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 "Nd4+N (Normal potion formula)"

output Nd6 named "Nd6, no rerolls"

output [highest of 2*N and Nd6] named "Nd6, minimum 2N"

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

output [highest of 2*N and ((2*N)d6+1)/2] named "2d6 halved, rounded up, minimum 2N"

Use d6 with customised distribution

As potions use d4 in pairs, you may modify standard d6 pairs to approximate the same results:

• one dice should have 2 pips removed from the 5 and 6 faces (they become "standard-looking" 3 and 4) - leading to a dice distribution of {1, 2, 3, 3, 4, 4} - call it the "good dice", and use green / white ones.


• the other one should have 4 pips removed from the 5 and 6 faces (they become "standard-looking" 1 and 2) - leading to a dice distribution of {1, 1, 2, 2, 3, 4} - call it the "bad dice", and use red / black ones.

That way, whenever you roll one of each, you obtain the exact same mean, maximum and minimum - and a very similar variance as 2d4.

Anydice comparison for a standard potion of healing:

output d{1, 2, 3, 3, 4, 4}+d{1, 1, 2, 2, 3, 4}+2

output 2d4+2

The tricky part is obviously to modify the dice so that they remain legible, and keep a good balance - I'd advise agains Tipp-ex on white dice, and instead look for good markers with colors close to the dice you use.

Alternative distributions may make things even more simple

If adding pips to existing dice does not discourage you, or you find cheap enough 12mm blank or custom dice - you may even consider alternate distributions to avoid the addition of a constant number. The following (non-exhaustive) list offers the same exact results as the previous one :

dice that range from 2 to 5:
output d{2, 3, 4, 4, 5, 5}+d{2, 2, 3, 3, 4, 5}

"very good dice" vs "very bad dice":
output d{3, 4, 5, 5, 6, 6}+d{1, 1, 2, 2, 3, 4}

rare extremes :
output d{3, 3, 4, 4, 5, 6}+d{1, 2, 3, 3, 4, 4}

Only downside : as the supreme potion does not follow the standard pattern, it would either need specific custom dice, or a +10 constant. I'd advise the latter.

If custom dice are not an option - flip them instead

Just define substitution rules on the potion's label, such as :

Black dice : flip the dice if it lands on a 5 or 6

White dice : flip the dice if it lands on a 1 or 2

Now they match the "very good dice" vs "very bad dice" distributions, and can be used for the same results.

• Healing potions get 1 of each


• Greater healing potions get 2 of each


• Superior healing potions get 4 or each


• Supreme healing potions either get 5 of each and a +10 - or you may use 8 good dice and 3 bad dice for approximately the same result.

How does it compare to other answers?

I couldn't resist the urge to aggregate the various methods listed here in a single anydice link (use comments to focus on a single potion when necessary - sorry I couldn't include rerolls!).
This method seems closest to the DMG calculation, but obviously can't beat Limari's proposal efficiency :)

This answer includes a frame challenge - Apologies if it not relevant due to your specific craft project :)

Roll paper instead of 6-sided-dice.

After all, a potion's potency is much more influenced by the way it has been prepared, than it is by the way it is drinked. You may roll d4 by yourself before, note the result on a paper, roll it and put it in your vial.

Cons : Players don't get to roll dice. Part of the fun goes away.

Pros : Several, actually :

• Proper Nd4+N formulas are respected


• Surprise remains


• Cheaper than buying dice


• Faster than rolling dice in game


• You are not limited to healing potions


• The content may differ from the label (adulterated healing potion, or even worse - poison?)


• Sometimes, the players won't be able to tell. Did this Superior Healing potion heal only 17 hit points because of a lack of luck, or because it was a Greater Healing Potion?

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
text{“twos” d6} & text{twos} & text{“ones” d6} & text{ones} & text{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 depending on their materials at hand, but color coding dice pairs and scribing the "twos" dice is one possible solution.

In fact, the excruciating table above is meant to motivate the idea of converting 2d6 into 1d4. It is not actually necessary to pair, e.g., these two green d6s into a single d4, and those two red d6s into another separate d4.

Instead, one can color code the "ones" dice vs the "twos" dice, i.e., black dice for "ones" and and red dice for "twos".

Then a roll of black = (1, 2, 3, 4) and red = (3, 4, 5, 6) converts to:

Black = (0 + 0 + 0 + 1) and red = (0 + 2 + 2 + 2) for a total of 1 + 6 + 4 + 4 = 15 (where the first 4 is necessary to convert to a d4 and the second 4 converts to d4+1)

Finally if you can find appropriately sized blank d6s which you believe are fair, you can skip all this numerical conversion stuff and just paint the appropriate numerals on the faces of the dice.

If these are always and forever going to be used for 1d4+1, just paint the "twos" dice with 1 and 3, and the "ones" dice with 1 and 2. You can just sum those up directly, and save yourself the hassle of adding those 4's every time.

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. (or, if you want to incorporate the +1 per d4: 2-3-5-4)

• Pro: perfectly models a d4 (or d4+1)


• Con: modified dice & possibly multiple rerolls

adding "coins"

As an alternative to modifying the dice, we can add some "coin tosses". Let's first look at what a coin could do: for each 5 and 6 one coin is tossed. A head modifies one number above 4 by -4, a tail by -2, turning the die to be either a 1-2-3-4-1-2 (head) or 1-2-3-4-3-4 (tails). Each distribution happens 50% of the time, so the resulting distribution is exactly as the d4.

• Pro: perfectly models a d4


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

But do we need a coin? No. Not actually. Let's look at the dice. If we color code them, so that each is a different color, we could make a circle system. Like for 5d4 we might write down "black-red-green-umbra-white(-black)". If one shows a 5 or 6, we look if the next die in the row is even (head) or odd (tails). Even and odd results are just like a coin and it's quite common to replace coin tosses with this.

Example: We roll 5d4 using the above color code and get b5 r2 g3 u6 w5. With even being -4 and odd -2 that gets turned into b1 r2 g3 u2 w3. That's 11.

• pro: perfectly models a d4, no modified items or extra things needed


• con: the need to mark down the color-code along the dice tube.