I want to check my solutions for this problem.
Klaus and Anna throw tetrahedrons (bodies with 4 triangles as faces) at the same time. Klaus throws 2 tetrahedrons and Anna throws one. What is the probability that half the sum of the numbers in Klaus's tetrahedron is greater than the number in Anna's tetrahedron. Set up a suitable model.
For the model $(\Omega, F, P)$ I thought
Let
$k_i=$ Half of the sum of the numbers in Klaus's tetrahedron is i
$a_i=$ Anna has a i
$\Omega=\{(k_1,k_2,k_3,k_4) \} \times \{(a_1,a_2,a_3,a_4) \}$
We have $F=2^{\Omega}$ and $P(A)\left\{\begin{matrix} 1/16&A=k_1 \\ 3/16&A=k_2 \\ 3/16& A=k_3\\ 1/16& A=k_4\\ 1/16& A=a_i \end{matrix}\right.$
So what we want to find is $P(\text{1/2 the sum in Klaus's tetrahedron is greater than the number in Anna's tetrahedron})=\\= P(k_i\geq 1 \cap a_2)+P(k_i\geq 2 \cap a_2)+P(k_i\geq 3 \cap a_3)+P(k_i\geq 4 \cap a_4) \\ \overset{\text{since independent}}=\sum_{i=0}^4 P(k_i\geq 1)P(a_2)=5/64$
Am I right or have I done any mistake? I appreciate every help :)
Eric's point in the comments is correct, this misses the fact that Klaus can score a 1.5. Rather than try to model the averages as events in the probability space, it is perhaps easier to make the sample space
$$\Omega = \{1,2,3,4\}^3$$
(assuming the faces are numbered 1 through 4)
Then every event in the sample space has equal probability and we can regard the first two coordinates of an atomic event as the tetrahedron outcomes for Klaus and the third coordinate the outcome for Anna. This isn't the only way to model the problem but it might be easier.
For any atomic event $\langle x_1,x_2,x_3\rangle$ it is in the event $A$ if and only if $\frac{x_1+x_2}{2} > x_3$ so we need to count the number of events for which $x_1+x_2>2x_3$. Of course we could only have $x_3=1,2,3$, and we could find the number of atomic events by a relatively straight-forward enumeration. Then divide by the size of the sample space, 64.