Roll a die 3N times, win 2 euros if 5 or 6, lose 1 euro else. What is the probability that you leave with the same amount of money?

Question

We roll a die 3N times.
If it falls on 5,6 $\rightarrow$ + 2 \$
If it falls on 1,2,3,4 $\rightarrow$ - 1 \$

So we have $\frac{1}{3}$ chance to win 2 \$ and $\frac{2}{3}$ to lose 1 \$.

What is the probability that at the end of the session, you leave with the same amount of money ? Find the result by hand using Stirling's formula.

I said that it was $P(N \ \text{wins} \ | \ 2N \ \text{losses})$ which is equal to $$\frac{1}{6^{3N}} \cdot \binom{3N}{2N} = \frac{(3N)!}{6^{3N} \cdot(2N)! \cdot N!}$$
And then using Stirling's formula $N! = \sqrt{2 \pi N} \cdot (\frac{N}{e})^{N}$

I found $$\frac{1}{32^{N}} \cdot \sqrt{\frac{3}{4 \pi N}}$$
But I just can't compute $\frac{1}{32^{N}}$. Where am I wrong ? (Probably how I looked at the problem)

Answer 1

The $\frac{1}{6^{3N}}$ term is wrong. The probability of a win is 1/3, a loss 2/3. So the probability of a particular sequence of $N$ wins and $2N$ losses is $(1/3)^N(2/3)^{2N}$

Answer 2

$P(A)=\binom {3N}{2N}\cdot (\frac{1}{3})^N\cdot (\frac{2}{3})^{2N}=\frac{2^{2N}\cdot (3N)!}{(2N)!\cdot N!\cdot 3^{3N}}$ and then you should use Stirling's formula.