A basket contains 30 apples and 30 oranges. You draw two fruits at a time (without replacement) until the basket is empty -- that is 30 draws total. On each draw:
If your final score is strictly higher than the vendor's, you win $\
What is the maximum amount you should pay to enter this game so that you break even in expectation?
How to Think About It: Before doing any math, think about the symmetry. Apples and oranges play perfectly symmetric roles -- the only difference is the label. Every time you draw two apples (you score), there is a "mirror" scenario where those two fruits are oranges (vendor scores). So intuitively, your score and the vendor's score should have the same distribution. The interesting question is whether ties are possible and how likely you are to win versus lose.
Let $A$ = your score (number of apple-apple draws) and $V$ = vendor's score (number of orange-orange draws). Let $M$ = number of mixed draws. Then $A + V + M = 30$ (total draws). Each apple-apple draw uses 2 apples, each mixed draw uses 1 apple, so A + M = 30$ (all apples used). Similarly V + M = 30$ (all oranges used). Subtracting: A - 2V = 0$, so $A = V$ always.
Quick Estimate: Since $A = V$ with certainty, your score equals the vendor's score every single time. You never win, so the game pays $\$0$ with probability 1.
Approach: This is a pure counting/conservation argument -- no probability needed beyond observing the constraint.
Formal Solution:
Label each draw as AA (two apples), OO (two oranges), or AO (one of each). Let $A$, $V$, and $M$ be the counts of each type. We have three conservation laws:
$A + V + M = 30 \quad \text{(total draws)}$ $2A + M = 30 \quad \text{(total apples)}$ $2V + M = 30 \quad \text{(total oranges)}$
From the last two equations: A + M = 2V + M$, which gives $A = V$. This holds deterministically -- no matter what order the fruits come out, the number of apple-apple pairs always equals the number of orange-orange pairs.
Since $A = V$ with probability 1, you never have a strictly higher score than the vendor. The payoff is always $\$0$.
Answer: The fair value of this game is $\$0$. You should pay nothing to play.
This is a classic trick question that tests whether you jump into complicated probability calculations or step back and think about structural constraints. The key insight is conservation: you start with exactly 30 of each fruit, and every draw consumes exactly 2 fruits total. The number of apples consumed in apple-apple draws plus mixed draws must equal 30, and the same holds for oranges. These two constraints force the apple-apple count to equal the orange-orange count, regardless of the random order.
In quant interviews, problems like this test your instinct to look for deterministic structure before reaching for stochastic tools. Many candidates start computing hypergeometric probabilities when the answer follows from simple counting. The broader lesson: always check if there are conservation laws or symmetry arguments that pin down the answer before doing hard math.