Rolling $2$ dice: NOT using $36$ as the base?

Solution 1:

The key point is that if you distinguish the two dice all $36$ possibilities are equally likely. That is the only thing that allows you to convert number of possibilities to probability.

If you don't distinguish the two dice then there are only $21$ possibilities, but some of them are more likely than others -- and this issue gets more complicated the more dice you have. So knowing that there are $21$ possibilities doesn't give you a probability of $1/21$.

To extend your approach to a point where it more clearly doesn't work, suppose I change all the numbers other than $1$ to $2$s (so each die has $1,2,2,2,2,2$ on it). This clearly doesn't affect the probability of double-$1$, but now there are only three possibilities...

Solution 2:

Paint the dice different colors, say red and blue. Now (red 1, blue 2) is clearly different from (red 2, blue 1). But the dice don't know they're painted, because they're dice.

Solution 3:

Because to get $(1,1)$, both dice must show a $1$. To get a $1$ and a $2$, it could be either $(1,2)$ or $(2,1)$.

Here's another way to look at it ...

The probability of getting $(1,1)$ is $$\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}$$ Explanation: The dice are independent and each die has probability $\large{\frac{1}{6}}$ of showing a $1$.

To get a $1$ and a $2$, the first die must show either $1$ or $2$, and the second die must show whichever of $1,2$ did not show on the first die. Hence the probability of getting a $1$ and a $2$ is $$\frac{2}{6}\times\frac{1}{6} = \frac{2}{36}$$

Solution 4:

In order to throw (1, 1), the values of both dice (let's call them A and B) need to equal 1. Since these events are statistically independent of one another:

$$Prob[(1, 1)] = Prob[A = 1\,and\,B = 1] = Prob[A=1] Prob[B=1] = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$

If you specify that $(1, 2) = (2, 1)$, it essentially does not make a difference whether $A=1$ and $B=2$, or $A=2$ and $B=1$. In this case, the probability of throwing (1, 2) comes down to:

$$Prob[(1, 2)] = Prob[A = 1\,and\,B = 2] + Prob[A = 2\,and\,B = 1] = \frac{1}{36} + \frac{1}{36} = \frac{2}{36}$$

As you can see, the events $(1, 1)$ and $(1, 2)$ are not equiprobable: the former has a probability of 1/36, while the latter has a probability of 1/18.

There are six types of throws with equal numbers $(1, 1), (2, 2), \ldots, (6, 6)$ and fifteen types of throws with different numbers $(1, 2), (1, 3), \ldots, (5, 6)$. Since these events cover the total sample space, the sum of the probabilities equals:

$$6 \cdot \frac{1}{36} + 15 \cdot \frac{2}{36} = \frac{36}{36} = 1$$