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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?

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  • $\begingroup$ How do you argument (mathematically) that $B$ is bad? Because for $B$ you get an average of $1.25\cdot 0.7=0.875$ of your input but for $M$ you get $1.666\cdot 0.6=1$ as a reward? $\endgroup$
    – LegNaiB
    Sep 1, 2022 at 7:15
  • $\begingroup$ I am betting B amount on team B , If expected value is less than B why would I bet on that? That is why I am not betting on B $\endgroup$
    – bigstreet
    Sep 1, 2022 at 7:19
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    $\begingroup$ But scale the Match A solution up to {37.78 , 62.22} and the benchmark profit is $3.9 before selecting whole numbers for the solution set. $\endgroup$
    – S Spring
    Sep 16, 2022 at 0:11
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    $\begingroup$ @bigstreet Curious to know if you moved on to the next round or if ur approach was correct? Thanks $\endgroup$ Nov 13, 2023 at 9:12
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    $\begingroup$ Looks like an Arbitrage qn where they are trying to see how u hedge ur bets rather than EV in my opinion $\endgroup$ Nov 13, 2023 at 9:13

3 Answers 3

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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.

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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.

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    $\begingroup$ I agree. OP's question is very open ended and I suspect that the weight of a 'correct' answer is determined more by successful rhetoric than formal maths. In this case, a complete answer would then be very long. $\endgroup$
    – Snoop
    Sep 1, 2022 at 14:46
  • $\begingroup$ Hi Aaron, do you mind explaining the expected payoffs for me? Why is it $$ \left(P+ \frac{7}{4}P\right)0.4 + \left(M + \frac{2}{3}M\right)0.6+\left(B+\frac{1}{4}B\right)0.7+(C+3C)0.3 + R(unbet)$$ ? Why do you add all of them up? $\endgroup$
    – Sarah V.P
    Nov 27, 2022 at 8:09
  • $\begingroup$ @Snoop How did OP computed the expected values? Isn't in this case, the expected value for Pink is $(\frac{7}{4} + 1)\times 0.4 - 1\times 0.6 - 1\times 0.7 - 1\times 0.3$? $\endgroup$
    – Sarah V.P
    Nov 27, 2022 at 8:48
  • $\begingroup$ @SarahV.P Expected value is defined to be a sum of values weighted by their probabilities, which is why we add them. As for why that expression is correct: if we bet P dollars on team P, then with probability 0.4 they win and we get our bet back plus an additional 7/4 of it (as specified by the payout odds). If team P loses, then we get nothing, so there's an implied $0 \cdot 0.6$ in the P expression and a matching hidden 0 term for all other teams. $\endgroup$ Nov 28, 2022 at 12:41
  • $\begingroup$ @AaronMontgomery Hello Aaron, thanks for the reply. So The expected value for team P is $(\frac{7}{4}+1)×0.4+(0×0.6)$ for match A? I thought that if I bet 1 dollar on team P and team P loses, I get nothing but I will still lose that 1 dollar and hence we should also consider this outcome? Do we not need to factor this into account when calculating expected values? From my understanding expected value is multiplying each value of the outcome by its probability? Isn't losing that 1 dollar counts as an outcome? $\endgroup$
    – Sarah V.P
    Nov 28, 2022 at 15:43
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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 $11/3$.


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}

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  • $\begingroup$ Hi Rob, can you explain how OP computed the expected values? Isn't in this case, the expected value for Pink is $(\frac{7}{4} + 1)\times 0.4 - 1 \times 0.6 - 1 \times 0.7 - 1 \times 0.3$? Instead according to OP, the expected value is $(\frac{7}{4} + 1)\times 0.4$? Don't you have to include the other probabilities of you losing> $\endgroup$
    – Sarah V.P
    Nov 27, 2022 at 14:24

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