Average Die Results [Archive] - Wizards Community

Average Die Results [Archive] - Wizards Community

Post/Author/DateTime Post
DM_Newbie
09-04-05, 02:25 AM Can anyone here point me in the direction for the "average" die results? I think i am missing something (probably my brain), but the PHB speaks of "average die results" how do you figure them? Or, what are they for each die?
Escef
09-04-05, 02:40 AM The same way you compute any average, add up all numbers and divide by the quantity of entries. In this case, you add up the values of all the faces on the die and divide by the die's size.

Of course, the fastest and simplest way is to divide the size of the die by two and add 0.5.

For example, a d6.

(1+2+3+4+5+6)/6=21/6=3.5
(6/2)+0.5=3.5

K?
DM_Newbie
09-04-05, 02:48 AM thank you, as i mentioned, the gray matter needed seems to have fallen away in very dramatic, though pointless manner....Thank you.
Tyler Do'Urden
09-04-05, 03:27 AM Yeah, the average for an individual die is (die size+1) / 2. So average on a d8 would be 9/2, or 4.5. For multiple dice you multiply.

The probability curves are a little different though.

For a single die, the probability is linear (or at least it should be) a straight horizontal line.
For two dice, the probability is an upside-down V; straight line from the minimum up to the average, then back down to the maximum.
For three or more dice, the probability is an increasingly narrow bell curve.

I wrote a program a while back that generates probability curves an input dice configuration. It was neat.

'Course it might not be totally relevant to this discussion, but I'm sure somebody will find it interesting.
Sapp
09-04-05, 03:52 AM I find it interesting, Tyler.

However, I'm wondering why such a program would be necessary for non-loaded dice. Assuming that the dice are normal (not loaded), aren't the probability curves always flat?

Ah, nevermind. I see you mean for more than one dice.

I often ignore probability curves when trying to come up with ways to speed up large numbers of dice rolls.

For example, for a 11d6 DB Fireball, I would simply do two d6 rolls.
Take one of them, multiply by 10, and then add the second d6 result.

It leads to a flat distribution, but it's faster and does cover the entire range of possibilities (from 11 to 66).
Escef
09-04-05, 04:18 AM Ah, nevermind. I see you mean for more than one dice.

...

For example, for a 11d6 DB Fireball, I would simply do two d6 rolls.

Die is singular, dice is plural. And you can't get 11d6 out of a fireball.

Carry on.
Sapp
09-04-05, 04:32 AM Die is singular, dice is plural.

True. My bad. I often type without thinking.

And you can't get 11d6 out of a fireball.
Carry on.
I said "DB Fireball". As you might already have begun to suspect, I'm tired. I felt lazy so I abbreviated "Delayed-Blast" into "DB". I believe Delayed-Blast Fireballs can go up to 20d6.
Escef
09-04-05, 04:36 AM I said "DB Fireball". As you might already have begun to suspect, I'm tired. I felt lazy so I abbreviated "Delayed-Blast" into "DB". I believe Delayed-Blast Fireballs can go up to 20d6.

My bad, too. I didn't even see the "DB", if I had I would've figured it out.
Solaris
09-04-05, 04:55 AM For example, for a 11d6 DB Fireball, I would simply do two d6 rolls.
Take one of them, multiply by 10, and then add the second d6 result.

It leads to a flat distribution, but it's faster and does cover the entire range of possibilities (from 11 to 66).

No it doesn't.
Sapp
09-04-05, 04:57 AM Oh no, I insist. That was my bad too.

I normally spell out the entire names of spells and italicize them for legibility as well. The abrupt change could easily confuse people.

To be on topic, if people don't like the Flat Distribution Method (FDM) of rolling for large values that I mentioned earlier, you could also try the Spiked Distribution Method (SDM) for quick dice-rolling.

Using the same Delayed-Blast Fireball example from earlier, we quickly calculate the average ((11+66)/2 = 38.5) and subtract the average of a single die roll from that (to get 38.5 - 3.5 = 35). Then add the result of a single d6.

The range of results goes from 36 to 41. This averages out to 38.5, which is the correct average for 11d6. However, this does not cover the range of possibilities (from 11 to 66). It is, however, more consistent than the FDM mentioned above, in which the probability for rolling 11 is as high as the probability of rolling a 38. With this method, you're definitely more likely to make consistent rolls.

Random Vars:
diceRoll = 11d6
singleRoll = 1d6

Constants:
Avg(diceRoll) = 38.5
Avg(singleRoll) = 3.5

Spiked Distribution Method (SDM):
Avg(diceRoll) - Avg(singleRoll) + singleRoll = yourResult

A combination of the SDM and FDM might be a good way to simulate Normal Distribution Method (where you just roll the entire 11d6).
You roll a die and note whether its odd or even. Or just flip a coin.
Use the result to determine whether you use SDM or FDM.

The resulting distribution from this Hybrid Distribution Method (HDM) would be something like this:

___----____

The middle values, ranging from 36 to 41, would be more likely, but every possibility (from 11 to 66) is still covered by the HDM.
Sapp
09-04-05, 04:58 AM No it doesn't.
Prove it. Provide a counterexample.

EDIT: Actually, I found one myself. NEVERMIND. Haha. I forgot that this method should only work with d10s.
Solaris
09-04-05, 05:02 AM Prove it. Provide a counterexample.

17.

Also 18, 19, 20, 27, 28, 29, 30, etc.
Sapp
09-04-05, 05:05 AM 17.

Also 18, 19, 20, 27, 28, 29, 30, etc.
Yes, my bad, again. DB Fireball was a bad example. This only works with d10s. For d6, a more rigorous method is needed. Give me some time to work this out. It is an interesting point.

I must have a way to simulate large dice rolls (without electronic equipment).

EDIT:
Perhaps the second dice should be replaced by a d10. This allows values ranging from 11 to 70. The values 67, 68, 69, and 70 demand re-rolls, but they happen so infrequently that it doesn't add too much time, in comparison to rolling 11 d6's and adding them up.

It might be beneficial in comparison when you're dealing with 20d8, or something like that. You'd been to buy 20 different d8s, then sum the results after rolling them. Instead, you could roll a d8, multiply by 19 and add the result to a d10, and hope to avoid the numbers 161 and 162. Haha. (Though re-rolling wouldn't be as bad).
Solaris
09-04-05, 05:15 AM This only works with d10s.

No it doesn't.
Sapp
09-04-05, 05:18 AM No it doesn't.
This one, admittedly, has me stumped.
Perhaps I'm too tired. Care to prove and provide counterexamples again?
Solaris
09-04-05, 05:22 AM This one, admittedly, has me stumped.
Perhaps I'm too tired. Care to prove and provide counterexamples again?

20, 30, etc.
Sapp
09-04-05, 05:25 AM 10, 20, 30, etc.
Actually, I'm fairly certain that it is you who is incorrect this time.

For an 11d10 roll. You take a d10 roll and multiple by one less than the number of rolls (11-1 = 10), to get a range from 10 to 100, in multiples of 10.

10,20,...,100.

You add the second d10 roll to get a range from 1 to 10.

Added to the values 10,20,...,100, you can expect to get 11 (min) to 110 (max).

Values such as 20 are achieved with a roll of 1 on the first die (which is multiplied by 10 to get 10) and a roll of 10 on the second.
10+10 = 20.

Similar methods are used to get 30, 40... 110.

You shouldn't be expecting to get a 10 from a 11d10 roll anyway.

Yes, it is confusing, but it's clever in its own little way, and I'm sure it would save time if done often enough. (Who DOESN'T do a lot of dice rolling? Haha). Besides, it's always fun to play around with Discrete Probability.
Solaris
09-04-05, 05:27 AM You're right, of course. Sorry.
Sapp
09-04-05, 05:32 AM You're right, of course. Sorry.
....... No need for apologies. I was the one who made the first mistake (a significant one too that makes a distribution that ignores many values of rolls).

(How is it that I'm always made to look like an ass, even when I'm not trying. Don't get me wrong, I do try, Often... but this isn't one of those times...too tired to try.)

I'll try to come up with a more efficient simplification/simulation tonight (one that's easy to remember and clearly makes sense). This problem should put me to sleep.
Dark Stryke
09-04-05, 10:31 AM Does anybody still roll the dice in D&D?! I mean, sure, it's a bit faster in the long run to use FDM or SDM, but I just enjoy rolling the dice! Also, it's funny to see the look on my players' faces when I begin scooping up large numbers of d6 and cackling madly whilst describing what the evil wizard is doing.
Gilean
09-04-05, 11:04 AM The probabilities of small numbers of dice, or with small enough a number of sides, can be taken directly from Pascal's triangle. When there are too many dice, or they are too big (the formula is somewhere in my notes, ask if interested enough). Then it gets a bit more complicated.
Sapp
09-04-05, 01:02 PM @ Dark Stryke
I'm not completely discounting that option. I would roll large amounts of dice if it's a very important roll that no one wants simplified. However, it would be nice to create a simple method/algorithm to simulate large dice rolls that can be done "mostly in your head," without the aid of a programmable calculator or laptop nearby. I like computers; I just don't like having to depend on them during gaming (which is kind of ironic, considering that most people play games on their computer nowadays).

I'm sure there are a few others who'd like a simple, effective way to replace 20d6 rolls and such.

...

To get a Normal Distribution similar to what we get by rolling massive amounts of dice, we could try doing something fancy with a estimate standard deviation dependent on the number of dice and the number of sides of each dice... Eh... I see why it might be nice to have a probability distribution render such as the one Tyler Do'Urden has. I guess I'll work on that tonight or tomorrow night.

Post/Author/DateTime	Post
DM_Newbie 09-04-05, 02:25 AM	Can anyone here point me in the direction for the "average" die results? I think i am missing something (probably my brain), but the PHB speaks of "average die results" how do you figure them? Or, what are they for each die?
Escef 09-04-05, 02:40 AM	The same way you compute any average, add up all numbers and divide by the quantity of entries. In this case, you add up the values of all the faces on the die and divide by the die's size. Of course, the fastest and simplest way is to divide the size of the die by two and add 0.5. For example, a d6. (1+2+3+4+5+6)/6=21/6=3.5 (6/2)+0.5=3.5 K?
DM_Newbie 09-04-05, 02:48 AM	thank you, as i mentioned, the gray matter needed seems to have fallen away in very dramatic, though pointless manner....Thank you.
Tyler Do'Urden 09-04-05, 03:27 AM	Yeah, the average for an individual die is (die size+1) / 2. So average on a d8 would be 9/2, or 4.5. For multiple dice you multiply. The probability curves are a little different though. For a single die, the probability is linear (or at least it should be) a straight horizontal line. For two dice, the probability is an upside-down V; straight line from the minimum up to the average, then back down to the maximum. For three or more dice, the probability is an increasingly narrow bell curve. I wrote a program a while back that generates probability curves an input dice configuration. It was neat. 'Course it might not be totally relevant to this discussion, but I'm sure somebody will find it interesting.
Sapp 09-04-05, 03:52 AM	I find it interesting, Tyler. However, I'm wondering why such a program would be necessary for non-loaded dice. Assuming that the dice are normal (not loaded), aren't the probability curves always flat? Ah, nevermind. I see you mean for more than one dice. I often ignore probability curves when trying to come up with ways to speed up large numbers of dice rolls. For example, for a 11d6 DB Fireball, I would simply do two d6 rolls. Take one of them, multiply by 10, and then add the second d6 result. It leads to a flat distribution, but it's faster and does cover the entire range of possibilities (from 11 to 66).
Escef 09-04-05, 04:18 AM	Ah, nevermind. I see you mean for more than one dice. ... For example, for a 11d6 DB Fireball, I would simply do two d6 rolls. Die is singular, dice is plural. And you can't get 11d6 out of a fireball. Carry on.
Sapp 09-04-05, 04:32 AM	Die is singular, dice is plural. True. My bad. I often type without thinking. And you can't get 11d6 out of a fireball. Carry on. I said "DB Fireball". As you might already have begun to suspect, I'm tired. I felt lazy so I abbreviated "Delayed-Blast" into "DB". I believe Delayed-Blast Fireballs can go up to 20d6.
Escef 09-04-05, 04:36 AM	I said "DB Fireball". As you might already have begun to suspect, I'm tired. I felt lazy so I abbreviated "Delayed-Blast" into "DB". I believe Delayed-Blast Fireballs can go up to 20d6. My bad, too. I didn't even see the "DB", if I had I would've figured it out.
Solaris 09-04-05, 04:55 AM	For example, for a 11d6 DB Fireball, I would simply do two d6 rolls. Take one of them, multiply by 10, and then add the second d6 result. It leads to a flat distribution, but it's faster and does cover the entire range of possibilities (from 11 to 66). No it doesn't.
Sapp 09-04-05, 04:57 AM	Oh no, I insist. That was my bad too. I normally spell out the entire names of spells and italicize them for legibility as well. The abrupt change could easily confuse people. To be on topic, if people don't like the Flat Distribution Method (FDM) of rolling for large values that I mentioned earlier, you could also try the Spiked Distribution Method (SDM) for quick dice-rolling. Using the same Delayed-Blast Fireball example from earlier, we quickly calculate the average ((11+66)/2 = 38.5) and subtract the average of a single die roll from that (to get 38.5 - 3.5 = 35). Then add the result of a single d6. The range of results goes from 36 to 41. This averages out to 38.5, which is the correct average for 11d6. However, this does not cover the range of possibilities (from 11 to 66). It is, however, more consistent than the FDM mentioned above, in which the probability for rolling 11 is as high as the probability of rolling a 38. With this method, you're definitely more likely to make consistent rolls. Random Vars: diceRoll = 11d6 singleRoll = 1d6 Constants: Avg(diceRoll) = 38.5 Avg(singleRoll) = 3.5 Spiked Distribution Method (SDM): Avg(diceRoll) - Avg(singleRoll) + singleRoll = yourResult A combination of the SDM and FDM might be a good way to simulate Normal Distribution Method (where you just roll the entire 11d6). You roll a die and note whether its odd or even. Or just flip a coin. Use the result to determine whether you use SDM or FDM. The resulting distribution from this Hybrid Distribution Method (HDM) would be something like this: ___----____ The middle values, ranging from 36 to 41, would be more likely, but every possibility (from 11 to 66) is still covered by the HDM.
Sapp 09-04-05, 04:58 AM	No it doesn't. Prove it. Provide a counterexample. EDIT: Actually, I found one myself. NEVERMIND. Haha. I forgot that this method should only work with d10s.
Solaris 09-04-05, 05:02 AM	Prove it. Provide a counterexample. 17. Also 18, 19, 20, 27, 28, 29, 30, etc.
Sapp 09-04-05, 05:05 AM	17. Also 18, 19, 20, 27, 28, 29, 30, etc. Yes, my bad, again. DB Fireball was a bad example. This only works with d10s. For d6, a more rigorous method is needed. Give me some time to work this out. It is an interesting point. I must have a way to simulate large dice rolls (without electronic equipment). EDIT: Perhaps the second dice should be replaced by a d10. This allows values ranging from 11 to 70. The values 67, 68, 69, and 70 demand re-rolls, but they happen so infrequently that it doesn't add too much time, in comparison to rolling 11 d6's and adding them up. It might be beneficial in comparison when you're dealing with 20d8, or something like that. You'd been to buy 20 different d8s, then sum the results after rolling them. Instead, you could roll a d8, multiply by 19 and add the result to a d10, and hope to avoid the numbers 161 and 162. Haha. (Though re-rolling wouldn't be as bad).
Solaris 09-04-05, 05:15 AM	This only works with d10s. No it doesn't.
Sapp 09-04-05, 05:18 AM	No it doesn't. This one, admittedly, has me stumped. Perhaps I'm too tired. Care to prove and provide counterexamples again?
Solaris 09-04-05, 05:22 AM	This one, admittedly, has me stumped. Perhaps I'm too tired. Care to prove and provide counterexamples again? 20, 30, etc.
Sapp 09-04-05, 05:25 AM	10, 20, 30, etc. Actually, I'm fairly certain that it is you who is incorrect this time. For an 11d10 roll. You take a d10 roll and multiple by one less than the number of rolls (11-1 = 10), to get a range from 10 to 100, in multiples of 10. 10,20,...,100. You add the second d10 roll to get a range from 1 to 10. Added to the values 10,20,...,100, you can expect to get 11 (min) to 110 (max). Values such as 20 are achieved with a roll of 1 on the first die (which is multiplied by 10 to get 10) and a roll of 10 on the second. 10+10 = 20. Similar methods are used to get 30, 40... 110. You shouldn't be expecting to get a 10 from a 11d10 roll anyway. Yes, it is confusing, but it's clever in its own little way, and I'm sure it would save time if done often enough. (Who DOESN'T do a lot of dice rolling? Haha). Besides, it's always fun to play around with Discrete Probability.
Solaris 09-04-05, 05:27 AM	You're right, of course. Sorry.
Sapp 09-04-05, 05:32 AM	You're right, of course. Sorry. ....... No need for apologies. I was the one who made the first mistake (a significant one too that makes a distribution that ignores many values of rolls). (How is it that I'm always made to look like an ass, even when I'm not trying. Don't get me wrong, I do try, Often... but this isn't one of those times...too tired to try.) I'll try to come up with a more efficient simplification/simulation tonight (one that's easy to remember and clearly makes sense). This problem should put me to sleep.
Dark Stryke 09-04-05, 10:31 AM	Does anybody still roll the dice in D&D?! I mean, sure, it's a bit faster in the long run to use FDM or SDM, but I just enjoy rolling the dice! Also, it's funny to see the look on my players' faces when I begin scooping up large numbers of d6 and cackling madly whilst describing what the evil wizard is doing.
Gilean 09-04-05, 11:04 AM	The probabilities of small numbers of dice, or with small enough a number of sides, can be taken directly from Pascal's triangle. When there are too many dice, or they are too big (the formula is somewhere in my notes, ask if interested enough). Then it gets a bit more complicated.
Sapp 09-04-05, 01:02 PM	@ Dark Stryke I'm not completely discounting that option. I would roll large amounts of dice if it's a very important roll that no one wants simplified. However, it would be nice to create a simple method/algorithm to simulate large dice rolls that can be done "mostly in your head," without the aid of a programmable calculator or laptop nearby. I like computers; I just don't like having to depend on them during gaming (which is kind of ironic, considering that most people play games on their computer nowadays). I'm sure there are a few others who'd like a simple, effective way to replace 20d6 rolls and such. ... To get a Normal Distribution similar to what we get by rolling massive amounts of dice, we could try doing something fancy with a estimate standard deviation dependent on the number of dice and the number of sides of each dice... Eh... I see why it might be nice to have a probability distribution render such as the one Tyler Do'Urden has. I guess I'll work on that tonight or tomorrow night.