What is the probability you will have more low than high numbers AND more odd than even numbers if a 6 sided dice is thrown many times? I recently encountered the below question:

You roll a fair die a very large number of times. What is the probability you will have more low (123) than high (456) numbers AND ALSO have more odd (135) than even (246) numbers?

I assumed that P(low number) = P(high number) = P(even number) = P(odd number)

P(low number) and all the other ones mentioned above would normally be equal to 1/2.

However, the question above is asking for "more" numbers on one side. I don't get how to go about it from here. How do you calculate the probability for more on one side? Would it just be (1/2)^2 = 1/4? How would you attack this question?
P.S I think it is somewhere between 1/3 and 1/4.
Thanks a lot.
 A: Let $N13$ be the number of 1's and 3's added up, $N2$ be the number of 2's, $N46$ be the number of 4's and 6's and let $N5$ be the number of 5s.
The event we are looking at is that simultaneously
$$N13 + N2 > N46 + N5$$
and
$$N13 + N5 > N46 + N2$$
You see that whenever $N13 = N46$ the event will not occur because then $N2$ would have to be simultaneously less than and more than $N5$. Thinking a little bit harder we see that the event can also not occur when $N13 < N46$. All in all we find that the event of interest can be rewritten as:
$$|N2 - N5| < N13 - N46$$
(note that the left hand side has absolute bars and the right hand side does not).
Now how to get a grip on the probability distributions of these quantities?
$N2$ follows a binomial distribution with expected value $N/6$ (where $N$ is the total number of throws) but for large $N$ we can we can approximate it with a normal distribution with mean $N/6$ and variance $(5/36)N$, see Wikipedia.
The same holds for $N5$ and so the quantity
$$X := N2 - N5$$
(without absolute bars) follows (approximately) a normal distribution with expected value 0 and variance $(10/36)N = (5/18)N$.
Similarly the distributions of N13 and N46 can be approximated by normal distributions with expectation (1/3)N and variance $(2/9)N$ so that
$$Y := N13 - N46$$
approximately follows a normal distribution with expectation $0$ and variance $(4/9)N = (8/18)N$.
So we are left with a new question:

Suppose $X$ is normally distributed with mean $0$ and variance $V$ and $Y$ is normally distributed with mean $0$ and variance $(8/5)V$ what is the probability that $|X| < Y$?

Maybe I'll edit in some answer to this new question also, but perhaps you can already solve it yourself.
Remarks:

*

*the normal approximation is of course just that: an approximation that gets better as $N$ gets larger. One could also obtain an exact solution that would depend on $N$ in a complicated way, hopefully someone else will take the time to write that out.


*I think we can assume that $X$ and $Y$ are independent but this is not completely obvious since $N13, N2, N46$ and $N5$ are not independent: given any three of them we can compute the fourth by subtracting the other three from $N$. So if $X$ and $Y$ are indeed independent a separate argument for that is needed and if they are not, the new question suddenly becomes a lot harder. I will think about this a bit more.
