Which wheel has most chance of winning a roulette game? Player A get to choose the first wheel. Player B must choose a wheel from the 2 others.
Both players spin their wheel.

Just in case the image doesn't work :
$$\text{wheel 1 }( 3,4,8 )$$
$$\text{wheel 2 } ( 1,5,9 )$$
$$\text{wheel 3 } ( 2,7,6 )$$
First off, I figured there are $3^3$ possibilities.
Then I went and wrote down each possibilities...
I have that the first wheel has a 8/27 chance to win
wheel 2 has 11/27
wheel 3 has 8/27.
So player A has an advantage if he chose wheel #2.
Is it the right way to solve this problem ? Was I suppose to use the choose or permutation notation ?
Edit : The player that spins the highest number win the game.
This is a fictive game.
Edit2 : For example, player A decide to choose wheel 1. Player B then choose wheel 3.
Player A spin his wheel and get 4. Player B spin and get a 7.
Player B wins.
 A: We just need to patiently calculate probabilities. It will take a while. But you will be doing most of the work.  And the final result will be very interesting. To read much more about the proble, look at the Wikipedia article on Non-transitive Dice.
Case 1: Suppose that player A chooses wheel $1$.  Let us see what B should do.
Suppose that B chooses wheel $2$. what is the probability that B wins?
Maybe A's wheel will produce a $3$. That has probability $\frac{1}{3}$. Given that A's wheel produces a $3$, with probability $\frac{2}{3}$ B's wheel produces a higher number, so she wins.
Maybe A's wheel will produce a $4$. That has probability $\frac{1}{3}$. 8Given* that this happened, again, with probability $\frac{2}{3}$, B wins.
Maybe A's wheel will produce an $8$. That has probability $\frac{1}{3}$. Then with probability $\frac{1}{3}$, B wins.
So if A chooses wheel $1$ and B chooses wheel $2$, the probability B wins is $$\frac{1}{3}\cdot\frac{2}{3}+\frac{1}{3}\cdot\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{3}=\frac{5}{9}.$$
Suppose that now that A chooses wheel $1$ and B chooses wheel $3$. The same sort of calculation shows that the probability B wins is $\frac{1}{3}\cdot\frac{2}{3}+\frac{1}{3}\cdot\frac{2}{3}+\frac{1}{3}\cdot\frac{0}{3}=\frac{4}{9}$.
So it is clear that if A chooses wheel $1$, then B should choose wheel $2$ to maximize her probability of winning. And her probability of winning will be bigger than $\frac{1}{2}$. that is, B will have an edge.
Case 2: Suppose A chooses wheel $2$? What should B do? What is B's probability of winning if she chooses wisely?  You can do the calculations. You will find that in that case, B should pick wheel $3$, and that if she does, she will have probability $\frac{5}{9}$ of winning. 
Case 3: Suppose A chooses wheel $3$? What should B do?  Again, you can do the calculations.
Remark: After you do the calculations, you will notice that whatever choice A makes, B can choose a wheel that on average will beat A.  This is a very surprising result: In a sense, wheel $2$ is "better" than wheel $1$.  And wheel $3$ is better than wheel $2$. But wheel $1$ is better than wheel $3$!  
A: You need to pit each wheel against each of the others.
Suppose that Player A chooses the first wheel and Player B the second. The $3^2=9$ (equally likely) possible outcomes are $\langle 3,1\rangle,\langle 3,5\rangle,\langle 3,9\rangle,\langle 4,1\rangle,\langle 4,5\rangle,\langle 4,9\rangle,\langle 8,1\rangle,\langle 8,5\rangle$, and $\langle 8,9\rangle$. Player A wins $4$ of them, and Player B wins $5$.
If Player A chooses the first wheel and Player B the third, on the other hand, the outcomes are $\langle 3,2\rangle,\langle 3,7\rangle,\langle 3,6\rangle,\langle 4,2\rangle,\langle 4,7\rangle,\langle 4,6\rangle,\langle 8,2\rangle,\langle 8,7\rangle$, and $\langle 8,6\rangle$, and Player A wins $5$ of them. Thus, if Player A chooses the first wheel, Player B should choose the second, and Player A’s probability of winning will be $\frac49$.
You only have to work out one more calculation like this, for the second wheel against the third, and you’ll be able to figure out what happens if Player A chooses the second wheel or the third. 
A: here is how I see it.
wheel 1 vs. wheel 2
if wheel 1 rolls 3, chance of winning is 1/3
if wheel 1 rolls 4, chance of winning is 1/3
if wheel 1 rolls 8, chance of winning is 2/3
probability that wheel 1 beats wheel 2 is therefore 4/9
wheel 1 vs. wheel 3
if wheel 1 rolls 3, chance of winning is 1/3
if wheel 1 rolls 4, chance of winning is 1/3
if wheel 1 rolls 8, chance of winning is 3/3
probability that wheel 1 beats wheel 3 is therefore 5/9
wheel 2 vs. wheel 3
if wheel 2 rolls 1, chance of winning is 0/3
if wheel 2 rolls 5, chance of winning is 1/3
if wheel 2 rolls 9, chance of winning is 3/3
probability that wheel 2 beats wheel 3 is therefore 4/9
conclusion
we have a "rock paper scissors" situation where wheel 1 beats wheel 3 beats wheel 2 beats wheel 1, all with probability 5/9. Therefore  player B has an advantage,but needs to choose the wheel that beats the wheel player A picked.
