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I am planning to create some custom monsters for my games. I know how to determine a monster's average hit points, however I don't know how you determine the decomposition in hit dice.

For example if I create something similar to a Goblin I am able to tell it should have 7 hit points but I don't know how 2d6 is determined.

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The DMG outlines the process for creating monsters as a DM.

There are two ways to decide how much HP your monster has:

Method 1: The CR Table

You can start with the monster's expected challenge rating and use the Monster Statistics by Challenge Rating table to determine an appropriate number of hit points. The table presents a range of hit points for each challenge rating. (DMG p.276)

In this method, you are going to do some math to go from the hit point range of your monster to the number of hit die your creature has.

In this case you use the table "Monster Statistics by Challenge Rating" (DMG p. 274) to look at the line with your expected challenge rating and see the range of hit points in that table.

Make sure at this point you know the size of your monster because that is what determines the size of the monster's hit die.

I am going to quote this AngryGM article because it explains this part of the process well:

For example, I might be designing a creature and need to roll between 80 and 100 hit points. If the creature is small and it has a Constitution modifier of +1, I can actually figure out exactly the dice code I need to get in that range. Follow the logic. Small creatures roll a d6 for hit points. So this creature is going to roll 1d6+1 some number of times to determine its hit points. What’s the average roll? Well, the average on 1d6 is 3.5 (half of 6 plus a half), so the average of 1d6+1 is 3.5+1 or 4.5. So, if I take, say, 85 and divide it by 4.5, I get 18.8. That means I need to roll 1d8+1 [ed: actually 1d6+1] about 18 times to get in the ballpark of 85. In this case, if I multiply 4.5 times 18, I get 81. Perfect. So, this creature has 81 (18d6+18) hit points.

After going through the above process you should now have the number of hit dice your creature possesses.

Method 2: Assigning Hit Dice

Alternatively, you can assign a number of Hit Dice to a monster, then calculate its average hit points. Don't worry if the hit points aren t matching up with the expected challenge rating for the monster. Other factors can affect a monster's challenge rating, as shown in later steps, and you can always adjust a monster's Hit Dice and hit points later on. (DMG p.276)

This one is very easy: you simply assign the monster's number of hit dice and make sure that the CR comes out the way you want by adjusting other features of the monster further down the line.

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The size of the die is determined by monster size, and the number of die is determined by CR.

According to AngryGM's very useful monster building series, the size of the hit die is determined by the size of the creature:

The die you roll is determined entirely by the creature’s size. Small creatures ALWAYS use a d6. Large creatures ALWAYS use a d10. Notice this is true even of humans with apparent class features (MM 342-350) and when adding class levels to a creature (DMG 283).

You can find a table with this information for all sizes on MM 7.

The constitution modifier is something you determine while you're building the monster.

The number of hit die to roll, and your ultimate HP target, is determined by your desired defensive CR (DMG 274), which tells you how much HP your monster should have.

Thus, you start with the amount of HP your monster should have, and then work backward to calculate how many hit die your monster should have. Here's an example, coming again from the same AngryGM article:

For example, I might be designing a creature and need to roll between 80 and 100 hit points. If the creature is small and it has a Constitution modifier of +1, I can actually figure out exactly the dice code I need to get in that range. Follow the logic. Small creatures roll a d6 for hit points. So this creature is going to roll 1d6+1 some number of times to determine its hit points. What’s the average roll? Well, the average on 1d6 is 3.5 (half of 6 plus a half), so the average of 1d6+1 is 3.5+1 or 4.5. So, if I take, say, 85 and divide it by 4.5, I get 18.8. That means I need to roll 1d8+1 [ed: actually 1d6+1] about 18 times to get in the ballpark of 85. In this case, if I multiply 4.5 times 18, I get 81. Perfect. So, this creature has 81 (18d6+18) hit points.

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  • \$\begingroup\$ Except, if you want between 80 and 100 hp, then you should be aiming for an average of 90, which is 20*4.5. So your creature should have 20d6+20. Which will give you a range of 40 to 140. \$\endgroup\$ – Benubird Jan 16 '18 at 12:43
  • \$\begingroup\$ @Benubird That's true (assuming it has CON +1), but that's true of all the MM monsters too. Plus, the chances that you'll roll all 1's (or 6's) on 20 d6's is something like 1 in 10^16... \$\endgroup\$ – Icyfire Jan 16 '18 at 16:18
  • \$\begingroup\$ This answer would be improved by noting that this makes the actual HD the monster uses basically irrelevant in gameplay and monster creation; it is vestigial. \$\endgroup\$ – Yakk Jan 17 '18 at 22:42
  • \$\begingroup\$ @Yakk Not if the monster takes a short rest and wants to spend some hit dice to recover hp. \$\endgroup\$ – the dark wanderer Jan 19 '18 at 6:37
  • \$\begingroup\$ @the Yes that is a core mechanic whose lack would be sorely missed \$\endgroup\$ – Yakk Jan 19 '18 at 11:19
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For example if I create something similar to a Goblin I am able to tell it should have 7 hit points but I don't know how 2d6 is determined.

7 (2d6) is just a form of shorthand.

As you probably know, "2d6" means "two six-sided dice". The "7" is just the average of the rolls. There's a simple equation you can use to determine the average of dice rolls: $$Dice_{ValueAverage} = \frac{Dice_{Quantity}+(Dice_{Sides}*Dice_{Quantity})}2.$$ Does it seem complicated? Try this, then: $$x = \frac{y+zy}2.$$

"x" is the average you are looking for. "y" is how many dice you are rolling. "z" is how many sides are on the dice.

In the case of 2d6, the equation looks like this: $$7 = \frac{2+(6*2)}2$$ $$7 = \frac{2+12}2$$ $$7 = \frac{14}2$$

So, that's the explanation of how the hit points are determined. 7 is the average value of 2d6. This means that a goblin cursed by the gods will have as few as 2 hp and one that was blessed by the gods will have as much as 12 hp. (Not much of a blessing, if you ask me...)

If you want to retroactively determine hit dice from a preferred average hit points, you just work backwards. Let's say you want a monster with an average of 27 hit points. Generally, the way to do it is by working with a size chart. If you'd rather determine the size later, though, you'll be left with 3 dice choices: d4, d8, and d10. Any other dice value will result in an average value different than 27. Here's the way of finding out the dice values, though: $$27 = \frac{27*2}2$$ $$27 = \frac{54}2$$ Because 27 is the average value, you need to multiply it by 2 in order to get the middle value. This means "(y+zy)" should equal 54. Because we usually round down half-values in D&D, "27.5" is considered the same as "27" for our purposes, expanding the final value of "(y+zy)" to ultimately include either 54 OR 55, but no value higher or lower than these two. (If you wish to see the math for the result of 55, I can add it at your request, but it really is just a matter of following these same steps with "27.5" instead of "27".) $$27 = \frac{y+zy}2$$ Here, we reach variables. Because we know "y" has to be the number of faces on a dice, we don't have to worry about y being greater than 20 or less than 4. Because we know "x" is the number of dice being rolled, we don't have to worry about x being a negative number. This makes our large list of possible outcomes very much more contained. The number of Dice Sides (z) we have for our creature's hit dice can only be 4 (d4), 6 (d6), 8 (d8), 10 (d10), 12 (d12), and 20 (d20). This gives us 12 equations to test. (Because we need a final value of 27 OR 27.5.)

$$27 = \frac{y+(4*y)}2$$ $$27.5 = \frac{y+(4*y)}2$$ $$27 = \frac{y+(6*y)}2$$ $$27.5 = \frac{y+(6*y)}2$$ $$27 = \frac{y+(8*y)}2$$ $$27.5 = \frac{y+(8*y)}2$$ $$27 = \frac{y+(10*y)}2$$ $$27.5 = \frac{y+(10*y)}2$$ $$27 = \frac{y+(12*y)}2$$ $$27.5 = \frac{y+(12*y)}2$$ $$27 = \frac{y+(20*y)}2$$ $$27.5 = \frac{y+(20*y)}2$$

While I admit this looks intimidating, it's really not that difficult to work through. Just take it one step at a time through each equation. After all, at this point, you just need to find "y".

Equation 1: $$27 = \frac{y+(4*y)}2$$

You need to get rid of the "2" that it is being divided by, so multiply both sides by 2.

$$27*2 = (\frac{y+(4*y)}2)*2$$ $$54 = y+(4*y)$$ $$54 = y+4y$$

"4y" is just another way of saying "y+y+y+y", so let's simplify this together. $$54 = y+y+y+y+y$$ $$54 = 5y$$

Now, just divide 5 from both sides. $$54/5 = 5y/5$$ $$10.8 = y$$ Well, it looks like that wasn't quite right. Let's try the steps again, but this time let's work with 27.5.

Equation 2: $$27.5 = \frac{y+(4*y)}2$$

Multiply both sides by 2. $$27.5*2 = (\frac{y+(4*y)}2)*2$$ $$55 = y+(4*y)$$ $$55 = y+4y$$

Simplify. $$55 = y+y+y+y+y$$ $$55 = 5y$$

Divide 5 from both sides. $$55/5 = 5y/5$$ $$11 = y$$

As you can see, we got "11" as our result. If we replace "y" in the original equation with "11", we get the following:

$$27.5 = \frac{11+4(11)}2$$ $$27.5 = \frac{11+44}2$$ $$27.5 = \frac{55}2$$ $$27.5 = 27.5$$

With this, we find that by using "x=27.5" (which rounds down to 27), "y=11", and "z=4" (to represent a 4-sided dice), it takes 11d4 to get the average of 27.5. I'm going to continue the other equations for examples.

Equation 3: $$27 = \frac{y+(6*y)}2$$ $$27*2 = (\frac{y+(6*y)}2)*2$$ $$54 = y+(6*y)$$ $$54 = y+6y$$ $$54 = y+y+y+y+y+y+y$$ $$54 = 7y$$ $$54/7 = 7y/7$$ $$7.7 = y$$

Equation 4: $$27.5 = \frac{y+(6*y)}2$$ $$27.5*2 = (\frac{y+(6*y)}2)*2$$ $$55 = y+(6*y)$$ $$55 = y+6y$$ $$55 = y+y+y+y+y+y+y$$ $$55 = 7y$$ $$55/7 = 7y/7$$ $$7.85 = y$$ Result: You cannot get an average of 27 or 27.5 by using only 6-sided dice.

Equation 5: $$27 = \frac{y+(8*y)}2$$ $$27*2 = (\frac{y+(8*y)}2)*2$$ $$54 = y+(8*y)$$ $$54 = y+8y$$ $$54 = y+y+y+y+y+y+y+y+y$$ $$54 = 9y$$ $$54/9 = 9y/9$$ $$6 = y$$ Result: You only need 6d8 in order to get an average of 27. Equation 6 is unnecessary.

Equation 7: $$27 = \frac{y+(10*y)}2$$ $$27*2 = (\frac{y+(10*y)}2)*2$$ $$54 = y+(10*y)$$ $$54 = y+10y$$ $$54 = y+y+y+y+y+y+y+y+y+y+y$$ $$54 = 11y$$ $$54/11 = 11y/11$$ $$4.9 = y$$

Equation 7: $$27.5 = \frac{y+(10*y)}2$$ $$27.5*2 = (\frac{y+(10*y)}2)*2$$ $$55 = y+(10*y)$$ $$55 = y+10y$$ $$55 = y+y+y+y+y+y+y+y+y+y+y$$ $$55 = 11y$$ $$55/11 = 11y/11$$ $$5 = y$$ Result: You only need 5d10 in order to get an average of 27.5.

I could be a completionist, but I think you get the point. It's just basic algebra after this point. As long as you understand the math, it's pretty easy to retro-engineer any average HP value you want.

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    \$\begingroup\$ Your formula is fine, but I find it easier when teaching people that struggle with math to give the following rule of thumb: every 2dX has an average roll of X+1. This works because the average of any homogenous combination of dice is the same as the sum of its components' averages however you decide to cut up the combination, and indeed if you plug '2' as the quantity into your formula you'll see that it simplifies nicely to 'die size +1'. May be worth mentioning. \$\endgroup\$ – the dark wanderer Jan 16 '18 at 7:57
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    \$\begingroup\$ @thedarkwanderer rounding down is almost always the rule in 5e and I could not find any exception that would make this any different so I rolled your edit back. \$\endgroup\$ – Rubiksmoose Jan 16 '18 at 21:43
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    \$\begingroup\$ @thedarkwanderer, your rule is only applicable to 2dX, though. unless you're shooting for a monster with an average of 5, 7, 9, 11, 13, or 21 HP, that rule isn't very useful for this. Besides, anybody who has a spreadsheet program could just make a spreadsheet for this and find the answer without doing the math themselves. It doesn't take a genius to open Excel and copy and paste this formula, replacing the variables with other cell locations. That's what I do and I'm a bonafide idiot. lol \$\endgroup\$ – Sora Tamashii Jan 17 '18 at 17:02
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    \$\begingroup\$ Additionally, @thedarkwanderer, This is outright stated in the basic rules, "There’s one more general rule you need to know at the outset. Whenever you divide a number in the game, round down if you end up with a fraction, even if the fraction is one-half or greater." (PBR V0.2, pg. 4) The only exception to this I know of is Arcane Recover, "The spell slots can have a combined level that is equal to or less than half your wizard level (rounded up)," (PBR, pg. 31). I can't recall any other instance where my players have been able to round up. \$\endgroup\$ – Sora Tamashii Jan 17 '18 at 17:13
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    \$\begingroup\$ @thedarkwanderer, Um... this might sound rude, but I don't mean it as such: Do you know what "averages" are? (a+b+c+...)/[number of values added together] Averages are inherently division. The only time you "round up" is if a specific instance of the rules say to do so. Please, cite where the rules indicate rounding up in regards to dice. You round up when rolling for a d3 by rolling a d6 and the fixed value alternative for rolling an HD on Level Up is rounded up, but that is it in regards to dice. Nowhere else in the PHB is rounding dice up mentioned. \$\endgroup\$ – Sora Tamashii Jan 17 '18 at 18:43
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So, first of all, the size of the die is determined by the size of the creature (see MM pg 7). That just leaves the number of dice.

While there are a number of mathematical techniques that can be used to calculate that, when teaching people who struggle with math I give the following rule of thumb:

Every 2dX has an average roll of X+1. To find the average of a collection of dice, just find the average of each part and add them up. So, for 2d6 we know that that's just 6+1=7, but for 4d6 we'd have to do 7*2=14 (because we have 2 groups of 2d6).

Since you want to work backwards, starting with the average hit points and ending with a number of dice, you can just plug in the size of the die and then divide.

So let's say you want a 38 hp Large creature. That's a d10 hit die so every 2d10 adds up to 11. 38/11 is 3 and a middle-sized bit, so we use 7d10. If the remainder felt small, we'd use 6d10. If it felt big, we'd use 8d10.

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