Probability of throwing a 1 or a 2 on two dice

Question

I'm reading the book, Against the Gods: A Remarkable Story of Risk, by Peter L. Bernstein and in the chapter titled "A Renaissance Gambler", he talks about the probability of throwing a 1 or a 2 using two dice.

He explains how intuitively it seems like the probability with two dice should be 2/3 since throwing a 1 or a 2 on just one die is 1/3. However he writes that the probability is actually 5/9 because:

"When throwing two dice, there is one chance out of nine that a 1 or a 2 will come up on both dice on the same throw, but the probability of a 1 or a 2 on either die has already been accounted for; hence we must deduct 1/9 probability from 2/3. Thus, 1/3 + 1/3 - 1/9 = 5/9"

I totally get how throwing a 1 or 2 on two dice is 5/9 (just write down all the possible outcomes) but what I don't understand is when he says that the probability of throwing 1's or 2's on both dice is a 1/9 probability. Shouldn't that be a 1/18 probability? (1/6 * 1/6) + (1/6 * 1/6) = 1/18?

Is there a flaw in my thinking or is this an error in the book?

Answer 1

You get $\frac{1}{6}\frac{1}{6}$ for throwing two ones, the same for throwing two twos, but you also have the option of throwing one on the first dice and two on the other, or the other way around, so you need to discard $4\frac{1}{6}\frac{1}{6}=\frac{1}{9}$

Answer 2

Well, you have four outcomes to account for.

1: No die gets 1 or 2.

2: First die gets a 1 or 2, the other does not.

3: First does not, second does.

4: Both do.

These are all the possible outcomes, and so the probability of each of these sum up to 1.

So what's the probability of option 1?

Well, we have to get 3, 4, 5, or 6 on both dice. This has a $\frac46 \cdot \frac46 = \frac49$ chance of happening.

Now, the opposite of "no die gets 1 or 2" is "at least one die gets 1 or 2", which is the outcome we want to calculate.

The probability of the opposite outcome is $1-\text{(probability of original outcome)}$.

So the probability of getting at least one die with 1 or 2, is $1 - \frac49 = \frac59$.

This could all be written more mathematically proper, but I'm just trying to write in the same tone that the question was written in.