So I have this question and I was wondering what is meant by 'convex combination of vectors' and as a consequence how this zero sum game is reduced in the way the answer sheet states.
What does it mean that the last column is dominated by a convex combination of the first 2 columns?
And what does it mean that player 2 will play a convex combination of the first two columns?
Can you explain how to get to the final form of the game with the underlying concept of convex combination of two vectors?
I have learning disabilities and the simplest and clearest answer would be greatly appreciated!
Thanks!
I'll try to make this answer as clear as possible, but please let me know if there is something that needs more clarification.
A convex combination of two vectors (let's say $v$ and $u$) mean that you can pick two numbers $a,b$ between 0 and 1 that are 1 combined (such as $\frac{1}{2}, \frac{1}{2}$ or $\frac{2}{5},\frac{3}{5}$) and add the vectors together using these numbers; the vector $w$: $w = a\cdot v + b\cdot u$ (for example $w = \frac{1}{2}\cdot v + \frac{1}{2}\cdot u$) is a convex combination of the vectors $u$ and $v$.
In this case, the solution says that a combination of the first and second column gives you a column that has lower values than the last one; in this case, you could try $a=0.4$ and $b=0.6$, i.e. $w = 0.4*\begin{pmatrix}6\\1\end{pmatrix} + 0.6*\begin{pmatrix}0\\5\end{pmatrix}$. The first value in this column would be $0.4*6 + 0.6 * 0 = 2.4$ and the second would be $0.4*1 + 0.6*5 = 3.4$. In both cases this is lower than the fourth column ($2.4 \leq 3$ and $3.4 \leq 4$). This is not the only convex combination that works, but it's just an example.
That means that the player is better of playing the first column 40% of the time and the second column 60% of the time, compared to playing the fourth column. So he will do that instead, and the fourth column is not an option to him.