While at first glance your reasoning seems to make sense ("green faces are more likely .. so the more green faces, the more likely that is"), it can easily be shown that there must be something fishy with it.
Take it to its logical conclusion: suppose you throw that dice $1000$ times .. which is the most likely outcome? You can get $0$ green faces, .. or $1$, or $2$ ... all the way up to $1000$. According to your reasoning, out of these $1001$ possible outcomes, getting all $1000$ green faces is the most likely.
OK, but think about it ... not getting any red face?!? That's bordering on absolutely incredible! Getting a red face is not that much more unlikely ... indeed, what if you had a coin that was slightly biased (say, $50.1$%) towards heads, and you flip it $1000$ times ... don't you think it would be crazy to think that the most likely outcome is to get all heads?!
Now, in your case, we have $\frac{2}{3}$ vs $\frac{1}{3}$ ... so sure, there is a bias towards getting green faces ... but again not all that great of a bias ... and indeed getting all $1000$ green faces when flipping $1000$ times should be really, really unlikely.
Indeed, just using your common sense, the probability of getting $1000$ green faces is certainly way smaller than $1$ in $1001$, and given that there are $1001$ possible outcomes, it should be clear that the most likely outcome is in fact not the outcome of getting all $1000$ times a green face when flipping that coin $1000$ times.
OK, so your reasoning does not work. But why not? Where does your reasoning go wrong?
Well, you forget about the fact that there is only and exactly $1$ way to get all green faces: you need to get a green face every time! However, there are $1000$ ways to get $999$ green faces, and $1$ red face, as the red face can be the first throw, or the second, or the third .... So, already you can see that getting $999$ green faces and $1$ red face would be far more likely than getting $1000$ green faces.
So far the intuitions. Mathematically what is going is this. Look at the binomial formula:
$$P(X=N)={M \choose N} \cdot p^N \cdot (1-p)^{N-M}$$
OK, so sure, getting a green face is more likely than getting a red face, i.e. $p > 1-p$, and so if you just look at the part:
$$p^N \cdot (1-p)^{N-M}$$
then that indeed will be higher, the higher $X$ is, and indeed will be highest when $X=M$
However, this does not mean that you are most likely to get 1000 green faces, because you forget about the other part
$${M \choose N}$$
which is the number of ways to get the outcome. And again, for $X=M$, there is only one way ... but for smaller $X$'s there can be many more ways.
So, you get an interesting interplay: yes, the higher $X$ is, the higher the right part of the formula, but if $X$ gets too high, then the left part will shrink.
Now, if you do the math, it turns out that the most likely outcome is an outcome that reflects the probabilities $p$ and $1-p$, i.e. about two thirds green and one third red. So, options A and D are a bit 'out of whack' in their respective proportions, and it would have to be between B and C ... which one? The explanation provided gives the answer
Best Answer
Well, A throws it more often than B in the following cases:
B throws 1, A throws 2,3,4
B throws 2, A throws 3,4
B throws 3, A throws 4
And now you go and check for the all-together probability of these events. For example:
B throws 1, A throws 2,3,4:
$\mathbb{P}(B=1) = \frac{2}{6}$ and $\mathbb{P}(A\in \{2,3,4\}) = \mathbb{P}(A \neq 1) = \frac{4}{6}$. And finally $\mathbb{P}(B=1 \cap A \neq 1) = \mathbb{P}(B=1) \cdot \mathbb{P}(A \neq 1) = \frac{2}{9}$. Note that I used that the events are independent.