Game theory expected value We play a game involving two players. Each player calls a number 1 or 2. If the sum of these numbers are odd (i.e. equal to 3), then player 1 gets 3 points and player 2 loses 3 points. If the sum of these numbers are even (equal to 2 or 4), then player 1 loses 2 or 4 points and player 2 wins 2 or 4 points. (If the sum = 4, then they win/lose 4, if sum = 2 they win/lose 2). 
Each player has a probability with which they play a 1. Let's call it $p$ for player 1 and $q$ for player 2. The question is: Is there a $p \in [0, 1]$ such that, for all $q \in [0,1]$, the expected value for Player 1 is positive? Conversely, is there a $q \in [0, 1]$ such that, for all $p \in [0,1]$, the expected value for Player 2 is positive?
My work so far:
The expected value for Player 1 can be calculated as follows:
$E_1(V) = 3[p(1-q) + q(1-p)] - 2pq - 4(1-p)(1-q)$
$E_1(V) = 7p + 7q - 12pq - 4$
On the other hand, since Player 1 and Player 2 just trade money, we can say that $E_2(V) = 12pq + 4 - 7p - 7q$
However, I don't know what to do with these equations from here.
 A: I will start with the second question, the first one works with the very same method.
You can start by simply writing down what you want to have, and see if you can prove it or find a counterexample. We want:
$E_2(V)\geq 0 \Leftrightarrow\\
12pq-7p-7q\geq 4\Leftrightarrow\\
p(12q-7)\geq 4+7q>0\quad \forall q\in [0,1].$
At this point, can you think of a value for $q$ which ruins all the fun for player 2? Meaning, the inequality cannot hold?
A: Consider the first question first. $E_1$ is a linear and hence monotonic function of $q$, so to decide whether it’s always positive, you just need  consider its values at the extremes $q\in\{0,1\}$. These are $E_1(q=0)=7p-4$ and $E_1(q=1)=3-5p$. The first is positive for $p\gt\frac47$, and the second is positive for $p\lt\frac35$. Since $\frac47\lt\frac35$, there is indeed an (open) interval $\left(\frac47,\frac35\right)$ for $p$ such that $E_1(q)$ is positive for all $q\in[0,1]$.
Since the expected value is symmetric in $p$ and $q$ and $E_2=-E_1$, the second question is equivalent to the first question with “positive” replaced by “negative”. Since the corresponding inequalities $p\lt\frac47$ and $p\gt\frac35$ cannot be fulfilled simultaneously, there is no such value in this case.
A: An amusing trick is to write $$12E_2(V)=(12p-7)(12q-7)-1.$$ So you need a $q$ such that $(12p-7)(12q-7)\geq 1$ for all $p.$ But if $p=7/12,$ then the product on the left is always $0.$ So there is no such $q.$
This also shows that we can get $E_1(V)\geq 0$ for all $q$ by picking $p=\frac7{12}.$

It takes a little more work to find all values $p$ which work.
We need $p$ such that:
$(12p-7)(12q-7)\leq 1,$ for all $q\in[0,1].$
Now $-7\leq 12q-7\leq 5.$ For non-negative $12p-7,$ you can ensure this if $0\leq 12p-7\leq \frac{1}{5}.$ This give $\frac{7}{12}\leq p\leq \frac{3}{5}.$
For negative $12p-7,$ we require $0>12p-7\geq \frac{-1}{7},$ or $\frac{7}{12}>p>\frac{4}{7}.$
So, for player $1,$ any value $p\in\left[\frac{4}{7},\frac{3}{5}\right]$ will work for all $q.$

The first step is a common trick to play when seeing an expression like: $axy+bx+cy+d.$ Multiplying by $a,$ we get:
$$a(axy+bx+cy+d)=(ax+c)(ay+b) +ad-bc$$
This can quite often simplify a problem. It is a generalization of completing the square.
A: Well once you have the expected  values, the rest of the calculations are easy. We want to know for what fixed $p, E_1(V) \ge 0$. Solving this inequality gives
$$q(7-12p)+7p-4 \ge 0$$
$$q \ge \frac{4-7p}{7-12p}$$.
Then we want to know when $\frac{4-7p}{7-12p} = 0$. This happens when $p = \frac{4}7$. Similar for the other calculation. 
