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I want to make tabletop potions of healing out of corked vials (test tubes) filled with dice. Each vial would be labeled and filled based on the potion it represents. For example, a vial for a potion of greater healing would contain four d4's and be labeled with a +4 bonus; when administering such a potion, the player would just dump the vial and total the dice plus the bonus, yielding the correct 4d4+4 result, without having to fiddle with their own dice. The vials are meant to speed up play, act as physical reminders that a player has a potion available, and look super cute (assume that these intentions are inviolable and that this craft project is serious business).

It's hard to find enough d4's to pull off this craft project, and test tubes are normally too small for standard 16mm dice anyway, so I've considered using miniature 12mm d6's instead, which are much easier to come by in blocks of large quantities for cheap. The problem is that d6's are slightly more swingy than d4's, and I don't want to grossly deviate from the math underlying potions.

How can I approximate the rolls for each potion of healing while avoiding swingy results? By swingy results I mean unexpectedly low or high totals or a distribution that violates conventions for how healing works in the game.

The following are the restrictions on a valid solution:

• Only d6 dice can be used. It's a physical constraint of the problem.


• Each vial must contain a constant number of dice to dump and roll for the result, without requiring additional dice that weren't in the vial.


• Basic mental math like addition and subtraction is fine.


• Rerolling below a minimum total or similar rules of thumb are fine if they are simple.

Answers telling me to use the average instead of rolling, to roll with online tools, to find tinier d4's or bigger test tubes, to buy a commercially available set of d4-filled vials, or the like aren't solutions. I promise you this question doesn't suffer from an XY problem. The restrictions are inherent to the nature of the craft project, a very serious and important craft project.

For bonus, corresponding AnyDice formulas would be nice but not vital.

Exact solution: Replace each "d4+1" with "d6, reroll 1 and 6"

Each type of healing potion (except the supreme; see below) is some multiple of "d4+1", so a formula that emulates this with a single d6 would be ideal. d4+1 produces a uniform distribution from 2 to 5 inclusive, so a simple rule to emulate this is to roll a d6 and keep re-rolling it until you don't roll a 1 or 6. So, for example, a potion of greater healing, normally 4d4+4, would become 4d6, rerolling all 1s and 6s. Since this rule exactly reproduces the distribution of a d4+1 roll, potions rolled using this rule will behave identically to normal potions rolled using d4s.

As a side bonus, the new formula doesn't involve a modifier, so you don't need to worry about labeling different size potions with different modifiers. This means that you don't necessarily need a different label for each potion size. Just fill it with the appropriate number of dice and make sure people know the rerolling rules.

Note that unlike the other potions, the supreme potion of healing breaks the pattern of having the number of dice equal the modifier. Instead of 10d4+10, it heals for 10d4+20. To apply this rule to the supreme potion, split this into (10d4+10)+10, then replace (10d4+10) with 10d6. This yields 10d6+10 as the new formula for a potion of superior healing.

Quicker but less precise solution: replace each "d4+1" with "d6"

If rerolling some dice, potentially multiple times, is too slow, you could just roll the d6s once and be done with it. It just so happens that "d4+1" and "d6" have the same average roll (3.5), so this will heal for the same amount on average. However, this approach clearly increases the variance, which means that these potions will be more "swingy".

There are a couple of ways to partially mitigate this "swinginess" without using the per-die reroll method described above. First, you could simply say that the potion never heals for less then the normal minimum, regardless of the roll. For example, a greater healing potion would heal a minimum of 8 hit points, even if you roll 4 1s. The minimum amount is easy to calculate: it's twice the number of dice (except for supreme potions, as mentioned above). Alternatively, if the roll is less than the normal minimum, reroll all the potion's dice at once. This is still quicker than selectively rerolling specific dice, since you can scoop them all up at once. Using either of these methods, the rolls will still be a bit more swingy than the standard formula using d4s, but at least you'll never heal for less than what would normally be possible.

You probably don't want to impose the corresponding limit on the high end, since the math is a bit harder, and I doubt your players are going to complain about the rare cases where potions heal too much.

Similar mean & variance: Replace each "d4+1" with "2d6/2"

There is a pretty good way to get a similar mean and variance to the normal potion formula using only d6s, with no rerolls: for each d4+1 in the potion, roll two d6s and take the average. In other words, Nd4+N becomes (2*N)d6/2. This has exactly the same mean and only a slightly higher variance, so the distributions are quite similar. The downside of this solution is obvious, of course: you have to roll and add up twice as many dice.

Note: Since you're dividing by two, you'll need to choose a rounding rule. I recommend rounding up, since otherwise the mean will actually be 0.25 lower than a normal potion, and while we all know that this tiny amount is insignificant in the grand scheme, that probably won't stop your players from complaining about how you "nerfed" their potions. As you might expect, rounding up has the opposite effect, giving a mean that is 0.25 higher.

As with the previous rule, you can still apply the variant that any roll lower than the normal minimum potion roll is treated as that minimum (e.g. 8 for greater healing potions).

AnyDice formulas

Here are AnyDice formulas for all of the above solutions:

 Set N to 2 for regular healing potion, 4, for greater, 8 for superior 

N: 4



output Nd4+N named "d4+1 (Normal potion formula)"

output Nd6 named "d6, no rerolls"

output ((2*N)d6+1)/2 named "2d6 halved, rounded up"

This is an excruciatingly correct method for deriving 1d4 from 2d6... and I do mean excruciatingly: Take inspiration from how 2d10 are converted to a 1d100 percentile throw. More specifically, treat one of the d6 throws as a "ones" die, and another of the d6 throws as a "twos" die (as opposed to a "tens" die for a percentile throw.)

But wait, how is a d6 treated as a "ones" die? By convention: 1, 2, or 3 are treated as 0, while 4, 5, or 6 are treated as 1.

But wait, how is a d6 treated as a "twos" die? By convention: 1, 2, or 3 are treated as 0, while 4, 5, or 6 are treated as 2.

Then add the results, which will range from 0 to 3. Add 1 to get a conventional 1d4 or 2 to get your 1d4+1.

But wait, what the heck are you talking about, how does this even work?

Here, see this table, which by casual inspection shows us that the distribution of 1s, 2s, 3s, and 4s is uniform and thus functionally equivalent to 1d4, without needing to re-roll.

begin{array}{|c|c|c|c|}
hline
"twos" d6 & twos & "ones" d6 & ones & 1d4 \ hline
1 & 0 & 1 & 0 & 1\ hline
1 & 0 & 2 & 0 & 1\ hline
1 & 0 & 3 & 0 & 1\ hline
1 & 0 & 4 & 1 & 2\ hline
1 & 0 & 5 & 1 & 2\ hline
1 & 0 & 6 & 1 & 2\ hline
2 & 0 & 1 & 0 & 1\ hline
2 & 0 & 2 & 0 & 1\ hline
2 & 0 & 3 & 0 & 1\ hline
2 & 0 & 4 & 1 & 2\ hline
2 & 0 & 5 & 1 & 2\ hline
2 & 0 & 6 & 1 & 2\ hline
3 & 0 & 1 & 0 & 1\ hline
3 & 0 & 2 & 0 & 1\ hline
3 & 0 & 3 & 0 & 1\ hline
3 & 0 & 4 & 1 & 2\ hline
3 & 0 & 5 & 1 & 2\ hline
3 & 0 & 6 & 1 & 2\ hline
4 & 2 & 1 & 0 & 3\ hline
4 & 2 & 2 & 0 & 3\ hline
4 & 2 & 3 & 0 & 3\ hline
4 & 2 & 4 & 1 & 4\ hline
4 & 2 & 5 & 1 & 4\ hline
4 & 2 & 6 & 1 & 4\ hline
5 & 2 & 1 & 0 & 3\ hline
5 & 2 & 2 & 0 & 3\ hline
5 & 2 & 3 & 0 & 3\ hline
5 & 2 & 4 & 1 & 4\ hline
5 & 2 & 5 & 1 & 4\ hline
5 & 2 & 6 & 1 & 4\ hline
6 & 2 & 1 & 0 & 3\ hline
6 & 2 & 2 & 0 & 3\ hline
6 & 2 & 3 & 0 & 3\ hline
6 & 2 & 4 & 1 & 4\ hline
6 & 2 & 5 & 1 & 4\ hline
6 & 2 & 6 & 1 & 4\ hline
end{array}

But wait, how do I get this to work for 4d4+4? Use more dice. The remaining logistics are left as an exercise for the querent.

Several ways actually!

modified dice

let's get some blank dice. Now we mark two sides opposite each other as blank for reroll. And then we mark the other 4 sides with 1 to 4, using 1-2-4-3 as the order when spinning the die around the reroll axis.

• Pro: perfect models a d4


• Con: modified dice & possibly multiple rerolls

adding coins

As an alternative to modify the dice, we can add some coin tosses: for each 5 and 6 one coin is tossed. A head modifies one number above 4 by -4, a tail by -2.

• Pro: perfect models a d4


• Con: the need for additional coin tosses up to the number of dice.