PDA

View Full Version : Science Fun with Dice - Statistics

Yora
2014-05-18, 03:27 PM
There's a thread in the roleplaying forum, but instead of spreading out my questions all over the place, I want to start a new one here.

For my current D&D campaign, we created characters by rolling 2d6+6 six times for each characters ability scores.
One character got to 8s, while another one got two 17s and one 16. Now I was wondering what the chances for both of these results are.
8 is the lowest possible number, so the probability for a single 8 is simply 2.78%.
For the high stats, I also want to include possible even better scores, so that means the chance for a "17 or higher" is 8.33% and for "16 or higher" is 16.67%. Getting even more 8s or 16s is not given any relevance for this question. Those cases are to be included in the end results.

Now how would I calculate the chances for both results? ([8, 8, x, x, x, x]; [17+, 17+, 16+, x, x, x])

warty goblin
2014-05-18, 03:44 PM
Each 2d6+6 roll is independent of the others, so the joint distribution is just the product of the marginals; that is

P(X1 = x1, X2 = x2, ... X6 = x6) = P(X1 = x1)*P(X2 = x2)*....*P(X6 = x6)

Where X1 is a random variable corresponding to the value of the first 2d6+6 roll, and so forth.

Now for the values you aren't specifying, you are essentially just saying 'this roll is anything', which has probability one. So

P(X1 = 8, X2 = 8, X3 = {any possible value}, ..., X6 = {any possible value})
= P(X1 = 8)*P(X2 = 8)*P(X3 = {any possible value})*...*P(X6 = {any possible value})
= P(X1 = 8)*P(X2 = 8)*1*1*...*1
= P(X1 = 8)^2 (since the distributions are identical)
= 0.00077284

The probability of the second set of scores can be found using identical logic. The probability of the two sets of character statistics is just the product of these probabilities. It will be exceedingly low.

Alternatively, and identically, just find the product of the probabilities of the events you are interested in. Adding the additional {any possible value} events does not change the answer, since all the rolls are independent, and hence their presence in the joint distribution is irrelevant.

(This is also a probability question, not a statistical one. Pretty much nobody does dice statistics)

ChristianSt
2014-05-18, 05:17 PM
Each 2d6+6 roll is independent of the others, so the joint distribution is just the product of the marginals; that is

P(X1 = x1, X2 = x2, ... X6 = x6) = P(X1 = x1)*P(X2 = x2)*....*P(X6 = x6)

Where X1 is a random variable corresponding to the value of the first 2d6+6 roll, and so forth.

Now for the values you aren't specifying, you are essentially just saying 'this roll is anything', which has probability one. So

P(X1 = 8, X2 = 8, X3 = {any possible value}, ..., X6 = {any possible value})
= P(X1 = 8)*P(X2 = 8)*P(X3 = {any possible value})*...*P(X6 = {any possible value})
= P(X1 = 8)*P(X2 = 8)*1*1*...*1
= P(X1 = 8)^2 (since the distributions are identical)
= 0.00077284

The probability of the second set of scores can be found using identical logic. The probability of the two sets of character statistics is just the product of these probabilities. It will be exceedingly low.

Alternatively, and identically, just find the product of the probabilities of the events you are interested in. Adding the additional {any possible value} events does not change the answer, since all the rolls are independent, and hence their presence in the joint distribution is irrelevant.

(This is also a probability question, not a statistical one. Pretty much nobody does dice statistics)

It is kinda late here and I'm on my phone, so I'm not giving a long answer, but I wanted only to say that this is wrong.

You need to factor in how you can rearrange the stats. Using the approach you use, the probability for having two 8's wouldn't depend on the number of stats which is just false.

warty goblin
2014-05-18, 06:46 PM
It is kinda late here and I'm on my phone, so I'm not giving a long answer, but I wanted only to say that this is wrong.

You need to factor in how you can rearrange the stats. Using the approach you use, the probability for having two 8's wouldn't depend on the number of stats which is just false.

That depends on the exact question being asked. My answer is correct for the first and second elements being 8, irrespective of the values of the other rolls. As you point out, it is incorrect for the total number of eights rolled. The difference being that what I gave is an answer about the probability of a joint distribution taking a value in a particular set in a six-dimensional space, the second is a question about the value of a sum of indicators of values of a finite sequence of random variables.

So if the OP is in fact interested not in the probability of rolling two eights, followed by four whatevers, but the probability of rolling two eights out of six rolls, the answer is as follows.

For the case of two eights, we can simply use a binomial distribution with n = 6 and p = P(2d6+6 = 8). Note that this will be the probability of exactly two 8s, not a result with at least two eights. Get that just by doing 1 - (P(two 8s) + P(one 8) + P(zero 8s))

The second case is more complicated, and can be done as follows: We have six pairs of dice. Write (n, k) for 'n choose k'. Then we have (6, 2) 17+ and (4, 1) 16+, and (3, 3) 15 or less results. There is 1 way to get an 18, 2 to get 17 and 3 to get 16, so getting at least 17 can happen 3 ways and at least 16 can happen 6 ways. Getting 15 or less can happen 30 ways. The numerator is then [(6,2)*(3)*(4,1)*(6)*(3,3)*(30)]. The denominator is all possible values for six pairs of dice, which is (36)^5. Divide and there's your answer.

As before, the probability of both occurring is simply the product of the probabilities.

Razanir
2014-05-18, 08:22 PM
The first one is easy. We're just looking at a binomial distribution, which measures the number of successes in a set of independent trials. Here, we have 6 trials and the probability of a success (rolling an 8, as horrible of a score as that actually is) is 1/36. However, instead of calculating the probability of 2-6 successes, I'll find the probability of 0-1 successes and subtract it from 1. (Just because it's a slightly easier calculation)

P(No score is an 8) = (35/36)^6 = .8445
P(1 score is an 8) = 6*(1/36)*(35/36)^5 = .1448

P(2 or more scores are 8) = .0107

The second one is more complicated, because we have 3 cases. 8-15, 16, and 17-18 (probabilities are 5/6, 1/12, 1/12). This time I won't bother going through all the calculations, and just explain what I'm doing. We want P(2 or more are 17-18, and no more than 3 are 8-15). However, there are a lot of cases for that, so I'll work around it.

1 - P(0-1 are 17-18) - P(4-6 are 8-15) + P(0-1 are 17-18 and 4-6 are 8-15)

Now I have two binomial probabilities (much friendlier to calculate), and even though I still have a multinomial, it's at least much fewer cases.

1 - .9169 - .9377 + .8875 = .0329 (possibly with rounding error, and assuming all my math was right)