#5: Odder Ways of Rolling Dice

[RPGnet Columns]

http://www.rpg.net/columns/rollthebones/rollthebones5.phtml

Summary:

Sovereign Stone, Jadeclaw, and other odder die-rolling systems

Go to the column (http://www.rpg.net/columns/rollthebones/rollthebones5.phtml) for more information.

[Paul DuPont]

Thanks for your column as ussual. I especially liked some of the distinctions and comparisons you made between the median and variable die systems. It is interesting how they can share certain traits when used in comparative/mathless roll systems.

I also had a new idea inspired by this column so I think it only fair that I share it with you. To handle tasks of variable chaos (i.e. more or less predictable results), a single die results of up to 12 could be generated by three different methods. The most chaotic method would be rolling 1D12, slightly less chaotic 2D6 and even more predictable 3D4. This also has the effect that more 'predictable' results don't suffer from very low results but can still reach those very high results with a decreasing percentage. I am not sure how far the variability differs from one to the other but the commonly available dice all add up to the same maximum value and you only ever have to add three small numbers at most. The restriction of getting lower results with less variable actions can be a feature, it can also be altered by subtracting 1 or 2 from the die roll so the restriction is on the high end. What do you think?

[Mataxes]

I've got two systems I'd be interested in seeing you look at. First is the Step System (used in Earthdawn).

The other is a system with two (simultaneous) tests -- one that checks against the skill (say, roll skill rating or lower on a d20), and one that checks against environmental penalties -- say, poor visibility or something like that. (Let's say, on the second test, penalties are accumulated on a 10-point scale, and the test must roll higher than the penalty on a d10.) Both tests must succeed in order for the action to succeed.

[Careful Plum]

As an aside, the roll m dice and add the n highest is the normal task resolution in Legend of the Five Rings (always using d10). As Attribute plus Skill are rolled, and Attribute kept, this tends to very heavily emphasize attributes (as you can see from the article ;) )

[torbenm]

I also had a new idea inspired by this column so I think it only fair that I share it with you. To handle tasks of variable chaos (i.e. more or less predictable results), a single die results of up to 12 could be generated by three different methods. The most chaotic method would be rolling 1D12, slightly less chaotic 2D6 and even more predictable 3D4. This also has the effect that more 'predictable' results don't suffer from very low results but can still reach those very high results with a decreasing percentage. I am not sure how far the variability differs from one to the other but the commonly available dice all add up to the same maximum value and you only ever have to add three small numbers at most. The restriction of getting lower results with less variable actions can be a feature, it can also be altered by subtracting 1 or 2 from the die roll so the restriction is on the high end. What do you think?

One thing to note is that not only is 2d6 less variable than 1d12, but it has a higher average (7 versus 6.5). Similarly, 3d4 has average 7.5. So when you go to more dice, you not only avoid low rolls, you also get a better average, so this would tend to make people choose this unless they really need a high result.

Up to and including 8, 2d6 has a higher or equal chance of achieving the target, so only for requirements of 9 or more would you roll 1d12. With 3d4, you actually have a higher chance of rolling at least 9 than you do on 2d6!

Getting the same average requires subtracting 0.5 from 2d6, which is a bit silly (and would equal subtracting 1 if all target numbers are integers). 3d4-1 has the same average as 1d12, though, so using this as a less chaotic alternative to 1d12 would work fine. You cut off both the lowest and highest possible values and you reduce the spread from 3.45 to 1.94. If you want three choices, the next step could be 5d2-1, which ranges from 4 to 9 with an average of 6.5 (like 1d12 or 3d4-1) but has a spread of only 1.12.

[torbenm]

I've got two systems I'd be interested in seeing you look at. First is the Step System (used in Earthdawn).

The other is a system with two (simultaneous) tests -- one that checks against the skill (say, roll skill rating or lower on a d20), and one that checks against environmental penalties -- say, poor visibility or something like that. (Let's say, on the second test, penalties are accumulated on a 10-point scale, and the test must roll higher than the penalty on a d10.) Both tests must succeed in order for the action to succeed.

The Step System is another system where the ability determines the dice type. Unlike those I mentioned in the article, the dice are open-ended: If you roll the maximum value, you add one more dice of the same type, and rolling max on this adds yet another and so on.

Additionally, the progression is different: After going from 1d4 to 1d12, the next steps are d6+d6, d8+d6, d10+d6, d10+d8, d10+d10, d10+d12 and d20+d4 (or d12+d12, depending on version).

An open-ended dN has average N(N+1)/(2(N-1)), so the average progresses as follows:

3.333, 4.2, 5.143, 6.111, 7.091, 8.4, 9.345, 10.311, 11.254, 12.222, 13.202, 14,086 (or 14.182)

In most cases, the average increases by slightly less than 1, but when going from 1d12 to 2d6, the average increases by a bit over 1.3. So this step is more significant than the others -- not only do you eliminate the possibility of rolling 1, you also gain more in average than in other steps.

As with other systems where ability determines dice rolls, the spread increases with skill (until you go from one to two dice, where it drops), but since the chance of rerolling decreases with larger dice, the spread decreases (slightly) relative to the average, unlike the case of single dice without reroll.

Your system where you must roll under skill on a d20 and over a penalty on a d10 makes the penalty dominant when skill is high and skill dominant when it is low. For example, if the penalty is 8 (so you must roll 9 or 10 on the d10), you can never get more than 20% chance of success, no matter how skilled you are.

In general, if penalty is P and skill is S, the chance of success is (10-P)*(S-1)/200, as the two rolls are independent.

It is more interesting if you make both roll under (or both roll over) and allow the player to select which dice go to skill and which to penalty _after_ they are rolled.

The effect is that highly skilled characters can usually use the highest of the two dice for the skill, so the lowest can be used for the penalty. So the rolls are no longer independent.

Let us say that both use a d20 and roll under, so penalties range from 2 (hardest) to 20 (easiest). If X is the least of the skill and the penalty and Y is the higher of those, the chance of success is (X-1)^2/400 + (X-1)*(Y-X)/200.

[Paul DuPont]

One thing to note is that not only is 2d6 less variable than 1d12, but it has a higher average (7 versus 6.5). Similarly, 3d4 has average 7.5. So when you go to more dice, you not only avoid low rolls, you also get a better average, so this would tend to make people choose this unless they really need a high result.

Up to and including 8, 2d6 has a higher or equal chance of achieving the target, so only for requirements of 9 or more would you roll 1d12. With 3d4, you actually have a higher chance of rolling at least 9 than you do on 2d6!

Getting the same average requires subtracting 0.5 from 2d6, which is a bit silly (and would equal subtracting 1 if all target numbers are integers). 3d4-1 has the same average as 1d12, though, so using this as a less chaotic alternative to 1d12 would work fine. You cut off both the lowest and highest possible values and you reduce the spread from 3.45 to 1.94. If you want three choices, the next step could be 5d2-1, which ranges from 4 to 9 with an average of 6.5 (like 1d12 or 3d4-1) but has a spread of only 1.12.

I can see how the change in average would definitely be a problem if the choice was 'free'. This might have an interesting effect on a roll under system where there are three distinct levels of character ability. Apprentices would rolls 3D4, Journeyman would roll 2D6, and Master would roll 1D12. Additional levels of skill mastery may add more D12s and allow the player to pick the lowest die. Hmmm... that could be interesting.

Back to a roll-over system, this could be a similar mechanic to Take-10 and Take-20 in the D-20 system. But with more minute effects and retains some variability. Example: Bowman (or gunner) takes an extra round aiming to roll 2D6 instead of 1D12, two rounds might grant the 3D4, but three rounds would not grant any better roll options. It seems appropriate and compatible in this circumstance, unless the math does other things I haven't noticed.

[torbenm]

All this talk of using different number of dice to get different spreads for the same range makes me wish that you could get dice showing 0..n-1 instead of 1..n (like most d10s actually do). This way, the sum of N dice always have a range starting at 0, and if different combinations of dice give the same maximum value, they also have the same average (equal to half the maximum value). Example: A d10 would have values from 0 to 9 and, hence, an average of 4.5. A d4 would range from 0 to 3 with average of 1.5, so 3d4 would range from 0 to 9 and have an average of 4.5.

It would probably be better to name such dice by their maximum value rather than the number of sides, so a die ranging from 0 to 3 would be d03, z3 (for zero-based die) or some such. So the above two examples would be written as, for example, z9 and 3z3.

[jeff0]

A z6, as you're calling it, isn't all that uncommon. You can find them by searching for '0-5 dice'. I also found some z16s at Gencon in 2006. They're labelled in hexadecimal, 0-F. I don't recall the name of the company that sold them.

A (somewhat awkward) way of getting at something similar would be to pair each dN with a Fudge die (1dF = 1d3-2 = 1z3-1). 1dF + 1dN gives the same distribution as 1z3 + 1zN. Each pair of dice would have a goofy pyramid distribution with 'steps' at 0,1,N, and N+1 and a flat top in between 1 and N.

On a math-geeky note:

I didn't know there was such a nice formula for open-ended rolls. It's interesting that there's a factor of (N+1)N/2 in there. I wonder if there is an interpretation of the formula that relates to binomial coefficients.

[torbenm]

I didn't know there was such a nice formula for open-ended rolls. It's interesting that there's a factor of (N+1)N/2 in there. I wonder if there is an interpretation of the formula that relates to binomial coefficients.

I doubt you can get binomial coefficients into this. Binomials occur when you roll several dice (the binomial function of n and k is the number of ways you can get k heads when flipping n coins).

The sum of 1...n is n*(n+1)/2, so the average of a normal dn is n*(n+1)/2n = (n+1)/2. An open-ended dn has an average Ao determined by the equation

Ao = A + 1/n*Ao

where A is the average of a normal dn, since you roll a normal dn (with an average of A) and with a probability of 1/n add an open-ended dn (with average Ao). This solves to Ao(n-1) = A*n, so Ao = A*n/(n-1). Since A = (n+1)/2, Ao = n*(n+1)/2(n-1). Note that the average of an open-ended die doesn't depend on _which_ value causes reroll. So, a d6 with reroll on 6 has the same average as a d6 with reroll on 1 (but a different spread).

If we regard zero-based dice, the sum of 0...n is also n*(n+1)/2, but since there are n+1 sides on an zn, the average is n/2. An open-ended zn has 1/(n+1) chance of reroll (assuming reroll on one value), so the equation for the average is

Ao = n/2 + 1/(n+1)*Ao

which gives Ao = (n+1)/2.

As you can see, the formulas become a lot simpler when you use zero-based dice.

[adaen]

Hi Torben,

Might you be able to provide a formula for the hi_3dX and lo_3dX scenarios. Is there a generalized formula for 1st highest, 2nd highest, 3rd highest, etc. for ndx systems. I've been working with these via brute force, but a formula would save a lot of time. Thanks for the very excellent column!

Best,

~AoB

[smascrns]

Hi Torben,

Might you be able to provide a formula for the hi_3dX and lo_3dX scenarios. Is there a generalized formula for 1st highest, 2nd highest, 3rd highest, etc. for ndx systems. I've been working with these via brute force, but a formula would save a lot of time. Thanks
Yes, that formula would realy rock.

[torbenm]

Hi Torben,

Might you be able to provide a formula for the hi_3dX and lo_3dX scenarios. Is there a generalized formula for 1st highest, 2nd highest, 3rd highest, etc. for ndx systems. I've been working with these via brute force, but a formula would save a lot of time. Thanks for the very excellent column!


There is a formula for the mth lowest of ndx being equal to y. We can find it in this way:

If the mth lowest die is equal to y, then the following things must be true:

1. There are k1 dice strictly less than y.

2. There are k2 dice equal to y.

3. 0<=k1<m, m<=k1+k2<=n

Note that properties 1 and 2 are not independent.

There are (x-y+1)^(n-1) * (y-1)^k1 * {n:k1} combinations out of x^n where there are k1 dice strictly less than y. {n:k1} = n!/(k1! * (n-k1)!). So the probability is (x-y+1)^(n-1) * (y-1)^k1 * {n:k1} / (x^n).

The remaining n-k1 dice are known to be between y and n, so we can treat them as d(x-y+1)+(y-1), and k2 of these must be equal to y, or equivalently as d(x-y+1) that must be equal to 1 Out of the (x-y+1)^(n-k1) combinations, there are (n-y)^(n-k1-k2) * {n-k1:k2} where k2 dice are equal to 1 so the probability is (n-y)^(n-k1-k2) * {n-k1:k2} / ((x-y+1)^(n-k1))

The total probability of having k1 dice less than y and k2 dice equal to y is, hence,

(x-y+1)^(n-1) * (y-1)^k1 * {n:k1} / (x^n) * (n-y)^(n-k1-k2) * {n-k1:k2} / ((x-y+1)^(n-k1))

= (x-y+1)^(k1-1) * (y-1)^k1 * {n:k1} / (x^n) * (n-y)^(n-k1-k2) * {n-k1:k2}

We must now add these for all the possible k1 and k2:

Sum k1 = 0 to m-1 (
Sum k2 = m-k1 to n-k1 (
(x-y+1)^(k1-1) * (y-1)^k1 * {n:k1} / (x^n) * (n-y)^(n-k1-k2) * {n-k1:k2}
))

So, not exactly a simple formula.

Note that the probability for the mth highest of ndx being equal to y is the same as the probability of the mth lowest of ndx being equal to (n-y+1).

[adaen]

Thanks very much, TM. You rock (as always).

Respectfully yours,

~Adaen of Bridgewater

[Wratts]

A third problem I see with rerolls is that they make the probability distribution uneven: You can get holes in the distribution (i.e., impossible results) or places where the probability drops sharply but then stays nearly constant for a while. For example, if you roll a d20 and reroll on each 10 or 20, adding the new results to the original (as in Torg), you can never get results that divide evenly by 10, but all other results are possible. Also, every result from 1 to 9 are equally likely at 5% and results between 11 and 19 are all at 5.25%, but then there is a sharp drop to 0.5125% for results between 21 and 29.
This is actually the only sound argument you can bring against "re-rolls" (assuming that you mean all sorts of re-rolls); the others are then truly just personal opinions. Sure, you said it's personal bias in advance, but it didn't seem well thought out enough to warrant such a large mention. Basically, your example aside, re-rolling 1d6 6 times or rolling a pool of 6d6 in which you look for the highest/lowest result amounts to the same thing (aside from resolution speed). I think you poorly overgeneralized what "re-rolling" can be used for. Me, aside from the speed aspect, would not put down its practical use in games. Just like increasing pool sizes, dice modifiers, or whatever, practically speaking it's another solid way of giving players the illusion of control in a system where things are randomized to a certain degree.

For example, as I don't know Torg, I'm unclear whether you mean to simply "re-roll" dice as in to edit a result by trying again, or whether you mean what I've come to know as "exploding dice" (where dice with a certain result -- i.e. 6 on a 1d6 -- are re-rolled and summed up with the original result). I'm assuming you meant the latter, in which I agree that it's a bit of a pain to use (thus giving more to your point about them detracting system speed). However, in defense of such systems, one could easily say that they are actually truly "random" systems, rather than many of the systems you suggest to be preferable that are in all actuality simply "more predictable" systems.

I love this column, but this little insert in this article about "re-rolls" read a bit too biased and rushed for my taste, and was a tad bit unfitting with the solid rest I've read so far.

[torbenm]

This is actually the only sound argument you can bring against "re-rolls" (assuming that you mean all sorts of re-rolls); the others are then truly just personal opinions. Sure, you said it's personal bias in advance, but it didn't seem well thought out enough to warrant such a large mention. Basically, your example aside, re-rolling 1d6 6 times or rolling a pool of 6d6 in which you look for the highest/lowest result amounts to the same thing (aside from resolution speed). I think you poorly overgeneralized what "re-rolling" can be used for. Me, aside from the speed aspect, would not put down its practical use in games. Just like increasing pool sizes, dice modifiers, or whatever, practically speaking it's another solid way of giving players the illusion of control in a system where things are randomized to a certain degree.

For example, as I don't know Torg, I'm unclear whether you mean to simply "re-roll" dice as in to edit a result by trying again, or whether you mean what I've come to know as "exploding dice" (where dice with a certain result -- i.e. 6 on a 1d6 -- are re-rolled and summed up with the original result). I'm assuming you meant the latter, in which I agree that it's a bit of a pain to use (thus giving more to your point about them detracting system speed). However, in defense of such systems, one could easily say that they are actually truly "random" systems, rather than many of the systems you suggest to be preferable that are in all actuality simply "more predictable" systems.

I did mean what you call "exploding dice". I defined what I meant with re-rolls in an earlier column, so I didn't think of making this explicit.

I love this column, but this little insert in this article about "re-rolls" read a bit too biased and rushed for my taste, and was a tad bit unfitting with the solid rest I've read so far.

I did warn that it was personal opinion and I quite expected lots of people to disagree with me. I don't mind that and I don't say that these people are wrong. After all, personal preferences are not universal.

While I tend to stick to factual stuff, I do have opinions, and I don't want to hide them. I don't mind discussing them either, though.

You will see a bit of opinion in the next column, but I hope not so controversial as my dislike of accumulative re-rolling appears to be.