Probability for getting -1 in a game of rolling die In a game, a fair die is rolled. If the result is 1 or 2, you can get 2 points. If the result is 3 or 4, you get -1 points. If the result is 5 or 6, you get 0 points. Now you can roll the die for 6 times, what is the probability that your final score is -1. The answer provided is $\frac {22}{243}$.
I can only think of 2 possible situations, which are getting 3/4 once and 5/6 for the rest, and getting 1/2 once, three 3/4 and two 5/6, so that the final score is -1. But my answer is wrong. 
 A: There are $3$ values that can come up on each roll so there are $3^6 = 729$ total rolls you can get. 
How many ways are there to get $-1$?
You can have one $-1$ and the rest $0$. Since the $-1$ can occur on any of the six rolls there are $6$ ways to get that. 
If you have two $-1$s it is impossible to get a final score of $-1$ because there is no way to get $1$. 
If you have three $-1$s you can get a score of $-1$ by rolling a $2$ on another roll and $0$ on the rest. Thus there are $6\choose 3$ ways to roll the $-1$s and for each of those there are $3$ ways to roll the $2$ for a total of $6\choose 3$ $*$ $3=60$ ways to get three $-1$s. 
If you roll $4, 5$ or $6$ negative $1$s there is no way to get a final score of $-1$.
Therefore, there are $66$ ways to get $-1$ and the probability is $\frac{66}{729}=\frac{22}{243}$.
A: 
Hint: $−1=0−1=1−2=2−3$  –  stochasticboy321

Let $\vec X=(x,y,z)$ mean the event of: $x$ "-1", y "0", and $z$ "+1" from the roll of the dice.
This is a multinomial distribution; which is analogous to a binomial distribution: $$\mathsf P(\vec X=(x,y,z))=\binom{6}{x,y,z}{(\frac 1 3)}^6$$
You want to measure $\mathsf P(\vec X\in\{(1,5,0),(2,3,1),(3,1,2)\})$ , the probability of obtaining one more -1 die than +1 die.
