Dice roll probability

John wrote:
> Has anyone ever sat down and worked out a formula or method to determine the
> probability that rolling N dice will produce a score above or below X?

My way isn't very sophisticated and involves a lot of tally marks. For
instance, to figure 1d, you mark rows 1 through 6 once. To figure 2d,
take 1d and on a new graph, for every tally on 1d, mark that number
plus 1 through 6 on the new graph. (So one tally on 1 produces one
tally on each of 2 through 7, two tallies on 3 would produce two
tallies on 4 through 9, etc.) Iterate as necessary, although I've only
done this up to 3 dice. That gives you the probability that you'll roll
e.g. 11 on 3d. To figure out the probability of rolling 11 or above,
count all tallies for 11 and above and divide by the total number of
tallies (6^3 = 216).

I'm sure I'm re-inventing some portion of probability theory here, and
I'd never use this method beyond a maximum of 4 dice, but... well, it's
a method.

-Max
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John wrote:
> Has anyone ever sat down and worked out a formula or method to determine the
> probability that rolling N dice will produce a score above or below X?
>
> I've tried to come up with one, as it sounds rather useful, but I keep
> having to refer to things I haven't looked at for ten years. So before I
> reinvented the wheel I thought I'd ask if anyone has already done the work
> for me.

There's a table with the probabilities for 3d6 in GURPS Basic 3rd
Edition Revised. I cannot recall whether it is also in 4E. No
mathematics, just the frequencies (including the ones for rolling
equal-to-or-less) expressed as percentages.

--
Peter Knutsen
sagatafl.org
 >> Stay informed about: Dice roll probability 

On 15 Jun 2006 09:43:41 -0700, "Max Wilson" <wilson.max DeleteThis @gmail.com>
wrote:

>John wrote:
>> Has anyone ever sat down and worked out a formula or method to determine the
>> probability that rolling N dice will produce a score above or below X?
>
>My way isn't very sophisticated and involves a lot of tally marks. For
>instance, to figure 1d, you mark rows 1 through 6 once. To figure 2d,
>take 1d and on a new graph, for every tally on 1d, mark that number
>plus 1 through 6 on the new graph. (So one tally on 1 produces one
>tally on each of 2 through 7, two tallies on 3 would produce two
>tallies on 4 through 9, etc.) Iterate as necessary, although I've only
>done this up to 3 dice. That gives you the probability that you'll roll
>e.g. 11 on 3d. To figure out the probability of rolling 11 or above,
>count all tallies for 11 and above and divide by the total number of
>tallies (6^3 = 216).
>
>I'm sure I'm re-inventing some portion of probability theory here, and
>I'd never use this method beyond a maximum of 4 dice, but... well, it's
>a method.
>
>-Max

that definitely qualifies as doing it the hard way. probability
theory isn't that tough really. shadowrun for instance and their dice
pools gets a bit hairy granted but most of it's pretty simple.
 >> Stay informed about: Dice roll probability 

John wrote:
> Has anyone ever sat down and worked out a formula or method to determine the
> probability that rolling N dice will produce a score above or below X?
>
> I've tried to come up with one, as it sounds rather useful, but I keep
> having to refer to things I haven't looked at for ten years. So before I
> reinvented the wheel I thought I'd ask if anyone has already done the work
> for me.
>
> John

I've been thinking about this problem a bit, and I don't think there is
a simple formula that you can plug N and X into to get the desired
results. I'm not a math whiz by any means, but it looks to me like the
formula for rolling any number is going to be different depending on
how many dice.

For example, The chance of rolling N on N number of dice is going to be
1/6^N, because there is only one way to roll this number (all 1's).

The chance of rolling N+1 on N dice is going to be N/6^N, because no
matter how many dice you have the only way to roll it is to get one 2
and all the rest 1's, and the 2 can occur on any of the dice. But you
can't just put an X in there for the number you're rolling against
because the formula you use for a given number depends on the number of
dice. Your chance of rolling a 3 on 2 dice is 2/6^2, but on 3 dice it's
1/6^3, and obviously you can't do it on 4 or more dice.
 >> Stay informed about: Dice roll probability 

I once wrote a M-Basic program that graphed dice rolls, even open ended so I
could see the results of a few thousand rolls, very interesting differences
between say 3d6+2 and a d20. I even at one stage added to hit rolls
calculated against a armour class, as I was designing a sytem that used one
d20 roll for to hit and damage, you had a base damage plus the amount you
hit by. I was trying to modify cyberpunk to speed up combat and make a .22
pistol actually dangerous.


"Scooter the Mighty" <Greyguy3.RemoveThis@hotmail.com> wrote in message
news:1151697072.204558.117250@y41g2000cwy.googlegroups.com...
>
> John wrote:
>> Has anyone ever sat down and worked out a formula or method to determine
>> the
>> probability that rolling N dice will produce a score above or below X?
>>
>> I've tried to come up with one, as it sounds rather useful, but I keep
>> having to refer to things I haven't looked at for ten years. So before I
>> reinvented the wheel I thought I'd ask if anyone has already done the
>> work
>> for me.
>>
>> John
>
> I've been thinking about this problem a bit, and I don't think there is
> a simple formula that you can plug N and X into to get the desired
> results. I'm not a math whiz by any means, but it looks to me like the
> formula for rolling any number is going to be different depending on
> how many dice.
>
> For example, The chance of rolling N on N number of dice is going to be
> 1/6^N, because there is only one way to roll this number (all 1's).
>
> The chance of rolling N+1 on N dice is going to be N/6^N, because no
> matter how many dice you have the only way to roll it is to get one 2
> and all the rest 1's, and the 2 can occur on any of the dice. But you
> can't just put an X in there for the number you're rolling against
> because the formula you use for a given number depends on the number of
> dice. Your chance of rolling a 3 on 2 dice is 2/6^2, but on 3 dice it's
> 1/6^3, and obviously you can't do it on 4 or more dice.
>
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