Probability when two dice are rolled and the minimum of a rolled number is two

Question

I am trying to figure out the probability when two dice are rolled and the minimum of a rolled number is equal to two.

So, $x_1$ is the outcome of dice 1 and $x_2$ is the outcome of dice 2. And $y= \min(x_1,x_2).$ What is the probability that $y=2$?

So the answer I came up with is $\frac8{36}=0.2222222$, since there are only $8$ cases where $2$ is the minimum value. It seems that I should be getting a value over $0.25$. What am I doing wrong? Thanks in advance!

Answer 1

There are actually 9 cases where the minimum is two:

1. $(2, 3)$
2. $(2, 4)$
3. $(2, 5)$
4. $(2, 6)$
5. $(3, 2)$
6. $(4, 2)$
7. $(5, 2)$
8. $(6, 2)$
9. $\mathbf{(2, 2)}$

I'm not sure why you expect to get an answer over $0.25$, but the correct answer is $9/36 = 0.25$.

Answer 2

It appears that you’ve overlooked the case that both dice came up $2$. However, you can compute this probability without explicitly counting possible results. The probability that the minimum of two die rolls is equal to $2$ is equal to the probability that both rolls are at least $2$ less the probability that they are both at least $3$, namely $$\left(\frac56\right)^2-\left(\frac46\right)^2 = \frac 14.$$ This same reasoning applies to any number of dice being rolled, for which enumerating outcomes explicitly could be impractical, and to any target minimum value.