Question: Suppose there are 2 matches being played. First match is between player A and player B. Second match is between player C and player D. There are betting odds for each player.
In a scenario like this, how much money would you bet on each time if you have $100 (can choose to not bet some money)? (multiple correct answers)
My attempt: I'm unsure how to attempt this problem, but I've calculated the EV of betting 1 dollar on each of the players, and this is what I got:
E[betting on A] = -.1, E[betting on B] = .2, E[betting on C] = 1/8, E[betting on D] = 0
So this means that if I bet 100 dollars on C, I'll have the highest EV. But, that's extremely risky because I can lose all of my money. How do I go about hedging my bets, and making the smartest bet? I'm pretty sure this question is often used in interviews.
You can choose your bets to ensure a profit. Suppose we bet $x$ on $C$ and $100-x$ on $D$. If $C$ wins we have $\frac 92x$. If $D$ wins we have $\frac 43(100-x)$. If we want no risk at all, we can set these equal. $$\frac 92x=\frac 43(100-x)\\ \frac{35}6x=\frac 43\cdot 100\\ x=\frac 8{35}\cdot 100 \approx 22.86$$ Either way we wind up with $\frac {36}{35}\cdot 100$. You can do the same analysis for betting the other match and see how you come out.
Whether this is the best approach depends on your risk tolerance. Note that I did not use the stated probability of $C$ winning anywhere. It is just because the payoff odds were set incorrectly that I could guarantee a profit.