Quick Dice Probability Question
- Thread starter Paul_Klein
- Start date
But I think the 1 in 500 can be done if the confirm is 1 followed by a third roll which is in 1-16.
Paul_Klein
Explorer
I hate you all.
Don't get me wrong, I appreaciate you answering my questions, but your profeciency at math makes me green with envy
Don't get me wrong, I appreaciate you answering my questions, but your profeciency at math makes me green with envy
Herremann the Wise
First Post
Just confirming Zappo's math - 1 followed by a 1 followed by a 1 to 16. (0.0025 x 0.8 = 0.002)
How are you working out your critical fumbles anyway?
Best Regards
Herremann the Wise
How are you working out your critical fumbles anyway?
Best Regards
Herremann the Wise
Paul_Klein
Explorer
It is for Star Wars d20. Each weapon (blaster weapons at least) have a number of shots. Blaster Pistol carries 100, while a Rifle carries 50, etc...
Instead of keeping track of each shot fired (yeah right!), but still wanting someone's power pack coming up empty now and then, I figured I would use critical fumbles. And by using different numbers for each different weapon's stats, I can accuratly simulate how many shots per power pack and the correct probability (per dice roll) where one could wind up empty.
Instead of keeping track of each shot fired (yeah right!), but still wanting someone's power pack coming up empty now and then, I figured I would use critical fumbles. And by using different numbers for each different weapon's stats, I can accuratly simulate how many shots per power pack and the correct probability (per dice roll) where one could wind up empty.
Zappo
Explorer
Keeping track of shots? What I do is...
- when I shoot, put a small tick with a pencil somewhere on the sheet
- at the end of the fight, count the ticks and update the number
- every time you have 5 ticks, strike them, so that you can count them faster
This is fast, accurate and easy. BTW, in Star Wars weapons have so many charges, and charges are so cheap, that I've found they just don't run out in any normal fight, assuming that the characters just remember to change the battery when they have time.
- when I shoot, put a small tick with a pencil somewhere on the sheet
- at the end of the fight, count the ticks and update the number
- every time you have 5 ticks, strike them, so that you can count them faster
This is fast, accurate and easy. BTW, in Star Wars weapons have so many charges, and charges are so cheap, that I've found they just don't run out in any normal fight, assuming that the characters just remember to change the battery when they have time.
S
shurai
Guest
It's a known fact that any real probability can be simulated with any method of random number generation. So, there is a way to use your regular old D&D dice to simulate, say, a d23 or a d47.
I wrote a game system once that utilized this fact.
Mmm, sweet probability. It's actually very intuitive as mathematics goes: it all behaves according to observable phenomena. It's easy too, especially if it's just rolling dice; all you have to do is fit your head around a few basic concepts and then the rest works itself out.
-S
I wrote a game system once that utilized this fact.
Mmm, sweet probability. It's actually very intuitive as mathematics goes: it all behaves according to observable phenomena. It's easy too, especially if it's just rolling dice; all you have to do is fit your head around a few basic concepts and then the rest works itself out.
-S
This is a very good question:
How to simulate the odds of an outcome by rolling dice. If one-in-ten, throw a d10. If one-in-500, throw a few dice, such that the s/r for every die multiplied together = 500. Here, r = # acceptable die results and s = # of sides. So in this example, 20/1*20/1*20/16 = 500.
This approach fails as an exact method when the # for the one-in-# odds you're trying to simulate is prime (cannot be represented as a product of smaller positive integers), and > 20.
Why? Quick proof: For n dice,
let the product of s1/r1*s2/r2*...*sn/rn = S/R. S and R must both be positive integers since we're talking dice here. Now require that S/R = target prime number. Then S = R*(target prime number).
Forget about what R meant originally (the product of acceptable rolls for all dice rolled)--just know that it's a positive integer. Do recall what S means: the product of die-#-of-sides, for all dice rolled. But we're saying here that S has factors of R and the target prime number. R has more factors beyond r1, r2, ... rn, unless these are all prime.
The target prime #, however, cannot be factored further by definition. And since the target prime >20, it cannot be represented as a D&D die. Ergo, you'd need to roll at least one die with the target prime number of sides to simulate your odds.
In other words, for a 1-in-47, you really would need to roll at least one d47.
Harder to prove, but still true (I think), is that the requirement of target prime > 20 is too stringent. Really, it only needs to be greater than 5, the highest prime factor of #-of-sides among the D&D dice: d20, d12, d10, d8, d6, & d4.
This can be extended to cases where the odds against (not in the true parlance of odds, where 1:47 means 1/48) are multiples of non-prime and prime number(s), where at least one prime is not a factor of 20, 12, 10, 8, 6, or 4.
However, rolling a 1-in-47 can be approximated with D&D dice, and I suspect this is what shurai means by "simulated". Furthermore any good random # generator can get at this very simply. (Of course, there is no such thing as a perfect random # generator--but then, are the dice we roll perfect either?)
Details.
What's a little harder is this:
Really there are two outcomes for a 1 in 500 odds roll. Either you make your 1, or you don't. Success=1st outcome. Failure=2nd outcome. There are of course many more ways of failing.
But what if the odds are slightly better? Say, 3 in 500? Three is not a factor of 500, so the simplest expression of this ratio is 3/500 (or 3:497). How do you simulate that with dice?
Answer: the solution must satisfy s1/r1*s2/r2*...*sn/rn = 500/3
Is this correct? Seems so. One solution:
If you roll a 1 with d20, roll again. If 6 or less, roll again. If 8 or less, congratulations, you just made a lucky 3 in 500 shot.
20/1*20/6*20/8 = 500/3.
Lastly, I'll link a somewhat related thread because it was so satisfying. In it the question is asked: what is the probability of rolling a certain outcome (sum of dice), from n dice with s sides? Or more generally, an outcome from n1 dice with s1 sides + n2 dice with s2 sides + ...
Yikes.
probability of dice
How to simulate the odds of an outcome by rolling dice. If one-in-ten, throw a d10. If one-in-500, throw a few dice, such that the s/r for every die multiplied together = 500. Here, r = # acceptable die results and s = # of sides. So in this example, 20/1*20/1*20/16 = 500.
This approach fails as an exact method when the # for the one-in-# odds you're trying to simulate is prime (cannot be represented as a product of smaller positive integers), and > 20.
Why? Quick proof: For n dice,
let the product of s1/r1*s2/r2*...*sn/rn = S/R. S and R must both be positive integers since we're talking dice here. Now require that S/R = target prime number. Then S = R*(target prime number).
Forget about what R meant originally (the product of acceptable rolls for all dice rolled)--just know that it's a positive integer. Do recall what S means: the product of die-#-of-sides, for all dice rolled. But we're saying here that S has factors of R and the target prime number. R has more factors beyond r1, r2, ... rn, unless these are all prime.
The target prime #, however, cannot be factored further by definition. And since the target prime >20, it cannot be represented as a D&D die. Ergo, you'd need to roll at least one die with the target prime number of sides to simulate your odds.
In other words, for a 1-in-47, you really would need to roll at least one d47.
Harder to prove, but still true (I think), is that the requirement of target prime > 20 is too stringent. Really, it only needs to be greater than 5, the highest prime factor of #-of-sides among the D&D dice: d20, d12, d10, d8, d6, & d4.
This can be extended to cases where the odds against (not in the true parlance of odds, where 1:47 means 1/48) are multiples of non-prime and prime number(s), where at least one prime is not a factor of 20, 12, 10, 8, 6, or 4.
However, rolling a 1-in-47 can be approximated with D&D dice, and I suspect this is what shurai means by "simulated". Furthermore any good random # generator can get at this very simply. (Of course, there is no such thing as a perfect random # generator--but then, are the dice we roll perfect either?)
Details.
What's a little harder is this:
Really there are two outcomes for a 1 in 500 odds roll. Either you make your 1, or you don't. Success=1st outcome. Failure=2nd outcome. There are of course many more ways of failing.
But what if the odds are slightly better? Say, 3 in 500? Three is not a factor of 500, so the simplest expression of this ratio is 3/500 (or 3:497). How do you simulate that with dice?
Answer: the solution must satisfy s1/r1*s2/r2*...*sn/rn = 500/3
Is this correct? Seems so. One solution:
If you roll a 1 with d20, roll again. If 6 or less, roll again. If 8 or less, congratulations, you just made a lucky 3 in 500 shot.
20/1*20/6*20/8 = 500/3.
Lastly, I'll link a somewhat related thread because it was so satisfying. In it the question is asked: what is the probability of rolling a certain outcome (sum of dice), from n dice with s sides? Or more generally, an outcome from n1 dice with s1 sides + n2 dice with s2 sides + ...
Yikes.
probability of dice
Last edited:
Herremann the Wise
First Post
frisbeet,
Possible solution for your 1d47 problem:
I can understand where you are coming from. However the solution is simple yet perfect.
To roll a 1d47, roll a perfect, non-bias 1d100. If it's between 1 and 47, that's your number, if not re-roll until you get a result. There is a perfectly distributed chance of getting any number between 1 and 47.
However, because you don't want to have to re-roll most of the time, take 1d60 (A 1d6 and a 1d10) and use the same method (assume that the 6 on the 1d6 is a 0).
A somewhat crass solution but one that is perfect. Is this what you were referring to as approximated? You obviously know your stuff when it comes to probability and I'm sure this idea is something that you've already thought of. If I'm barking up the wrong tree (you only want to roll once?), please tell me.
As for the 3 in 500, make it 6 in 1000 and roll 1d1000.
Just going across to your link now, it should be interesting.
Best Regards
Herremann the Wise
Possible solution for your 1d47 problem:
I can understand where you are coming from. However the solution is simple yet perfect.
To roll a 1d47, roll a perfect, non-bias 1d100. If it's between 1 and 47, that's your number, if not re-roll until you get a result. There is a perfectly distributed chance of getting any number between 1 and 47.
However, because you don't want to have to re-roll most of the time, take 1d60 (A 1d6 and a 1d10) and use the same method (assume that the 6 on the 1d6 is a 0).
A somewhat crass solution but one that is perfect. Is this what you were referring to as approximated? You obviously know your stuff when it comes to probability and I'm sure this idea is something that you've already thought of. If I'm barking up the wrong tree (you only want to roll once?), please tell me.
As for the 3 in 500, make it 6 in 1000 and roll 1d1000.
Just going across to your link now, it should be interesting.
Best Regards
Herremann the Wise
