July 28th, 2010 at 5:35:35 PM
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I am working on trying to understand the probability of a few different systems of rolling d10 dice.

The first system is a variable number of dice (anywhere from 1 to 15) but success is when an 8, 9 or 10 is rolled. I was able to figure out the formula for that one as

1-(.7^n) (where n is the # of dice rolled)

The second system is a tricky one:

Again you are rolling D10s but the success is variable.. so for say 10 d10s rolled, the success might be 6 or higher, 7 or higher 8 or higher or 9 or higher.

I am not trying to come up with the answer, just the formula for plugging in my own numbers.

The 3rd system is also trickier:

Again rolling D10s. The number of dice is set (10 of them) the success is set ( 6 or higher) but the variable is the number of successes needed. 1 6 or higher, 2 6s or higher, 3 6s or higher, etc.

Again I am just trying to work out the formula for plugging in my own numbers.

Any hints or help toward an answer would be really appreciated.

The first system is a variable number of dice (anywhere from 1 to 15) but success is when an 8, 9 or 10 is rolled. I was able to figure out the formula for that one as

1-(.7^n) (where n is the # of dice rolled)

The second system is a tricky one:

Again you are rolling D10s but the success is variable.. so for say 10 d10s rolled, the success might be 6 or higher, 7 or higher 8 or higher or 9 or higher.

I am not trying to come up with the answer, just the formula for plugging in my own numbers.

The 3rd system is also trickier:

Again rolling D10s. The number of dice is set (10 of them) the success is set ( 6 or higher) but the variable is the number of successes needed. 1 6 or higher, 2 6s or higher, 3 6s or higher, etc.

Again I am just trying to work out the formula for plugging in my own numbers.

Any hints or help toward an answer would be really appreciated.

July 30th, 2010 at 11:34:53 AM
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As I understand it, "d10 dice" refers to a 10 sided die used in "Dungeons and Dragons" and similar RPG games. However, the die that I am familar with is numbered from 0 to 9, so a single roll producing a 10 is not possible. Are you referring to something else, or do you use the "0" as "10"?

Simplicity is the ultimate sophistication - Leonardo da Vinci

July 30th, 2010 at 12:08:17 PM
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I am sorry for the confusion, yes "0" is used as 10.

the sides of the die are: 1,2,3,4,5,6,7,8,9,0. It is assumed that the die is weighted correctly and that the rolls are truly random.

the sides of the die are: 1,2,3,4,5,6,7,8,9,0. It is assumed that the die is weighted correctly and that the rolls are truly random.

July 30th, 2010 at 12:38:32 PM
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Quote:jimuThe second system is a tricky one:

Again you are rolling D10s but the success is variable.. so for say 10 d10s rolled, the success might be 6 or higher, 7 or higher 8 or higher or 9 or higher.

If you roll 10 dice (with values 1 - 10), the minimum TOTAL of the 10 rolls is 10. Did you actually mean, "a TOTAL of 50 or higher, 60+, etc.?"

Simplicity is the ultimate sophistication - Leonardo da Vinci

July 30th, 2010 at 1:13:51 PM
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the number that I am looking for when I am rolling the D10s in this example is the number of individual d10s that roll 6 or higher, or 7 or higher etc.

for example:

I roll 100 d10s

I am looking for how many of them roll 6-10. so probability states that aproximatly 1/2 of the hundred dice, will have a number 6 and above "a Success", and 1/2 the dice will be 5 and under "a fail".

this works really well on large samples, but as with most probability problems, when you take your sample size down to only 10, things get a little skewed.

so, to restate the problem with a little bit more detail:

I roll 10 d10s. I am counting how many of them roll a certain number or higher. is there a formula that can show, how the probability of rolling dice and getting successes when I am looking for the numbers to be above a 3 is easier to roll than when I am looking for the numbers to be above a 4.

for example:

I roll 100 d10s

I am looking for how many of them roll 6-10. so probability states that aproximatly 1/2 of the hundred dice, will have a number 6 and above "a Success", and 1/2 the dice will be 5 and under "a fail".

this works really well on large samples, but as with most probability problems, when you take your sample size down to only 10, things get a little skewed.

so, to restate the problem with a little bit more detail:

I roll 10 d10s. I am counting how many of them roll a certain number or higher. is there a formula that can show, how the probability of rolling dice and getting successes when I am looking for the numbers to be above a 3 is easier to roll than when I am looking for the numbers to be above a 4.

August 1st, 2010 at 8:26:01 AM
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You can use the binomial distribution, where n is the number of dice, p is the probability of success with one die (rolling 8 or higher = success implies p = 0.3), and x (or k) is the number of successes you are looking for.

This will give you your formula for both scenarios 2 and 3. I don't want to reproduce the formula here, just check the CDF for the distribtion at: wikipedia .

You can use the excel function =BINOMDIST( x , n , p , TRUE )

This will give you your formula for both scenarios 2 and 3. I don't want to reproduce the formula here, just check the CDF for the distribtion at: wikipedia .

You can use the excel function =BINOMDIST( x , n , p , TRUE )

Wisdom is the quality that keeps you out of situations where you would otherwise need it