How much money do you bet on each team?

Question

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.

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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?

Answer 1

Some comments:

1. I think you have the calculations right, and you have discovered the correct fundamental truth (betting on $C$ and $P$ is a good idea, betting on $B$ is not).

2. Although I'm not a quant finance guy, I suspect that communication is important here. There are subtle details in your work that are implicitly correct, but missing, in your work. For instance, consider this line: "So By looking at this equation I concluded that betting on B is not a good strategy so $B=0$" -- You're right, but you should show that. It's better to do the arithmetic, see that your payoff is less than $B$, and point to that when concluding not to bet on $B$. This is worth perhaps one extra sentence over what you've written and shouldn't take much, but it improves your argument quite a lot. Similarly, in "Since P+M+C+R = 100 , So final problem has reduced to...", I don't think your steps are clear and they could be fleshed out more, even though they seem to be pointing you toward a correct conclusion.

3. I think you're definitely on the right track, but I think you're not done. Where will you put your dollars, and why?

What strikes me as the most important part of the prompt is this:

There is no single "correct" answer, but there are many "wrong" answers.

You could choose to just chase the maximum expected value (spoiler alert: put down 100 bucks on $C$), but maybe you ought to consider hedging your bets and explaining why that's useful. Or, maybe you do want to go for that highest expected value; in that case, you might just say that and briefly defend that choice. I think the defense of your choice is going to be more important than the allocation of dollars, if I'm reading the subtext correctly.

Answer 2

If you want to maximize the worst-case profit (ignoring the probabilities), you can maximize $$\min\{(7/4+1)P,(2/3+1)M\}+\min\{(1/4+1)C,(3/1+1)B\}-(P+M+C+B)$$ subject to $P+M+C+B \le 100$. The resulting optimal solution turns out to be $(P,M,C,B)=(37,61,0,0)$, yielding a worst-case profit of $11/3$.

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By request, here is the SAS code I used, with $(P,M,C,B)$ renamed as $(X_1,X_2,X_3,X_4)$:

proc optmodel;
   num n = 4;
   num odds {1..n} = [(7/4) (2/3) (1/4) (3/1)];

   var X {1..n} >= 0 integer;
   max MinProfit =
      min {j in 1..2} (odds[j] + 1) * X[j] 
    + min {j in 3..4} (odds[j] + 1) * X[j]
    - sum {j in 1..n} X[j];
   con Budget:
      sum {j in 1..n} X[j] <= 100;

   solve linearize;
   print X;
quit;

The resulting solution is: \begin{matrix} i & X_i \\ \hline 1 & 37 \\ 2 & 61 \\ 3 & 0 \\ 4 & 0 \\ \end{matrix}