a game is represented by the die you roll, as well as how many you roll. But instead of adding the resulting rolls together, you're just hoping for ONE of the dice to roll high.

So clearly a d6 is better than a d4 because 1/3 of the time you will roll higher than the d4 can roll in the first place. And a d8 is better than the d6, etc.

And clearly if you can roll 2d6 and take the highest that is (mathematically) better than 1d6, as now you have two chances for a preferable number.

All of that makes easy sense and my brain has no trouble coping with it.

Here's where it breaks down for me and I'm struggling with how to figure things:

Say a character rolls 3d8 and takes the highest result. How do I compare the relative strength of that roll to some other combination? Is 3d8 better than 2d10, on mathematical average? Is 3d8 better than 5d6?

How should I set up the equation to figure out the probability of getting any particular number as the high result in a given pool of dice?

This basic introduction to games with dice pools. I highly recommend taking a look at some of those systems - they're pretty robust and solid!

So, you're asking a pretty complicated question. The question of the (relative) value of multiple dice is complex and depends heavily on the context and what's happening.

Bear in mind, there are far more intelligent and educated people than I am around here; I'm a math tutor, but I only go up to highschool level, and even then some of it is rusty.

jdripley wrote:

Here's where it breaks down for me and I'm struggling with how to figure things:

Say a character rolls 3d8 and takes the highest result. How do I compare the relative strength of that roll to some other combination? Is 3d8 better than 2d10, on mathematical average? Is 3d8 better than 5d6?

How should I set up the equation to figure out the probability of getting any particular number as the high result in a given pool of dice?

So basically what you're asking for is statistical analysis, and the famous quote comes to mind:

Quote:
There are lies, <censored!> lies, and statistics

Statistics has its trends and truths, but it is a deceptive field for many, many reasons. That said, that quote is most likely referring to the ability of different people to take a look at the same statistics and walk away with differing conclusions - not quite the same thing, here... but the lesson that statistics can be "swingy" should be kept in mind.

So I'm uncertain of the underlying elements of your system, but (depending on what you're suggesting) having a single random number prooooobably isn't the best idea, because, at it's core, there really is no way (beyond extremely basic % chance based on number of faces) that you can determine what a single dice is going to do (and even the extremely basic % chance isn't exactly a good indicator, if your dice were crafted a little wonky; it's more of a "best guess" than anything else).

For clarity, if you're saying there are static DCs, it's a better idea to roll but keep 1, but still problematic; if you're suggesting multiple dice pools opposing each other, I'd recommend either static DCs or roll a bunch and keep two, generating more reliable statistics.

So, for example, on a d20 the probability of any given number coming up is roughly 5%; this is true whether you're aiming for a 3, a 20, or a 10: they all have the same probability every time (in theory). Or, as it says, here,

Quote:
If you roll a fair, 6-sided die, there is an equal probability that the die will land on any given side. That probability is 1/6. This means that if you roll the die 600 times, each face would be expected to appear 100 times. You can simulate this experiment by ticking the "roll automatically" button above.

Statistically, when you're looking at multiple dice values, things begin to enter into what's called a "bell curve" distribution - that is, certain median values are far more likely to appear than certain extreme values. That same link from before puts it this way:

Quote:

Now imagine you have two dice. The combined result from a 2-dice roll can range from 2 (1+1) to 12 (6+6). However, the probability of rolling a particular result is no longer equal. This is because there are multiple ways to obtain certain results. Let's use 7 as an example. There are 6 different ways: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1, whereas the result 2 can only be obtained in a single way, 1+1. This means you are 6 times more likely to achieve a 7 than you are to achieve a 2.

As the number of dice increases, the difference in probability between the most likely and least likely gets larger. The probability of rolling six sixes is 1 in 46,656! If you leave the experiment running for a while, you begin to see the bar chart take on a unmistakable shape - that of the binomial distribution.

You can see this same principle in action with how we used to roll ability scores via 3d6 - the average result (the expected result) was between 10 and 11 (the scores we give commoners; this is why, incidentally, in Pathfinder 1e monsters tend to have three odd scores and three even scores: they are presumed to have either 10s or 11s in their ability scores, just like commoners).

Fundamentally, in a dice pool mechanic, I'd suggest leaning into the bell-curve distribution method as your actual goal, here - that is adding dice together (but only to an extent). So, for example, since you're rolling dice, let's look at the examples you've given:

2d10: this value stretches from 2 to 20, with a median expected result of 11 (10+1=11; 11/2 = 5.5; 5.5*2=11)

3d8: this value stretches from 3 to 24, with a median expected result of 13 or 14 (8+1=9; 9/2 = 4.5; 4.5*3=13.5)

5d6: this value stretches from 5 to 30, with a median expected result of 17 or 18 (6+1=7; 9/2 = 3.5; 3.5*5=17.5)

So in that simplistic breakdown, obviously more dice are better. But if you limit the pool, the math changes dramatically.

Let's start with a well-known version of the "dice pool, keep the highest" mechanic: 5e's Advantage/Disadvantage.

You can immediately see the rather large statistical swing... and what's more, you can see that the effect both intensifies and fades towards the edges of that probability.

This other article points out that the value of the advantage/disadvantage changes depending on your target:

Quote:

Let’s take an example from the table. Assume you need to roll an 11 to succeed. With a straight d20, you have a 50% chance of success. With advantage, this goes up to 75%. That’s the equivalent of a +5 bonus to the roll, since you would also have a 75% chance of success if you only needed a 6 or better on a single d20. Pretty impressive!

On the flip side for the target of 11, disadvantage means you only have a 25% chance of success, equivalent to a -5 penalty to the roll (when you need a 16 or better on a d20, you also have a 25% chance of success).

So does that mean advantage/disadvantage is equivalent to +/- 5? Not all the time. In fact, it’s only that big when you need exactly an 11 on the die.

Let’s say you need a 15 on the die to succeed. With a single d20, you’ll only get this 30% of the time. With advantage, you’ll get it 51% of the time – about the same as you would get an 11 or better on a single d20. So advantage in this case is worth about a +4. Disadvantage, similarly, is about a -4: You only succeed 9% of the time with disadvantage, which is about the same as a single d20 with a target of 19.

At the extremes, advantage makes the least difference. If you need a natural 20 to hit, that’s only going to happen 5% of the time normally. Advantage ups your chance to 9.75% – equivalent to getting a +1. Disadvantage takes you chance down to 0.25%, or 1 in 400. That’s the chance of rolling back to back crits – not a common occurrence. But in terms of a modifier, it’s not much different from giving you a -1 to your roll when you need a 20 – it’s just about impossible.

Fundamentally, when you're aiming at a higher-than-average number (or lower-than-average number) the impact of having multiple dice fades.

That said, there's also obviously more of a chance with more dice in the dice pool.

So the question is, mathematically, is it better to have more lower dice or fewer higher dice when aiming for a high number?

It heavily depends on what you mean by a "high number."

Looking again at what you have here:

- 2d10
- 3d8
- 5d6

You have several choices. Obviously, adding more dice is better if you're keeping them all, but if you're talking about opposing the sets and keeping the highest, how does that suss out? Or just rolling and keeping the top one?

There are more elegant solutions, by far, but let's use the "brute force" method (aka the comparative chart method I linked before).

For a 2d10 there's a 10% chance that any given number comes up on the dice, but (doing the comparative chart from before) we see:

Quote:


Result: Roll Values: Total Chances
1: 1,1: one

2: 1,2 or 2,1: two

3: 1,3 or 2,3 or 3,3 or 3,2 or 3,1: five

4: 1,4 or 2,4 or 3,4 or 4,4 or 4,3 or 4,2 or 4,1: seven

5: 1,5 or 2,5 or 3,5 or 4,5 or 5,5 or 5,4 or 5,3 or 5,2 or 5,1: nine

6: 1,6 or 2,6 or 3,6 or 4,6 or 5,6 or 6,6 or 6,5 or 6,4 or 6,3 or 6,2 or 6,1: eleven

7: 1,7 or 2,7 or 3,7 or 4,7 or 5,7 or 6,7 or 7,7 or 7,6 or 7,5 or 7,4 or 7,3 or 7,2 or 7,1: thirteen

8: 1,8 or 2,8 or 3,8 or 4,8 or 5,8 or 6,8 or 7,8 or 8,8 or 8,7 or 8,6 or 8,5 or 8,4 or 8,3 or 8,2 or 8,1: fifteen

9: 1,9 or 2,9 or 3,9 or 4,9 or 5,9 or 6,9 or 7,9 or 8,9 or 9,9 or 9,8 or 9,7 or 9,6 or 9,5 or 9,4 or 9,3 or 9,2 or 9,1: seventeen

10: 1,10 or 2,10 or 3,10 or 4,10 or 5,10 or 6,10 or 7,10 or 8,10 or 9,10 or 10,10 or 10,9 or 10,8 or 10,7 or 10,6 or 10,5 or 10,4 or 10,3 or 10,2 or 10,1: nineteen

Total number of dice results are: nineteen+seventeen+fifteen+thirteen+eleven+nine+seven+five+two+one = 19+17+15+13+11+9+7+5+2+1 = 36+28+20+12+3 = 64+32+3 = 96+3 = 99. Well, that's conveniently close to a hundred! While it's inaccurate (and I'm sorry my fellow mathies, I really, really am) you can kiiiiiiiiinda presume that we're working with a basic % as a result (because we're looking to understand the gist, not be perfect).

So, of the total number of results, there's a 19% chance to hit a 10, a 17% chance to hit a 9, a 15% chance to hit an 8, and so on. This, then, means that there is an almost overwhelmingly increasingly-large change with a bias towards larger numbers, though, again, looking at it in context is important.

While getting a 1 is an extremely rare event, comparatively, if you're looking to beat it, you've got a 100% chance. A DC 2 provides a (nearly) 99% chance of success, a DC 3 a (nearly) 97% chance of success, a DC 4 a (nearly) 92% chance of success, a DC 5 a (nearly) 85% chance of success, a DC 6 a (nearly) 76% chance of success, a DC 7 a (nearly) 65% chance of success, a DC 8 a (nearly) 50% chance of success, a DC 9 a (nearly) 33% chance of success, and a DC 10 a (nearly) 19% chance of success.

We then do the same thing for d8s!

Quote:


Result: Roll Values: Total Chances
1: 1,1,1: one

2: 1,1,2 or 1,2,1 or 1,1,2: three

3: 1,1,3 or 2,1,3 or 3,1,3 or 3,1,2 or 3,1,1 or 1,2,3 or 2,2,3 or 3,2,3 or 3,2,2 or 3,2,1 or 1,3,3 or 2,3,3 or 3,3,3 or 3,3,2 or 3,3,1: fifteen

4: 1,1,4 or 2,1,4 or 3,1,4 or 4,1,4 or 4,1,3 or 4,1,2 or 4,1,1 or 1,2,4 or 2,2,4 or 3,2,4 or 4,2,4 or 4,2,3 or 4,2,2 or 4,2,1 or 1,3,4 or 2,3,4 or 3,3,4 or 4,3,4 or 4,3,3 or 4,3,2 or 4,3,1 or 1,4,4 or 2,4,4 or 3,4,4 or 4,4,4 or 4,4,3 or 4,4,2 or 4,4,1: twenty-eight
It's about here I realize I've made a big mistake.

5: 1,1,5 or 2,1,5 or 3,1,5 or 4,1,5 or 5,1,5 or 5,1,4 or 5,1,3 or 5,1,2 or 5,1,1 or 1,2,5 or 2,2,5 or 3,2,5 or 4,2,5 or 5,2,5 or 5,2,4 or 5,2,3 or 5,2,2 or 5,2,1 or 1,3,5 or 2,3,5 or 3,3,5 or 4,3,5 or 5,3,5 or 5,3,4 or 5,3,3 or 5,3,2 or 5,3,1 or 1,4,5 or 2,4,5 or 3,4,5 or 4,4,5 or 5,4,5 or 5,4,4 or 5,4,3 or 5,4,2 or 5,4,1 or 1,5,5 or 2,5,5 or 3,5,5 or 4,5,5 or 5,5,5 or 5,5,4 or 5,5,3 or 5,5,2 or 5,5,1: forty-five

6: 1,1,6 or 2,1,6 or 3,1,6 or 4,1,6 or 5,1,6 or 6,1,6 or 6,1,5 or 6,1,4 or 6,1,3 or 6,1,2 or 6,1,1 or 1,2,6 or 2,2,6 or 3,2,6 or 4,2,6 or 5,2,6 or 6,2,6 or 6,2,5 or 6,2,4 or 6,2,3 or 6,2,2 or 6,2,1 or 1,3,6 or 2,3,6 or 3,3,6 or 4,3,6 or 5,3,6 or 6,3,6 or 6,3,5 or 6,3,4 or 6,3,3 or 6,3,2 or 6,3,1 or 1,4,6 or 2,4,6 or 3,4,6 or 4,4,6 or 5,4,6 or 6,4,6 or 6,4,5 or 6,4,4 or 6,4,3 or 6,4,2 or 6,4,1 or 1,5,6 or 2,5,6 or 3,5,6 or 4,5,6 or 5,5,6 or 6,5,6 or 6,5,5 or 6,5,4 or 6,5,3 or 6,5,2 or 6,5,1 or 1,6,6 or 2,6,6 or 3,6,6 or 4,6,6 or 5,6,6 or 6,6,6 or 6,6,5 or 6,6,4 or 6,6,3 or 6,6,2 or 6,6,1: sixty-six
Yes, confirming the "brute force" method was a mistake. XD

7: 1,1,7 or 2,1,7 or 3,1,7 or 4,1,7 or 5,1,7 or 6,1,7 or 7,1,7 or 7,1,6 or 7,1,5 or 7,1,4 or 7,1,3 or 7,1,2 or 7,1,1 or 1,2,7 or 2,2,7 or 3,2,7 or 4,2,7 or 5,2,7 or 6,2,7 or 7,2,7 or 7,2,6 or 7,2,5 or 7,2,4 or 7,2,3 or 7,2,2 or 7,2,1 or 1,3,7 or 2,3,7 or 3,3,7 or 4,3,7 or 5,3,7 or 6,3,7 or 7,3,7 or 7,3,6 or 7,3,5 or 7,3,4 or 7,3,3 or 7,3,2 or 7,3,1 or 1,4,7 or 2,4,7 or 3,4,7 or 4,4,7 or 5,4,7 or 6,4,7 or 7,4,7 or 7,4,6 or 7,4,5 or 7,4,4 or 7,4,3 or 7,4,2 or 7,4,1 or 1,5,7 or 2,5,7 or 3,5,7 or 4,5,7 or 5,5,7 or 6,5,7 or 7,5,7 or 7,5,6 or 7,5,5 or 7,5,4 or 7,5,3 or 7,5,2 or 7,5,1 or 1,6,7 or 2,6,7 or 3,6,7 or 4,6,7 or 5,6,7 or 6,6,7 or 7,6,7 or 7,6,6 or 7,6,5 or 7,6,4 or 7,6,3 or 7,6,2 or 7,6,1 or 1,7,7 or 2,7,7 or 3,7,7 or 4,7,7 or 5,7,7 or 6,7,7 or 7,7,7 or 7,7,6 or 7,7,5 or 7,7,4 or 7,7,3 or 7,7,2 or 7,7,1: ninety-one
Al... most... there...

8: 1,1,8 or 2,1,8 or 3,1,8 or 4,1,8 or 5,1,8 or 6,1,8 or 7,1,8 or 8,1,8 or 8,1,7 or 8,1,6 or 8,1,5 or 8,1,4 or 8,1,3 or 8,1,2 or 8,1,1 or 1,2,8 or 2,2,8 or 3,2,8 or 4,2,8 or 5,2,8 or 6,2,8 or 7,2,8 or 8,2,8 or 8,2,7 or 8,2,6 or 8,2,5 or 8,2,4 or 8,2,3 or 8,2,2 or 8,2,1 or 1,3,8 or 2,3,8 or 3,3,8 or 4,3,8 or 5,3,8 or 6,3,8 or 7,3,8 or 8,3,8 or 8,3,7 or 8,3,6 or 8,3,5 or 8,3,4 or 8,3,3 or 8,3,2 or 8,3,1 or 1,4,8 or 2,4,8 or 3,4,8 or 4,4,8 or 5,4,8 or 6,4,8 or 7,4,8 or 8,4,8 or 8,4,7 or 8,4,6 or 8,4,5 or 8,4,4 or 8,4,3 or 8,4,2 or 8,4,1 or 1,5,8 or 2,5,8 or 3,5,8 or 4,5,8 or 5,5,8 or 6,5,8 or 7,5,8 or 8,5,8 or 8,5,7 or 8,5,6 or 8,5,5 or 8,5,4 or 8,5,3 or 8,5,2 or 8,5,1 or 1,6,8 or 2,6,8 or 3,6,8 or 4,6,8 or 5,6,8 or 6,6,8 or 7,6,8 or 8,6,8 or 8,6,7 or 8,6,6 or 8,6,5 or 8,6,4 or 8,6,3 or 8,6,2 or 8,6,1 or 1,7,8 or 2,7,8 or 3,7,8 or 4,7,8 or 5,7,8 or 6,7,8 or 7,7,8 or 8,7,8 or 8,7,7 or 8,7,6 or 8,7,5 or 8,7,4 or 8,7,3 or 8,7,2 or 8,7,1 or 1,8,8 or 2,8,8 or 3,8,8 or 4,8,8 or 5,8,8 or 6,8,8 or 7,8,8 or 8,8,8 or 8,8,7 or 8,8,6 or 8,8,5 or 8,8,4 or 8,8,3 or 8,8,2 or 8,8,1: one hundred twenty

9: N/A

10: N/A

For the record, this should be correct, even if I've made mistakes in the table. If it is not correct, please feel free to correct me.

So: 120+91+66+45+28+15+3+1 = 211+111+43+4 = 322+47 = 369

Now, basic algebra should allow us to take care of this and turn any given value into a percentile.

For our purposes, 369 = 100%

Ah, yes, good ol' 369-sections of a %. Really easy to break down, there. Thanks, Statistics.

That means that any given number value has a ~~0.271% value 100/369. But we don't really care about that in specific - we're asking how likely it is to beat big numbers. So, for reasons humanity has yet to fathom, I've decided to once again go through the DCs starting with the small numbers.

Once again, a DC of 1 is beaten 100% of the time. You simply can't roll less than this.

A DC of 2 gives us our first math: there are three chances to fail out of 369. So that's three of those 0.271% values, making a .813% chance to lose on a 2. So far this is better! We've less than a 1% chance to fail a DC 2!

Following a similar process, a DC of 3 means 4.065% failure rate (almost 96% chance success); a DC of 4 means a 7.588% failure rate (about 92% chance of success); a DC of 5 means a 12.195% failure rate (about 88% chance of success); a DC 6 means a 17.886% failure rate (about 82% chance of success); a DC 7 means a 24.661% failure rate (about a 75% chance of success); a DC 8 means a 32.52% failure rate (about a 66% chance of success). We now get into real trouble, however, as a 9 and a 10 are right out! It is no longer possible to attain a success for either of those.

This yields a very important lesson: if you're going "keep the top" and "high number" it is critical that you actually, you know, have a reachable number you can get to with your dice pool. Left strictly to the merits of what a d8 is able to achieve, and comparing it strictly to the merits of what a d10 can achieve, you get:

100%:100%
99%:99%
96%:97%
92%:92%
88%:85%
82%:76%
75%:65%
66%:50%
0%:33%
0%:19%

Just taking a rough statistical average of success, a 3d8 pool doing d8 things (ignoring the d10, for now) gets a whopping 87.25% success rate over-all; on the other hand, a 2d10 pool doing d10 things gets a 71.6% chance of success - clearly the loser in this regard. ... however, if you factor in the range of both of these numbers, you have a very different result: the 87.25% drops to a 69.8% - a few points lower than the 2d10, but several DCs are entirely impossible within that spread.

I'm not going to do it (because I got a life... you don't know!) but the exact same truth can be found when you roll a 5d6 pool and compare. The more dice you roll and only keep one, the more likely higher the relative number you achieve over-all; but using the lower dice places upper limits on the beneficial results.

So getting a high relative number on its own merits is more likely the more dice you use; but getting a high number relative to all the possible dice you could use means it loses out, quickly, as the above points out. Which means that, depending on how you determine how hard something is changes the answer to your questions.

While that doesn't exactly answer your question, I hope it helps. I'm not able, at present, to go through creating formula for varying dice pools, probability, and likely results beyond the brute force method above, but that should be a good start.

Here is a dice roller slash calculator.

Hope all that helps!


Vanykrye wrote:

General rule of dice: more dice equals more consistency.

3d8 v 5d6...you will more consistently be in the 4-6 range on the 5d6 option than the 3d8 option, but you have 0 chance for a 7-8, eliminating your ability to make a skill roll requiring that high of a roll.

Basically you need to balance your need for a consistently good chance of getting at least one die in the upper half of the range versus a peak number.

Without knowing anything about the system beyond your description above, I'd probably take the 3d8 over the 5d6, as 3 dice is still a good chance of getting both upper range numbers and higher peak numbers.

I also would take the 3d8 over 2d10, simply because it's easier to roll a high number with 3 dice than 2.

This seems contradictory, but the trick is looking for balance in consistency versus peak. More dice gives consistency, bigger dice equals peak.

Example, with some limited math: If you are consistently successful with a result of 4, then 5d6 is going to be better than 3d8. A d6 rolls a 4 or better 50% of the time, and you have 5 chances at that. That's 5*.5 = 2.5. the d8 rolls at least a 4 on 62.5% odds, but you only have 3 chances. 3*.625 = 1.875. 2.5 is bigger than 1.875, so the 5d6 is better. Now if the desired results is a 6, the answer changes quickly. A 6 on a d6 is only a 16.7% shot. 5*.167 = .833. on a d8 you still have a 37.5% chance per die. 3*.375 = 1.125, which is bigger, and therefore better.

Does that give you a better idea of what you would be looking for?

Or, you know, I could be ninja'd by a fire dragon with more eloquence. OH WELL, I'M LEAVIN' IT! XD


Wow, I really made a dumb mistake.

The two-value results are much larger than I displayed here.

For 2d10, there should be three results instead of two:
1,2
2,1
and 2,2

The nice thing is that this makes the first example exactly 100 out of 100%, so each number lines up really nicely. Also, the pattern is easy to follow (odd number up).

And for the 3d8, there should be seven results for a 2 value instead of three:

1,1,2
2,1,2
2,1,1
1,2,1
1,2,2
2,2,2
2,2,1

I really have no idea what I was doing, up there.

Also, I realize that my 3d8 math may be off due to missing some basic values presupposing the middle dice is the only one that has the acceptable number. Blarg.

This means that each entry should balloon in number of values significantly. That's what I get for trying to build a table with two children asking if they can use the computer to, "watch Super Why, yet"... alas!

So I did a whole bunch of work, but poorly, and with incorrect results.

WOOOOOOOOOOOOOOPS.

Still, the basic idea is solid, and if you follow that you can get the idea, even if the specifics are off - sorry about the wrong math! You should be able to correct my mistake with a little work and some time to examine your results. And with that: away!


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Pathfinder Rulebook Subscriber

Thank you both for your responses!

I can't really flesh out too much of what the rest of the "system" is like because... there isn't one. The idea that as you get more skilled you use bigger dice has been rattling around, and today the idea of adding a factor that can increase or decrease the dice pool floated through my brain as well.

But, here's what I was thinking about.

You'd be rolling your dice pool versus the obstacle's dice pool. So, say you're trying to hit with an attack, you'd have such and such pool based on how good you are, and your opponent would have such and such pool based on how good they are at not getting hit. Something like a really good weapon could add dice to your pool - same thing with good armor.

So for example say the hero has a middling quality weapon and moderate skill - they roll 3d8 to attack. And the enemy is pretty nimble but has poor quality armor, so they're only rolling 2d10 to defend against the hit. The target rolls their 2d10 and gets, say, a 3 and a 7, so they declare "7 for defense." The attacker gets, perhaps, a 2, 4 and a 6, so they say "6 for attack; I missed."

And then there would be a consumable resource similar to Hero Points from Pathfinder where you could reduce your dice pool by 1 to increase the size of die rolled... so our attacker could use that option to turn their 3d8 into 2d10, which excludes the possibility that the defender could roll an unreachable number... but makes it even odds on whether they hit (well slightly better than even odds if they need to meet or exceed the defender's number).

Then I got to thinking, that is really hard to calculate quickly whether or not it is even worth it to alter your dice pool in that way. And based on the REAMS of calculations so kindly posted above, I'd say that it would not be immediately apparent to any old player whether they were actually doing themselves a favor.

So... seems like a really bad idea. Maybe the consumable resource just flat out makes the roll easier by adding a dice AND stepping up the dice rolled.

.... And in case anybody is actually interested, the success/failure die roll previously in this system works this way:
Both characters have a certain value in attack and defense stats (this is a big robot game, so it's Sensors and Movement, which captures a targeting computer's ability to get a fix on a moving target). There are 3 possibilities:
Sensors are greater than Movement; roll 2d6 and hit on a 5
Sensors are equal to Movement; roll 2d6 and hit on a 7
Sensors are less than Movement; roll 2d6 and hit on an 9

Your typical machine has 2 Movement and 3 Sensors; thus your bog standard shot needs only a 5 to hit, which is fairly likely to land. You'd better do something interesting to make it harder on your opponent or you're gonna get hit frequently. However if the target is in a swift machine with 3 move, the odds are more even; and if that swifter machine's pilot opts to push their ride hard and run at max speed, they gain an extra Movement for the turn and now the enemy shooting them needs a 9, and the odds are much in their favor.

All of that works fine in the context of a tabletop war game where you want to blast enemy robots into smoking bits. But I'd love to tack on RPG elements, and RPG elements mean increasing detail. Presently the game has no way to account for the skill of the pilot, so I've been thinking on that, including the above mentioned dice pool approach.. what if 3 sensors meant a die pool of 3, and pilot skill impacted which dice you're rolling 3 of? What if your opponent was great at defending but is in a slow machine, so they roll a really good die, but only twice?

But I think that the current system is fairly elegant and my dice pool method is a giant headache...


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Probability for high/low rolls is actually a really easy topic, because it's just the "opposite" of rolling low.

For example, what's the probability of getting at least one 5 when rolling 5d6?

Just figure out all the probability of ALL of the dice getting 4 or less.
(4/6) * (4/6) * (4/6) * (4/6) * (4/6) = 0.1317.

The probability of getting at least one 5 is 1 - that.
0.8683.

What about 3d8?
(4/8) * (4/8) * (4/8) = 0.125. 1-that.
0.8750. You're actually a bit better off with 3d8.

What about 2d10?
(4/10) * (4/10) = 0.16. 1-that.
0.84.

So fancy that! To get at least one 5, you're best off with 3d8.


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pours gasoline around the thread


starts heaping several NAUGHTY(tm) calendars of dancing drow outside, distracting the the Freehold, then puts on fans so it evaporates, instead


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QUENCH MY THIRST WITH GASOLINE


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Tacticslion wrote:
starts heaping several NAUGHTY(tm) calendars of dancing drow outside, distracting the the Freehold, then puts on fans so it evaporates, instead

I already own those calendars, this thread must be purged.


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Freehold DM wrote:
Tacticslion wrote:
starts heaping several NAUGHTY(tm) calendars of dancing drow outside, distracting the the Freehold, then puts on fans so it evaporates, instead
I already own those calendars, this thread must be purged.

*hefts bat*

I heard you need something purified.


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jdripley wrote:


You'd be rolling your dice pool versus the obstacle's dice pool. So, say you're trying to hit with an attack, you'd have such and such pool based on how good you are, and your opponent would have such and such pool based on how good they are at not getting hit. Something like a really good weapon could add dice to your pool - same thing with good armor.

Going to throw a wrench into how you think about this:

Why do the additional dice have to be the same size?

Example, my attack skill is really high (d10), but I am using an average quality sword. Maybe the sword only gives me a (d6).

It still increases my pool (whatever die is highest wins), but now you don't have to make as many decisions and considerations about dice pool math (ie, better to go up a die, or expand pool). It becomes very simple, I always want to add the largest die possible from a source. Choosing between the d6 and d8 sword is easy.

Its easier to design, and easier to understand as a player.

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