What is the difference between rolling more dice versus fewer dice?3d6 vs a d20: What is the effect of a...

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What is the difference between rolling more dice versus fewer dice?


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Let's say there are two attacks: one rolls a 10d10 and the other rolls a 5d20. Assuming the modifier was the same, which rolls would have better chances of rolling average, and which would have better chances of rolling the minimum or maximum?



To specify I am aware that more dice would have a higher avg and min roll but both of those would be by a slim margin and I'm not sure how those would affect your "odds" of hitting each number.










share|improve this question











$endgroup$












  • $begingroup$
    As it stands, I do not understand this question. Ability checks (if that's what you mean) do not use such rolls, and....well, I don't know at all what you are talking about when you talk about the modifier being the same and having better odds of hitting (abilities aren't used to hit anything). Are you actually talking about D&D?
    $endgroup$
    – PJRZ
    20 hours ago












  • $begingroup$
    @PJRZ by ability I meant a magic spell technique or some other form of attack. fixed it.
    $endgroup$
    – Mage in the Barrel
    20 hours ago












  • $begingroup$
    @NautArch And they can do a much better job than we can. I've seent it!
    $endgroup$
    – Premier Bromanov
    20 hours ago






  • 3




    $begingroup$
    @NautArch I've seen mods move posts to Math. SE, so it might just require moderator attention. That is, you can flag it
    $endgroup$
    – Premier Bromanov
    20 hours ago








  • 5




    $begingroup$
    This might be intro-level for dice mechanics but it’s not off topic here. We take design questions and dice stats questions, including entry-level ones. As a general principle, when a Q is on topic on multiple sites, wherever the asker chose to post is the place for it to be.
    $endgroup$
    – SevenSidedDie
    18 hours ago


















15












$begingroup$


Let's say there are two attacks: one rolls a 10d10 and the other rolls a 5d20. Assuming the modifier was the same, which rolls would have better chances of rolling average, and which would have better chances of rolling the minimum or maximum?



To specify I am aware that more dice would have a higher avg and min roll but both of those would be by a slim margin and I'm not sure how those would affect your "odds" of hitting each number.










share|improve this question











$endgroup$












  • $begingroup$
    As it stands, I do not understand this question. Ability checks (if that's what you mean) do not use such rolls, and....well, I don't know at all what you are talking about when you talk about the modifier being the same and having better odds of hitting (abilities aren't used to hit anything). Are you actually talking about D&D?
    $endgroup$
    – PJRZ
    20 hours ago












  • $begingroup$
    @PJRZ by ability I meant a magic spell technique or some other form of attack. fixed it.
    $endgroup$
    – Mage in the Barrel
    20 hours ago












  • $begingroup$
    @NautArch And they can do a much better job than we can. I've seent it!
    $endgroup$
    – Premier Bromanov
    20 hours ago






  • 3




    $begingroup$
    @NautArch I've seen mods move posts to Math. SE, so it might just require moderator attention. That is, you can flag it
    $endgroup$
    – Premier Bromanov
    20 hours ago








  • 5




    $begingroup$
    This might be intro-level for dice mechanics but it’s not off topic here. We take design questions and dice stats questions, including entry-level ones. As a general principle, when a Q is on topic on multiple sites, wherever the asker chose to post is the place for it to be.
    $endgroup$
    – SevenSidedDie
    18 hours ago
















15












15








15


1



$begingroup$


Let's say there are two attacks: one rolls a 10d10 and the other rolls a 5d20. Assuming the modifier was the same, which rolls would have better chances of rolling average, and which would have better chances of rolling the minimum or maximum?



To specify I am aware that more dice would have a higher avg and min roll but both of those would be by a slim margin and I'm not sure how those would affect your "odds" of hitting each number.










share|improve this question











$endgroup$




Let's say there are two attacks: one rolls a 10d10 and the other rolls a 5d20. Assuming the modifier was the same, which rolls would have better chances of rolling average, and which would have better chances of rolling the minimum or maximum?



To specify I am aware that more dice would have a higher avg and min roll but both of those would be by a slim margin and I'm not sure how those would affect your "odds" of hitting each number.







dice statistics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 20 hours ago









Tuorg

1,7621720




1,7621720










asked 21 hours ago









Mage in the BarrelMage in the Barrel

1818




1818












  • $begingroup$
    As it stands, I do not understand this question. Ability checks (if that's what you mean) do not use such rolls, and....well, I don't know at all what you are talking about when you talk about the modifier being the same and having better odds of hitting (abilities aren't used to hit anything). Are you actually talking about D&D?
    $endgroup$
    – PJRZ
    20 hours ago












  • $begingroup$
    @PJRZ by ability I meant a magic spell technique or some other form of attack. fixed it.
    $endgroup$
    – Mage in the Barrel
    20 hours ago












  • $begingroup$
    @NautArch And they can do a much better job than we can. I've seent it!
    $endgroup$
    – Premier Bromanov
    20 hours ago






  • 3




    $begingroup$
    @NautArch I've seen mods move posts to Math. SE, so it might just require moderator attention. That is, you can flag it
    $endgroup$
    – Premier Bromanov
    20 hours ago








  • 5




    $begingroup$
    This might be intro-level for dice mechanics but it’s not off topic here. We take design questions and dice stats questions, including entry-level ones. As a general principle, when a Q is on topic on multiple sites, wherever the asker chose to post is the place for it to be.
    $endgroup$
    – SevenSidedDie
    18 hours ago




















  • $begingroup$
    As it stands, I do not understand this question. Ability checks (if that's what you mean) do not use such rolls, and....well, I don't know at all what you are talking about when you talk about the modifier being the same and having better odds of hitting (abilities aren't used to hit anything). Are you actually talking about D&D?
    $endgroup$
    – PJRZ
    20 hours ago












  • $begingroup$
    @PJRZ by ability I meant a magic spell technique or some other form of attack. fixed it.
    $endgroup$
    – Mage in the Barrel
    20 hours ago












  • $begingroup$
    @NautArch And they can do a much better job than we can. I've seent it!
    $endgroup$
    – Premier Bromanov
    20 hours ago






  • 3




    $begingroup$
    @NautArch I've seen mods move posts to Math. SE, so it might just require moderator attention. That is, you can flag it
    $endgroup$
    – Premier Bromanov
    20 hours ago








  • 5




    $begingroup$
    This might be intro-level for dice mechanics but it’s not off topic here. We take design questions and dice stats questions, including entry-level ones. As a general principle, when a Q is on topic on multiple sites, wherever the asker chose to post is the place for it to be.
    $endgroup$
    – SevenSidedDie
    18 hours ago


















$begingroup$
As it stands, I do not understand this question. Ability checks (if that's what you mean) do not use such rolls, and....well, I don't know at all what you are talking about when you talk about the modifier being the same and having better odds of hitting (abilities aren't used to hit anything). Are you actually talking about D&D?
$endgroup$
– PJRZ
20 hours ago






$begingroup$
As it stands, I do not understand this question. Ability checks (if that's what you mean) do not use such rolls, and....well, I don't know at all what you are talking about when you talk about the modifier being the same and having better odds of hitting (abilities aren't used to hit anything). Are you actually talking about D&D?
$endgroup$
– PJRZ
20 hours ago














$begingroup$
@PJRZ by ability I meant a magic spell technique or some other form of attack. fixed it.
$endgroup$
– Mage in the Barrel
20 hours ago






$begingroup$
@PJRZ by ability I meant a magic spell technique or some other form of attack. fixed it.
$endgroup$
– Mage in the Barrel
20 hours ago














$begingroup$
@NautArch And they can do a much better job than we can. I've seent it!
$endgroup$
– Premier Bromanov
20 hours ago




$begingroup$
@NautArch And they can do a much better job than we can. I've seent it!
$endgroup$
– Premier Bromanov
20 hours ago




3




3




$begingroup$
@NautArch I've seen mods move posts to Math. SE, so it might just require moderator attention. That is, you can flag it
$endgroup$
– Premier Bromanov
20 hours ago






$begingroup$
@NautArch I've seen mods move posts to Math. SE, so it might just require moderator attention. That is, you can flag it
$endgroup$
– Premier Bromanov
20 hours ago






5




5




$begingroup$
This might be intro-level for dice mechanics but it’s not off topic here. We take design questions and dice stats questions, including entry-level ones. As a general principle, when a Q is on topic on multiple sites, wherever the asker chose to post is the place for it to be.
$endgroup$
– SevenSidedDie
18 hours ago






$begingroup$
This might be intro-level for dice mechanics but it’s not off topic here. We take design questions and dice stats questions, including entry-level ones. As a general principle, when a Q is on topic on multiple sites, wherever the asker chose to post is the place for it to be.
$endgroup$
– SevenSidedDie
18 hours ago












4 Answers
4






active

oldest

votes


















48












$begingroup$

From what I can gather what you're asking, you want to know the probabilistic difference between rolling 10d10 and 5d20. You've rightly pointed out that each roll has the same maximum and that each has a better chance at rolling their given averages. They obviously have different minimums (10 vs 5), and so you want to know precisely how different the rolls are.



Using AnyDice.com we can calculate the probability very simply with the commands output 10d10 and output 5d20. And really that's all there is to it. The black line below represents 10d10, and the yellow line represents 5d20.



enter image description here



Generally speaking, when you have a greater number of smaller dice, your rolls are less "swingy". Meaning, there are better odds at rolling the "average". But, you have worse odds at rolling higher numbers.



Put another way: Look at this graph, it represents the odds that you will roll at least a given number. You can see in general it's better to roll 10d10 because you have greater odds at hitting a certain number until about 60, then 5d20 gives you better odds at hitting those values, but only slightly.



enter image description here



So with 5d20, you have higher odds at hitting a greater range of values, meaning that if you roll 5d20 often, you'll see more "swingy" results. But with 10d10, the odds are more stacked in the middle, meaning it should feel like you're hitting the "average" or the "middle" results more often.



But let's simplify. Lets look at output 2d10 vs output 1d20. Same idea as 10d10 vs 5d20. Here we can see that each value of 1d20 has an equal chance to be rolled. But with 2d10, the odds change because there are a greater number of rolls that represent the middle values (11). there's 10-1, 9-2, 8-3, 7-4, 6-5, 5-6, 4-7, 3-8, 2-9, and 1-10 representing 11. 10% of all the combinations are 11. But for higher values (20), there is only 10-10 representing that, which is only 1% of all possibilities.



enter image description here



Similarly, if you wanted to compare 1d100 to 5d20 and 10d10, you would see a flat probability: a 1% chance for each value between 1 and 100.



enter image description here






share|improve this answer











$endgroup$





















    18












    $begingroup$

    More dice lead to more average results



    There are a number of dice calculators around the internet to illustrate the probabilities of whatever combinations you'd like to see, but a general statement is that rolling a combination of dice and summing them for a result will increase the probability of average results and reduce the probability of extreme results as compared with rolling fewer dice.



    For a simple example consider 2d4-1 and 1d7 (this is for you @SevenSidedDie) both produce a number between 1 and 7. Your chance of a 1 with the 7-sider is 1 in 7 or about 14%. Your chance with the 2d4 is 1 in 16 (6.25%), because there are 16 different results possible, but only one of them is the 2 ones which give you a 1. On the other hand there are 3 ways to get 3: 1 & 3, 2 & 2, and 3 & 1, so 3 chances in 16 (18.75%), but still only 1 in 7 with the 7-sider.






    share|improve this answer











    $endgroup$





















      0












      $begingroup$

      I'll just give a decent example. Rolling 1d6 is comepletely random. Every part has only one way to be rolled, and thus there are an equivalent amount of ways to roll each number. Given that the die is not weighted, then each roll is random. Rolling 2d6 means you suddenly can roll 6 in multiple different ways. The greatest example is that you can only roll "2" one way via 2d6, two "1s". However, 3+3 = 6, 2+4 = 6, 1+5 = 6, 4+2 = 6, and finally 5+1 = 6, so that means there are five ways to roll a six.



      This is actually something relavent in the game Mekton. There is an optional rule people use to use 2d6 instead of 1d10. It was introduced in the liscensed spin off, (not created by tallorn games) Gundam Senki. I've been in a campaign where limited Senki rules were used instead of Metkon rules.



      Also, it means a critical success, and thus exploding dice, was limited to a 1/36 vs a 1/10 chance. This is actually desirable. I'm playing in a campaign that's ironically Gundam themed, and we have quite the instance rate of critting and then disabling various famous mobile suits. Half the Gundams were captured by Zeon by killing the pilot, including the full armor 7th gundam, Netix, and a ton of interesting stuff. With a 2d12 system, the chance of that happening drops drastically, even if you still use the 1d10 table to roll for critical damage, to 1/180 vs 1/50 (there are two ways knock out enemy mobile suits without killing). It's slightly less than that, as there's more ways to disable a mekton without called shots, but there's my comparison. Still, we've had not quite good luck to die to a ball randomly, like had happened to Dozle, who took one right in the face.






      share|improve this answer










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      Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $endgroup$





















        0












        $begingroup$

        Die rolls have mean equal to the average of the largest and smallest number so for a die with f faces (a "df"), the average is (1+f)/2 and the variance is equal to the mean times (f-1)/6; i.e. (f+1)(f-1)/12. The mean and variance of a sum of dice is the sum of the means and the sum of the variances respectively. If the dice are all the same (ndf), then the mean is n(f+1)/2 and the variance is n(f+1)(f-1)/12.



        The chance of rolling the minimum (or the maximum) is (1/f)^n - if you halve the faces but double the number of dice, the second case has a larger minimum but the same maximum, but the maximum will be more likely with the larger die.



        If you're adding more than a few dice, the probability of the nearest-to-average roll will be roughly 1.38/[f√n]; if you halve the number of faces and double the number of dice you move the average up by 1/2 for each die you started with, and the probability of rolling the nearest-to-average roll will increase by roughly 42%. (These numbers are a bit more accurate with more faces and more dice and not so accurate with few faces and few dice.)






        share|improve this answer









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          4 Answers
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          48












          $begingroup$

          From what I can gather what you're asking, you want to know the probabilistic difference between rolling 10d10 and 5d20. You've rightly pointed out that each roll has the same maximum and that each has a better chance at rolling their given averages. They obviously have different minimums (10 vs 5), and so you want to know precisely how different the rolls are.



          Using AnyDice.com we can calculate the probability very simply with the commands output 10d10 and output 5d20. And really that's all there is to it. The black line below represents 10d10, and the yellow line represents 5d20.



          enter image description here



          Generally speaking, when you have a greater number of smaller dice, your rolls are less "swingy". Meaning, there are better odds at rolling the "average". But, you have worse odds at rolling higher numbers.



          Put another way: Look at this graph, it represents the odds that you will roll at least a given number. You can see in general it's better to roll 10d10 because you have greater odds at hitting a certain number until about 60, then 5d20 gives you better odds at hitting those values, but only slightly.



          enter image description here



          So with 5d20, you have higher odds at hitting a greater range of values, meaning that if you roll 5d20 often, you'll see more "swingy" results. But with 10d10, the odds are more stacked in the middle, meaning it should feel like you're hitting the "average" or the "middle" results more often.



          But let's simplify. Lets look at output 2d10 vs output 1d20. Same idea as 10d10 vs 5d20. Here we can see that each value of 1d20 has an equal chance to be rolled. But with 2d10, the odds change because there are a greater number of rolls that represent the middle values (11). there's 10-1, 9-2, 8-3, 7-4, 6-5, 5-6, 4-7, 3-8, 2-9, and 1-10 representing 11. 10% of all the combinations are 11. But for higher values (20), there is only 10-10 representing that, which is only 1% of all possibilities.



          enter image description here



          Similarly, if you wanted to compare 1d100 to 5d20 and 10d10, you would see a flat probability: a 1% chance for each value between 1 and 100.



          enter image description here






          share|improve this answer











          $endgroup$


















            48












            $begingroup$

            From what I can gather what you're asking, you want to know the probabilistic difference between rolling 10d10 and 5d20. You've rightly pointed out that each roll has the same maximum and that each has a better chance at rolling their given averages. They obviously have different minimums (10 vs 5), and so you want to know precisely how different the rolls are.



            Using AnyDice.com we can calculate the probability very simply with the commands output 10d10 and output 5d20. And really that's all there is to it. The black line below represents 10d10, and the yellow line represents 5d20.



            enter image description here



            Generally speaking, when you have a greater number of smaller dice, your rolls are less "swingy". Meaning, there are better odds at rolling the "average". But, you have worse odds at rolling higher numbers.



            Put another way: Look at this graph, it represents the odds that you will roll at least a given number. You can see in general it's better to roll 10d10 because you have greater odds at hitting a certain number until about 60, then 5d20 gives you better odds at hitting those values, but only slightly.



            enter image description here



            So with 5d20, you have higher odds at hitting a greater range of values, meaning that if you roll 5d20 often, you'll see more "swingy" results. But with 10d10, the odds are more stacked in the middle, meaning it should feel like you're hitting the "average" or the "middle" results more often.



            But let's simplify. Lets look at output 2d10 vs output 1d20. Same idea as 10d10 vs 5d20. Here we can see that each value of 1d20 has an equal chance to be rolled. But with 2d10, the odds change because there are a greater number of rolls that represent the middle values (11). there's 10-1, 9-2, 8-3, 7-4, 6-5, 5-6, 4-7, 3-8, 2-9, and 1-10 representing 11. 10% of all the combinations are 11. But for higher values (20), there is only 10-10 representing that, which is only 1% of all possibilities.



            enter image description here



            Similarly, if you wanted to compare 1d100 to 5d20 and 10d10, you would see a flat probability: a 1% chance for each value between 1 and 100.



            enter image description here






            share|improve this answer











            $endgroup$
















              48












              48








              48





              $begingroup$

              From what I can gather what you're asking, you want to know the probabilistic difference between rolling 10d10 and 5d20. You've rightly pointed out that each roll has the same maximum and that each has a better chance at rolling their given averages. They obviously have different minimums (10 vs 5), and so you want to know precisely how different the rolls are.



              Using AnyDice.com we can calculate the probability very simply with the commands output 10d10 and output 5d20. And really that's all there is to it. The black line below represents 10d10, and the yellow line represents 5d20.



              enter image description here



              Generally speaking, when you have a greater number of smaller dice, your rolls are less "swingy". Meaning, there are better odds at rolling the "average". But, you have worse odds at rolling higher numbers.



              Put another way: Look at this graph, it represents the odds that you will roll at least a given number. You can see in general it's better to roll 10d10 because you have greater odds at hitting a certain number until about 60, then 5d20 gives you better odds at hitting those values, but only slightly.



              enter image description here



              So with 5d20, you have higher odds at hitting a greater range of values, meaning that if you roll 5d20 often, you'll see more "swingy" results. But with 10d10, the odds are more stacked in the middle, meaning it should feel like you're hitting the "average" or the "middle" results more often.



              But let's simplify. Lets look at output 2d10 vs output 1d20. Same idea as 10d10 vs 5d20. Here we can see that each value of 1d20 has an equal chance to be rolled. But with 2d10, the odds change because there are a greater number of rolls that represent the middle values (11). there's 10-1, 9-2, 8-3, 7-4, 6-5, 5-6, 4-7, 3-8, 2-9, and 1-10 representing 11. 10% of all the combinations are 11. But for higher values (20), there is only 10-10 representing that, which is only 1% of all possibilities.



              enter image description here



              Similarly, if you wanted to compare 1d100 to 5d20 and 10d10, you would see a flat probability: a 1% chance for each value between 1 and 100.



              enter image description here






              share|improve this answer











              $endgroup$



              From what I can gather what you're asking, you want to know the probabilistic difference between rolling 10d10 and 5d20. You've rightly pointed out that each roll has the same maximum and that each has a better chance at rolling their given averages. They obviously have different minimums (10 vs 5), and so you want to know precisely how different the rolls are.



              Using AnyDice.com we can calculate the probability very simply with the commands output 10d10 and output 5d20. And really that's all there is to it. The black line below represents 10d10, and the yellow line represents 5d20.



              enter image description here



              Generally speaking, when you have a greater number of smaller dice, your rolls are less "swingy". Meaning, there are better odds at rolling the "average". But, you have worse odds at rolling higher numbers.



              Put another way: Look at this graph, it represents the odds that you will roll at least a given number. You can see in general it's better to roll 10d10 because you have greater odds at hitting a certain number until about 60, then 5d20 gives you better odds at hitting those values, but only slightly.



              enter image description here



              So with 5d20, you have higher odds at hitting a greater range of values, meaning that if you roll 5d20 often, you'll see more "swingy" results. But with 10d10, the odds are more stacked in the middle, meaning it should feel like you're hitting the "average" or the "middle" results more often.



              But let's simplify. Lets look at output 2d10 vs output 1d20. Same idea as 10d10 vs 5d20. Here we can see that each value of 1d20 has an equal chance to be rolled. But with 2d10, the odds change because there are a greater number of rolls that represent the middle values (11). there's 10-1, 9-2, 8-3, 7-4, 6-5, 5-6, 4-7, 3-8, 2-9, and 1-10 representing 11. 10% of all the combinations are 11. But for higher values (20), there is only 10-10 representing that, which is only 1% of all possibilities.



              enter image description here



              Similarly, if you wanted to compare 1d100 to 5d20 and 10d10, you would see a flat probability: a 1% chance for each value between 1 and 100.



              enter image description here







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 20 hours ago

























              answered 20 hours ago









              Premier BromanovPremier Bromanov

              12.4k646110




              12.4k646110

























                  18












                  $begingroup$

                  More dice lead to more average results



                  There are a number of dice calculators around the internet to illustrate the probabilities of whatever combinations you'd like to see, but a general statement is that rolling a combination of dice and summing them for a result will increase the probability of average results and reduce the probability of extreme results as compared with rolling fewer dice.



                  For a simple example consider 2d4-1 and 1d7 (this is for you @SevenSidedDie) both produce a number between 1 and 7. Your chance of a 1 with the 7-sider is 1 in 7 or about 14%. Your chance with the 2d4 is 1 in 16 (6.25%), because there are 16 different results possible, but only one of them is the 2 ones which give you a 1. On the other hand there are 3 ways to get 3: 1 & 3, 2 & 2, and 3 & 1, so 3 chances in 16 (18.75%), but still only 1 in 7 with the 7-sider.






                  share|improve this answer











                  $endgroup$


















                    18












                    $begingroup$

                    More dice lead to more average results



                    There are a number of dice calculators around the internet to illustrate the probabilities of whatever combinations you'd like to see, but a general statement is that rolling a combination of dice and summing them for a result will increase the probability of average results and reduce the probability of extreme results as compared with rolling fewer dice.



                    For a simple example consider 2d4-1 and 1d7 (this is for you @SevenSidedDie) both produce a number between 1 and 7. Your chance of a 1 with the 7-sider is 1 in 7 or about 14%. Your chance with the 2d4 is 1 in 16 (6.25%), because there are 16 different results possible, but only one of them is the 2 ones which give you a 1. On the other hand there are 3 ways to get 3: 1 & 3, 2 & 2, and 3 & 1, so 3 chances in 16 (18.75%), but still only 1 in 7 with the 7-sider.






                    share|improve this answer











                    $endgroup$
















                      18












                      18








                      18





                      $begingroup$

                      More dice lead to more average results



                      There are a number of dice calculators around the internet to illustrate the probabilities of whatever combinations you'd like to see, but a general statement is that rolling a combination of dice and summing them for a result will increase the probability of average results and reduce the probability of extreme results as compared with rolling fewer dice.



                      For a simple example consider 2d4-1 and 1d7 (this is for you @SevenSidedDie) both produce a number between 1 and 7. Your chance of a 1 with the 7-sider is 1 in 7 or about 14%. Your chance with the 2d4 is 1 in 16 (6.25%), because there are 16 different results possible, but only one of them is the 2 ones which give you a 1. On the other hand there are 3 ways to get 3: 1 & 3, 2 & 2, and 3 & 1, so 3 chances in 16 (18.75%), but still only 1 in 7 with the 7-sider.






                      share|improve this answer











                      $endgroup$



                      More dice lead to more average results



                      There are a number of dice calculators around the internet to illustrate the probabilities of whatever combinations you'd like to see, but a general statement is that rolling a combination of dice and summing them for a result will increase the probability of average results and reduce the probability of extreme results as compared with rolling fewer dice.



                      For a simple example consider 2d4-1 and 1d7 (this is for you @SevenSidedDie) both produce a number between 1 and 7. Your chance of a 1 with the 7-sider is 1 in 7 or about 14%. Your chance with the 2d4 is 1 in 16 (6.25%), because there are 16 different results possible, but only one of them is the 2 ones which give you a 1. On the other hand there are 3 ways to get 3: 1 & 3, 2 & 2, and 3 & 1, so 3 chances in 16 (18.75%), but still only 1 in 7 with the 7-sider.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 20 hours ago

























                      answered 20 hours ago









                      TuorgTuorg

                      1,7621720




                      1,7621720























                          0












                          $begingroup$

                          I'll just give a decent example. Rolling 1d6 is comepletely random. Every part has only one way to be rolled, and thus there are an equivalent amount of ways to roll each number. Given that the die is not weighted, then each roll is random. Rolling 2d6 means you suddenly can roll 6 in multiple different ways. The greatest example is that you can only roll "2" one way via 2d6, two "1s". However, 3+3 = 6, 2+4 = 6, 1+5 = 6, 4+2 = 6, and finally 5+1 = 6, so that means there are five ways to roll a six.



                          This is actually something relavent in the game Mekton. There is an optional rule people use to use 2d6 instead of 1d10. It was introduced in the liscensed spin off, (not created by tallorn games) Gundam Senki. I've been in a campaign where limited Senki rules were used instead of Metkon rules.



                          Also, it means a critical success, and thus exploding dice, was limited to a 1/36 vs a 1/10 chance. This is actually desirable. I'm playing in a campaign that's ironically Gundam themed, and we have quite the instance rate of critting and then disabling various famous mobile suits. Half the Gundams were captured by Zeon by killing the pilot, including the full armor 7th gundam, Netix, and a ton of interesting stuff. With a 2d12 system, the chance of that happening drops drastically, even if you still use the 1d10 table to roll for critical damage, to 1/180 vs 1/50 (there are two ways knock out enemy mobile suits without killing). It's slightly less than that, as there's more ways to disable a mekton without called shots, but there's my comparison. Still, we've had not quite good luck to die to a ball randomly, like had happened to Dozle, who took one right in the face.






                          share|improve this answer










                          New contributor




                          Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$


















                            0












                            $begingroup$

                            I'll just give a decent example. Rolling 1d6 is comepletely random. Every part has only one way to be rolled, and thus there are an equivalent amount of ways to roll each number. Given that the die is not weighted, then each roll is random. Rolling 2d6 means you suddenly can roll 6 in multiple different ways. The greatest example is that you can only roll "2" one way via 2d6, two "1s". However, 3+3 = 6, 2+4 = 6, 1+5 = 6, 4+2 = 6, and finally 5+1 = 6, so that means there are five ways to roll a six.



                            This is actually something relavent in the game Mekton. There is an optional rule people use to use 2d6 instead of 1d10. It was introduced in the liscensed spin off, (not created by tallorn games) Gundam Senki. I've been in a campaign where limited Senki rules were used instead of Metkon rules.



                            Also, it means a critical success, and thus exploding dice, was limited to a 1/36 vs a 1/10 chance. This is actually desirable. I'm playing in a campaign that's ironically Gundam themed, and we have quite the instance rate of critting and then disabling various famous mobile suits. Half the Gundams were captured by Zeon by killing the pilot, including the full armor 7th gundam, Netix, and a ton of interesting stuff. With a 2d12 system, the chance of that happening drops drastically, even if you still use the 1d10 table to roll for critical damage, to 1/180 vs 1/50 (there are two ways knock out enemy mobile suits without killing). It's slightly less than that, as there's more ways to disable a mekton without called shots, but there's my comparison. Still, we've had not quite good luck to die to a ball randomly, like had happened to Dozle, who took one right in the face.






                            share|improve this answer










                            New contributor




                            Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              I'll just give a decent example. Rolling 1d6 is comepletely random. Every part has only one way to be rolled, and thus there are an equivalent amount of ways to roll each number. Given that the die is not weighted, then each roll is random. Rolling 2d6 means you suddenly can roll 6 in multiple different ways. The greatest example is that you can only roll "2" one way via 2d6, two "1s". However, 3+3 = 6, 2+4 = 6, 1+5 = 6, 4+2 = 6, and finally 5+1 = 6, so that means there are five ways to roll a six.



                              This is actually something relavent in the game Mekton. There is an optional rule people use to use 2d6 instead of 1d10. It was introduced in the liscensed spin off, (not created by tallorn games) Gundam Senki. I've been in a campaign where limited Senki rules were used instead of Metkon rules.



                              Also, it means a critical success, and thus exploding dice, was limited to a 1/36 vs a 1/10 chance. This is actually desirable. I'm playing in a campaign that's ironically Gundam themed, and we have quite the instance rate of critting and then disabling various famous mobile suits. Half the Gundams were captured by Zeon by killing the pilot, including the full armor 7th gundam, Netix, and a ton of interesting stuff. With a 2d12 system, the chance of that happening drops drastically, even if you still use the 1d10 table to roll for critical damage, to 1/180 vs 1/50 (there are two ways knock out enemy mobile suits without killing). It's slightly less than that, as there's more ways to disable a mekton without called shots, but there's my comparison. Still, we've had not quite good luck to die to a ball randomly, like had happened to Dozle, who took one right in the face.






                              share|improve this answer










                              New contributor




                              Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.






                              $endgroup$



                              I'll just give a decent example. Rolling 1d6 is comepletely random. Every part has only one way to be rolled, and thus there are an equivalent amount of ways to roll each number. Given that the die is not weighted, then each roll is random. Rolling 2d6 means you suddenly can roll 6 in multiple different ways. The greatest example is that you can only roll "2" one way via 2d6, two "1s". However, 3+3 = 6, 2+4 = 6, 1+5 = 6, 4+2 = 6, and finally 5+1 = 6, so that means there are five ways to roll a six.



                              This is actually something relavent in the game Mekton. There is an optional rule people use to use 2d6 instead of 1d10. It was introduced in the liscensed spin off, (not created by tallorn games) Gundam Senki. I've been in a campaign where limited Senki rules were used instead of Metkon rules.



                              Also, it means a critical success, and thus exploding dice, was limited to a 1/36 vs a 1/10 chance. This is actually desirable. I'm playing in a campaign that's ironically Gundam themed, and we have quite the instance rate of critting and then disabling various famous mobile suits. Half the Gundams were captured by Zeon by killing the pilot, including the full armor 7th gundam, Netix, and a ton of interesting stuff. With a 2d12 system, the chance of that happening drops drastically, even if you still use the 1d10 table to roll for critical damage, to 1/180 vs 1/50 (there are two ways knock out enemy mobile suits without killing). It's slightly less than that, as there's more ways to disable a mekton without called shots, but there's my comparison. Still, we've had not quite good luck to die to a ball randomly, like had happened to Dozle, who took one right in the face.







                              share|improve this answer










                              New contributor




                              Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              share|improve this answer



                              share|improve this answer








                              edited 13 hours ago





















                              New contributor




                              Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              answered 14 hours ago









                              Ryan HofmannRyan Hofmann

                              92




                              92




                              New contributor




                              Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.





                              New contributor





                              Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.






                              Ryan Hofmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.























                                  0












                                  $begingroup$

                                  Die rolls have mean equal to the average of the largest and smallest number so for a die with f faces (a "df"), the average is (1+f)/2 and the variance is equal to the mean times (f-1)/6; i.e. (f+1)(f-1)/12. The mean and variance of a sum of dice is the sum of the means and the sum of the variances respectively. If the dice are all the same (ndf), then the mean is n(f+1)/2 and the variance is n(f+1)(f-1)/12.



                                  The chance of rolling the minimum (or the maximum) is (1/f)^n - if you halve the faces but double the number of dice, the second case has a larger minimum but the same maximum, but the maximum will be more likely with the larger die.



                                  If you're adding more than a few dice, the probability of the nearest-to-average roll will be roughly 1.38/[f√n]; if you halve the number of faces and double the number of dice you move the average up by 1/2 for each die you started with, and the probability of rolling the nearest-to-average roll will increase by roughly 42%. (These numbers are a bit more accurate with more faces and more dice and not so accurate with few faces and few dice.)






                                  share|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Die rolls have mean equal to the average of the largest and smallest number so for a die with f faces (a "df"), the average is (1+f)/2 and the variance is equal to the mean times (f-1)/6; i.e. (f+1)(f-1)/12. The mean and variance of a sum of dice is the sum of the means and the sum of the variances respectively. If the dice are all the same (ndf), then the mean is n(f+1)/2 and the variance is n(f+1)(f-1)/12.



                                    The chance of rolling the minimum (or the maximum) is (1/f)^n - if you halve the faces but double the number of dice, the second case has a larger minimum but the same maximum, but the maximum will be more likely with the larger die.



                                    If you're adding more than a few dice, the probability of the nearest-to-average roll will be roughly 1.38/[f√n]; if you halve the number of faces and double the number of dice you move the average up by 1/2 for each die you started with, and the probability of rolling the nearest-to-average roll will increase by roughly 42%. (These numbers are a bit more accurate with more faces and more dice and not so accurate with few faces and few dice.)






                                    share|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Die rolls have mean equal to the average of the largest and smallest number so for a die with f faces (a "df"), the average is (1+f)/2 and the variance is equal to the mean times (f-1)/6; i.e. (f+1)(f-1)/12. The mean and variance of a sum of dice is the sum of the means and the sum of the variances respectively. If the dice are all the same (ndf), then the mean is n(f+1)/2 and the variance is n(f+1)(f-1)/12.



                                      The chance of rolling the minimum (or the maximum) is (1/f)^n - if you halve the faces but double the number of dice, the second case has a larger minimum but the same maximum, but the maximum will be more likely with the larger die.



                                      If you're adding more than a few dice, the probability of the nearest-to-average roll will be roughly 1.38/[f√n]; if you halve the number of faces and double the number of dice you move the average up by 1/2 for each die you started with, and the probability of rolling the nearest-to-average roll will increase by roughly 42%. (These numbers are a bit more accurate with more faces and more dice and not so accurate with few faces and few dice.)






                                      share|improve this answer









                                      $endgroup$



                                      Die rolls have mean equal to the average of the largest and smallest number so for a die with f faces (a "df"), the average is (1+f)/2 and the variance is equal to the mean times (f-1)/6; i.e. (f+1)(f-1)/12. The mean and variance of a sum of dice is the sum of the means and the sum of the variances respectively. If the dice are all the same (ndf), then the mean is n(f+1)/2 and the variance is n(f+1)(f-1)/12.



                                      The chance of rolling the minimum (or the maximum) is (1/f)^n - if you halve the faces but double the number of dice, the second case has a larger minimum but the same maximum, but the maximum will be more likely with the larger die.



                                      If you're adding more than a few dice, the probability of the nearest-to-average roll will be roughly 1.38/[f√n]; if you halve the number of faces and double the number of dice you move the average up by 1/2 for each die you started with, and the probability of rolling the nearest-to-average roll will increase by roughly 42%. (These numbers are a bit more accurate with more faces and more dice and not so accurate with few faces and few dice.)







                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 34 mins ago









                                      Glen_bGlen_b

                                      1,712815




                                      1,712815






























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