# probability of rolling two dice: independent or not?

In researching a probability question, I found my answer in this old Math stackexchange question here: If I roll two fair dice, the probability that I would get at least one 6 would be....

However, the answers posted assume that the dice rolls are independent events so that the probabilities can be multiplied.

I was just curious if it makes a difference in the calculation if: the dice are rolled at the same time, or if they are rolled one after another?

Thanks.

• It should not matter if they rolled in two different cities and centuries as long as they are identical dice, independent rolls and are unaffected by any external forces unique to the location / time. – user1952500 Jul 4 '17 at 17:30
• Hi. okay, that's what I thought! Thanks. – Thomas Moore Jul 4 '17 at 17:32

There's no difference between the two procedures (throwing simultaneously or one-by-one) as far as independence is concerned. To see this you might go back to the formal definition of independence.

Definition. (Statistical) independence of two events.

Two events $A$ and $B$ are independent if their joint probability equals the product of their respective probabilities, i.e. $$P(A \cap B)=P(A)P(B)$$

As we see, it's NOT like independence (statistical independence) implies the product rule, rather the concept of statistical independence is defined by the product rule.

It is easy to check from the joint probability distribution that throwing of two dices are statistically independent.

Assume that the die is fair $($i.e. each of the sides come up with equal probability of $\frac{1}{6})$. If we define $X$ to be the number we get from the $1$st die and $Y$ to be the same from $2$nd die, then $$P(X=i, Y=j)=\frac{1}{36}=\frac{1}{6} \cdot \frac{1}{6} = P(X=i)P(X=j),$$ for $i=1(1)6, ~j=1(1)6$.

We can use this to compute $P(X \in A, ~Y \in B)$ which comes out to be $P(X \in A)P(Y \in B)$.

Now check! Do you think that the joint distribution of $(X, Y)$ changes because the dice are thrown together or one-by-one? In fact, after the experiment is done, and the outcome is attached to a random variable which has a known probability distribution, does the physical (not mathematical!) procedure of the experiment matter at all?

Lastly, I should (informally) say that, in a discussion of statistics and probability, we only care about statistical independence, which is well-defined. In Philosophy, the notion of independence can be way too complicated for mathematicians to handle, so we don't bother about that!

Let's say you roll the first die and get 1. Then you expect that now the probability is 1/6. But 1/6 it's not the probability to get at least a 6 on 2 rolls. It is rather the Probability of getting at least a 6 on 2 rolls given that the first roll is not a 6. So it's like you forget about the first target (precisely memorylessness, if you know what I mean) and continue the experiment rolling the remaining die.

So if you try the option of rolling the first die and then the second, the thing that may change this probability is only the fact that you look at the first roll and realize the outcome.

Let's say you close your eyes and roll the first die. Nobody tells you the result, and keeping your eyes closed you make the second roll. Then you open your eyes and then realize the result of both dice; redo the experiment. This time you roll both dice and see the result. Even now you realize the result of both dice at the same time. Both experiment are equivalent in this case, and probability would be 11/36.

Hope it helps