Combinatorics and Probability with Dice

Question

In how many ways will the sum of the number of 1’s and 2’s (total of occurrences of 1's and 2's) equal the sum of the number of 3’s, 4’s, 5’s and 6’s, after n rolls, assuming n = 2i for some positive integer i? What is the probability of this occurring?

Answer 1

$n=2i$, where $i \in \Bbb N$, so we want $i$ occurrences of $1$'s and $2$'s, and $i$ occurrences of $3$'s, $4$'s, $5$'s and $6$'s.

For simplicity of notation, denote the number of occurrences of a number $x$ by $n_x$.

Lets do the $1$'s and $2$'s first.

$n_1$ can be anything from $0$ to $i$ (giving $i+1$ choices), but whatever the value of $n_1$, it must be that $n_2=i-n_1$. i.e. we have $i+1$ choices for the $1$'s and $2$'s.

Now lets look at the $3$'s, $4$'s, $5$'s and $6$'s.

Again, $n_3$ can be anything from $0$ to $i$ (giving $i+1$ choices). But this time, we have more choices for $n_4$ as it can go from $0$ to $i-n_3$. Similarly, once we have decided on $n_4$, $n_5$ can only go from $0$ to $i-n_3-n_4$. Finally, $n_6$ must be whatever is left, so $n_6=i-n_3-n_4-n_5$. Written out mathematically, the number of choices here is $$\sum_{n_3=0}^{i}\sum_{n_4=0}^{i-n_3}\sum_{n_5=0}^{i-n_3-n_4}1=\frac{(i+3)(i+2)(i+1)}{6}$$

So combining them together, we get a total of

$$(i+1)\frac{(i+3)(i+2)(i+1)}{6} $$ ways.

EDIT

OK, perhaps this way of going about it is pretty complicated. In fact, there is a general formula for choosing a total of $r$ items out of $n$ different types of items without heed to order, namely $$H^n_r=C^{n+r-1}_{r}$$

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As for the second part of the question, we really only care how many ${1,2}$ and ${3,4,5,6}$ occur. A ${1,2}$ has probability of $\frac 13$ and a ${3,4,5,6}$ has probability $\frac 23$. We want $i$ occurrences of each, so by the binomial formula, the probability is $$C^{2i}_i\biggl(\frac 13\biggr)^i\biggl(\frac 23\biggr)^i$$