From a trading interview at Optiver, I have a probability question that apparently has multiple correct answers. I would like to know if mine is correct.
You are given the opportunity to make money by betting a total of 100 bucks on the outcome of two simultaneous matches:
- Match A is between the Pink team and the Maroon team
- Match B is between the Brown team and the Cyan team
The Pink team's probability of victory is 40%. The Brown team's probability of victory is 70%. The betting odds are
- Pink: 7:4
- Maroon: 2:3
- Brown: 1:4
- Cyan: 3:1
How much money do you bet on each team? You do not have to bet all 100 bucks, but your bets must be whole numbers and the total of all five blanks (bets on the four teams and the unbet amount) must sum to 100. There is no single "correct" answer, but there are many "wrong" answers. As a reminder, a hypothetical team having 2:7 odds means that if you bet 7 on that team and they win, you get your 7 bucks bet back and win an additional 2 bucks.
My solution
Equation of expected payoff
$$ \left(P+ \frac{7}{4}P\right)0.4 + \left(M + \frac{2}{3}M\right)0.6+\left(B+\frac{1}{4}B\right)0.7+(C+3C)0.3 + R(unbet)
$$
By looking at this, I concluded that betting on $B$ is not a good strategy so $B = 0$. Now the problem has become to maximize the following
$$ 1.1P + M + 1.2C + R $$
Since $P+M+C+R = 100$, the final problem has reduced to maximizing $0.1 P + 0.2 C$. Is my approach correct?
Best Answer
This looks a bit like arbitrage betting, however we have the added knowledge of the odds of each outcome. I think the best risk adjusted approach is to bet on each outcome on a single game in a way such that, no matter the result, the expected P/L is the same. One game will have a better expected P/L so say we bet \$X on Pink, we should bet $100-X on Maroon. These are my calculations.
Pink vs Maroon game: Bet \$X on Pink (and $100-X on Maroon)
Returns if pink wins: (7/4)X - (100 - X) Returns if Maroon wins: (2/3)(100-X) - X
Now, we can multiply each of these returns, by the probability we realise these returns, and then equate them so no matter the result, our returns are the same:
0.4*((7/4)X - (100 - X)) = 0.6*((2/3)(100-X) - X)
===> (21/10)X = 80 ===> X = 38 (To nearest whole number as we have to bet whole amounts)
Expected P/L for Pink winning = 1.8 Expected P/L for Maroon winning = 2
Brown vs Cyan game: Bet \$X on Brown (and $100-X on Cyan)
Returns if Brown wins: (1/4)X - (100 - X) Returns if Cyan wins: (3)(100-X) - X
Now, we can multiply each of these returns, by the probability we realise these returns, and then equate them so no matter the result, our returns are the same:
0.7*((1/4)X - (100 - X)) = 0.3*((3)(100-X) - X)
===> (80/43)X = 160 ===> X = 77 (To nearest whole number as we have to bet whole amounts)
Expected P/L for Brown winning = -2.6 Expected P/L for Cyan winning = -2.4
In this case, we would want to bet \$38 on pink and $62 on Maroon.
I am not sure if this result is correct as it seems very conservative and unintuitive, but it seems roughly the correct approach to me. By betting on either side such that the expected profit and loss is the same regardless, we are guaranteeing profit rather than gambling on a certain result happening. For example, we could just bet \$100 on Cyan winning as the potential gain is massive. In this case we have a 30% chance of making \$300, however we also have a 70% chance of losing \$100. If we bet on this game infinitely many times, this would be a far better bet as the expected value of each bet is $20. However, as we can only bet once, I believe we want to be conservative and guarentee profit.