I'm designing a game (non-professionally) and now I need to calculate probabilities to see how the statistics work out.
But this got me a bit stumped.
How do you calculate how many dice you need to roll before the probability threshold of rolling one of each side is reached?
For example, how many 6-sided dice do one need to roll before the probability is >50% of rolling each of 1 through to 6?
Then >75%?
My gut feeling is around 13 and 20 respectively, but for design purposes it would be nice to be able to actually formulate and calculate this.
Math: probability of rolling one of each using 6-sided dice
Moderator: Alyrium Denryle
Re: Math: probability of rolling one of each using 6-sided d
This (Warning:PDF) is your friend, specifically the answer to question 5 on page 10. Short summary: 13 for 50% is about right, for 75% you're fine with a bit less, namely 18.Spoonist wrote:How do you calculate how many dice you need to roll before the probability threshold of rolling one of each side is reached?
For example, how many 6-sided dice do one need to roll before the probability is >50% of rolling each of 1 through to 6?
Then >75%?
My gut feeling is around 13 and 20 respectively, but for design purposes it would be nice to be able to actually formulate and calculate this.
Re: Math: probability of rolling one of each using 6-sided d
Your example:
Your first roll has a uniqueness probability of [1]
Your second has a uniqueness probability of [1 - 1/6 = 5/36] (there is a 1/6 probability you will roll die 1's result)
Your third roll has a uniqueness probability of [1 - 1/6 - 5/36 = 25/36] (there is a 1/6 probability you will roll die 1's result, a 5/6 probability that die 2's result is unique, and a 1/6 probability of rolling that unique roll, for a combined probability of those two probabilities multiplied together that you will roll die 2's unique result)
your nth roll has a uniqueness probability of [1 - {SUM FROM 1 TO n-1}[(1/6)*P_n]
This summation is probably something that you can turn into an empirical formula, but that sounds an awful lot like work to me, so I'd just use a computer to do it.
To get the probability that you will have six unique results, I'm not sure how to go about it. The probability out of fewer than 6 is 0, obviously. The probability out of six is the product of the uniqueness probabilities from 1 to 6. The probability out of n is the uniqueness probability of the nth, multiplied by the probability that five out of six are unique already. Where do we go from here?
Your first roll has a uniqueness probability of [1]
Your second has a uniqueness probability of [1 - 1/6 = 5/36] (there is a 1/6 probability you will roll die 1's result)
Your third roll has a uniqueness probability of [1 - 1/6 - 5/36 = 25/36] (there is a 1/6 probability you will roll die 1's result, a 5/6 probability that die 2's result is unique, and a 1/6 probability of rolling that unique roll, for a combined probability of those two probabilities multiplied together that you will roll die 2's unique result)
your nth roll has a uniqueness probability of [1 - {SUM FROM 1 TO n-1}[(1/6)*P_n]
This summation is probably something that you can turn into an empirical formula, but that sounds an awful lot like work to me, so I'd just use a computer to do it.
To get the probability that you will have six unique results, I'm not sure how to go about it. The probability out of fewer than 6 is 0, obviously. The probability out of six is the product of the uniqueness probabilities from 1 to 6. The probability out of n is the uniqueness probability of the nth, multiplied by the probability that five out of six are unique already. Where do we go from here?
Re: Math: probability of rolling one of each using 6-sided d
Sweeeeeet.M wrote:This (Warning:PDF) is your friend, specifically the answer to question 5 on page 10. Short summary: 13 for 50% is about right, for 75% you're fine with a bit less, namely 18.
Thanks a million, bookmarked and saved.