Calclulating Probability of a Distribution of a set of Two-Dice rolls

Question

I was calculating the odds of a set of dice rolls in a game of Catan recently. In 35 rolls, we rolled 0 3's, and 0 5's.

To calculate the probability of this, I calculated the probability of rolling 0 3's in 35 rolls, and multiplied by the probability of rolling 0 5's in 35 rolls:

P(3)^0 * P(not 3)^35 * P(5)^0 * P(not 5)^35 = (2/36)^0 * (34/36)^35 * (4/36)^0 * (32/36)^35 = 2.19 E -3

But then I thought: this calculated probability should equal the probability of rolling a "not 3 or 5", 35 times:

P(not 3 or 5)^35 = (30/36)^35 = 1.69 E -3

These two different approaches do not yield the same answer. I've struggled to reason through which approach is correct. I've thought that the answers don't agree because "zero" is the roll; that it makes sense to say "roll two (3 or 5)'s in ten rolls" (in which case you'd use the second formula) but, if talking about "zero", must you say zero 3's AND 5's, and use the first formula?

A commenter on another site says that "rolling 0 3's makes a 5 more likely, thus rolling a 3 and rolling a 5 are not independent". I am not sure that I buy that answer; rolling a 3 has no bearing on rolling a 5, thus how could they not be independent? I can roll 100 3's in a row, or none, and that has no bearing on whether my next roll will be a 5 or not, right? If someone could flesh out the concept a little more, I'd be very grateful- after all, Catan-bragging-rights are on the line.

Answer 1

Your first answer is "double counting" rolls. It is computing the probability of rolling 35 times with no 3's, followed by another 35 rolls with no 5's. (The sum of the exponents is 70, the total number of rolls.)

As far as

rolling a 3 has no bearing on rolling a 5, thus how could they not be independent?

This is true if you have two different rolls. But a single roll cannot be both 3 and 5. This may be what the other commenter meant.

Answer 2

Each session of a game involving repeated rolling of two dice will have its own eccentricities: many more 3's than 11's (theoretically, equally likely), more 4's than 6's (where 6's are expected more often), and so on ad infinitum. There is no point in assessing the the chances of details of individual outcomes; each game outcome has a small chance of occurrence. That's partly what makes games of chance interesting to play.

More meaningful probability computations might be to ask for probabilites of particular kinds of occurrences for such games in general. Here are two examples:

• (1) What is the probability of getting no dice rolls less than 6 in 35 rolls of a pair of fair dice?

• (2) The theoretical mean score on one roll of two dice is $\mu=7.$ What is the probability my average score in 35 rolls is less then 6.

(1) For any one roll of two dice, $P(X < 6) = P(X \le 5) = 10/36 = 5/18.$ The probability all 35 rolls will show less than 4 spots is very small $(5/18)^35 = 9.399613e-21.$

(2) The mean score on any one roll of a pair of fair dice is $\mu = 7$ with a variance of $\sigma^2=5.8333$ and standard deviation of $\sigma=2.4152.$

p = c(1,2,3,4,5,6,5,4,3,2,1)/36; x = 2:12
mu = sum(p*x);  mu
[1] 7
vr = sum(p*x^2) - mu^2;  vr
[1] 5.833333
sd = sqrt(vr);  sd
[1] 2.415229

Then the average score $\bar X$ on 35 rolls is $\mu_{\bar X} = 7$ with standard deviation $\sigma_{\bar X} = \sigma/\sqrt{n} = 2.415229/\sqrt{35} = 0.4082483.$ Furthermore, $\bar X$ is approximately normal with this mean and standard deviation. So the desired probability is $P(\bar X < 6) \approx 0.00715.$

pnorm(6, 7, 0.40825)
[1] 0.007153143