What I am asking is just for my personal curiosity.
I was thinking about the betting system (ex. football). For example let's consider just this possibilities to bet:
Win of Team A Draw Win of Team B
2,0 4,0 8,0
As you know, if you bet 10€ on Win of Team A, and it wins, you get 20€ back.
I was wondering if you could bet on all three possibilities (with the right distribution of money) and be sure to get a reward > 0€. My common sense said "Of course not", otherwise everyone could always win.
I don't have a lot of Math knowledge, but I tried to formalize the problem to see why, analitically, it is impossible to do so.
Here is my attempt, but I can't go no further:
Win of Team A multiplier: a (constant)
Draw multiplier: d (constant)
Win of Team B: b (constant)
Money to bet on Win of Team A: x (variable)
Money to bet on Draw: y (variable)
Money to bet on Win of Team B: z (variable)
The system I want to solve (Is it the right way to express "Always win betting onto the all three?")
$$\begin{cases} ax -(x + y + z) > 0 \\ dy - (x + y + z) > 0 \\ bz - (x + y + z)> 0 \\ x \neq 0 \\ y \neq 0 \\ z \neq 0 \end{cases}$$
This system (to me) means: if you bet for the Win of Team A and you betted for the Others, you get all your betted money back plus something won... (the same for Draw and Win of Team B).
Then:
$$\begin{cases} ax > (x + y + z) \\ dy > (x + y + z) \\ bz > (x + y + z) \\ x \neq 0 \\ y \neq 0 \\ z \neq 0 \end{cases}$$
And:
$$\begin{cases} \frac{y+z}{x} > a-1 \\ \frac {x+z}{y} > d-1 \\ \frac {x+y}{z} > b-1 \\ \end{cases}$$
At this point I know that my knowledge of math is finished. I tried to trasform the system but I really can't solve it. I never studied disequations with 3 variables (nor 2, sigh). How can i solve it and verify for which values of x, y, z it is solvable?
Suppose you have $n$ mutually exclusive outcomes with payoffs $p_1, \ldots, p_n$. (In your case $n=3$ and $p_1=2$, $p_2=4$, and $p_3=8$.) I'll show that you can "game the system" to always win precisely when $$S=\frac1{p_1}+\frac1{p_2}+\cdots+\frac1{p_n}<1.$$ This holds in your case, as $1/2+1/4+1/8<1$; the argument below shows how to place your bets and compute your guaranteed profit.
Assume you're going to distribute a bet $B$ over the $n$ outcomes, and assume an adversary is going to control the outcome after you place your bet. Thus you want to maximize (over all distributions of the bet) the minimum payoff (over all $n$ possible outcomes). We'll show that in the optimal strategy, the bet on outcome $i$ is proportional to $1/p_i$. (More precisely, the bet on outcome $i$ is $B/(S p_i)$, so that the sum of the bets is $B$.)
If the outcome is $i$, your gross payoff is $p_i\cdot B/(S p_i)=B/S$ which is actually independent of $i$. Since you've paid $B$ to play, your profit is guaranteed to be $B/S - B$, which is positive exactly when $1/S-1>0$, i.e. when $S<1$.
Let's see why this is true. Finding the optimal strategy is a linear programming problem, and you can do it with matrices and all that, but it's really easier. Concentrate on outcomes $i$ and $j$; suppose you've bet $x_i$ and $x_j$ on those outcomes. Your payoff for that investment of $x_i+x_j$ will be $$\min (x_i p_i, x_j p_j).$$ The maximum of this payoff, subject to $x_i+x_j$ being fixed, occurs when $x_i p_i=x_j p_j$. Therefore, no matter what your investment in other outcomes, if you haven't chosen $x_i:x_j$ in proportion to $1/p_i:1/p_j$, you can improve your payoff. Since $i$ and $j$ are arbitrary, the optimal strategy must set $x_i$ proportional to $1/p_i$; thus $$x_i=\frac B{p_i S}$$ so that the sum of the bets is $B$.
In your specific case, $1/S=1/(1/2+1/4+1/8)=8/7$, so your profit will be $B/7$ if you place a total bet $B$. For example, if you bet $4$, $2$, and $1$ respectively, where the payoffs are $2$, $4$ and $8$ respectively, your total bet is $7$, and you're guaranteed to win $8$.