Player A vs Player B.
- Bookie 1 offers 1.36 odds on player A winning.
- Bookie 2 offers 5.5 on player B winning.
We have $1000 in total to bet. How would you place your bets such that profit is maximized?
I have been told that this can be solved using linear programming, but I don't know how to set the problem up properly. Any ideas?
Since you are spending $\$1000$, you want to guarantee winning more than $\$1000$ regardless of which player wins.
In order to win $\$1000$ for Player A winning, you would need to place a bet with Bookie 1 of $\frac{1000}{1.36}=\$735.294$
In order to win $\$1000$ for Player B winning, you would need to place a bet with Bookie 2 of $\frac{1000}{5.5}=\$181.818$
So to guarantee a profit, you should bet between $\$735.30$ and $\$818.18$ with Bookie 1 and the rest with Bookie 2.
Your profit will be maximised if you bet everything on B and then B wins, giving a net gain of $\$ 4500$. But it is not guaranteed, and is probably unlikely. I do not think this is what you want
If instead you
then you bet $\$1000$ in total and
so either way a profit of just over $\$90$. You cannot make your guaranteed profit any higher than that, since increasing your bet on option means reducing your bet on the other
Bet $0 \leq x \leq 1000$ at the first bookie and $1000 - x$ at the second bookie. The goal is to maximize
$$\min \{ 1.36 x - 1000, 4500 - 5.5 x \}$$
which is the inverted "V" depicted below
The maximum is attained when
$$1.36 x - 1000 = 4500 - 5.5 x$$
The maximum is $\approx 90$, which is attained at $x \approx 802$.
There is no need to use linear programming. However, if you really, really do want to use linear programming, then solve the following linear program in $x$ and $t$
$$\begin{array}{ll} \text{maximize} & t\\ \text{subject to} & 1.36 x - 1000 \geq t\\ & 4500 - 5.5 x \geq t\\ & 0 \leq x \leq 1000\end{array}$$
In general, suppose that you want to wager amount $A$ and the odds on two possible outcomes are $o_1$ and $o_2$. Let us denote by $A_1$ the amount you wager on the first outcome (with odds $o_1$) and $A_2$ the amount you wager on the outcome $o_2$.
So you know that $$A=A_1+A_2.$$ You also know that depending on the outcome, your winnings will be \begin{align*} w_1&=o_1A_1\\ w_2&=o_2A_2 \end{align*}
In order to maximize the profit in each case, you want $w_1=w_2$. (In other words, you won to bet in the way that you win the same amount, independently on the outcome.) So you need $$o_1A_1=o_2A_2. \tag{1}$$
Now simply by algebraic manipulation you get \begin{align*} A&=A_1\left(1+\frac{o_1}{o_2}\right)\\ A&=A_1\frac{o_1+o_2}{o_2}\\ A_1&=\frac{o_2A}{o_1+o_2} \end{align*} and similarly $$A_2=\frac{o_1A}{o_1+o_2}.$$
Another way to see this is not notice that $(1)$ says that $\frac{A_1}{A_2}=\frac{o_2}{o_1}=\frac{1/o_1}{1/o_2}$, which basically says that you want to divide $A$ in the ratio $\frac1{o_1}:\frac1{o_2}$. Therefore \begin{align*} A_1&=\frac{\frac1{o_1}}{\frac1{o_1}+\frac1{o_2}}A, A_2&=\frac{\frac1{o_2}}{\frac1{o_1}+\frac1{o_2}}A. \end{align*} Notice that this cen be simplified to $$\frac{A_1}A =\frac{\frac1{o_1}}{\frac1{o_1}+\frac1{o_2}} =\frac{\frac1{o_1}}{\frac{o_1+o_2}{o_1o_2}} =\frac{o2}{o_1+o_2} $$ so both expressions give the same value, only expressed in a different way.
However, the second expression you can see more easily that you make profit if and only if $\frac1{o_1}+\frac1{o_2}<1$. Indeed, to make profit you need \begin{align*} A&<o_1A_1\\ A&<\frac{A}{\frac1{o_1}+\frac1{o_2}}\\ 1&<\frac{1}{\frac1{o_1}+\frac1{o_2}}\\ \frac1{o_1}+\frac1{o_2}&<1 \end{align*}
For the particular numbers you need, you simply plug $A=1000$, $o_1=1.36$ and $o_2=5.5$ into the above formula.
I will add also a link to the Wikipedia article about Arbitrage betting.
