Red and Black Card Game With Optional Stopping: How Much Would You Pay to Play?
Company: Soros
Role: Quantitative Researcher
Category: Statistics & Math
Difficulty: medium
Interview Round: Technical Screen
A shuffled deck of 52 cards has 26 red cards and 26 black cards. Cards are turned over one at a time without replacement. Each red card earns you \$1 and each black card loses you \$1. You may stop the game at any time and keep your current winnings; if you never stop, the game ends when the last card is turned. How much would you be willing to pay to play this game?
The report notes that there is no ready-made formula to apply and that the interviewer mainly wanted to see the line of reasoning, not necessarily a closed-form answer derived on the spot. Explain how you would find the value of the game and the strategy that achieves it, and turn that into the price you would pay.
```hint What the future depends on
Whatever you have won so far is yours if you stop. Ask which quantities alone determine what the rest of the game can still add.
```
```hint Start small
Work out by hand what the game is worth with one red and one black card left, then with two of each, before thinking about the full deck.
```
### Clarifying Questions
- May I stop before the first card is turned?
- Do I see each card as it is turned, so that I always know how many red and black cards remain?
- Is the deck uniformly shuffled, with no information about the order?
- Is the price meant for a risk-neutral player who can play many times, or for a single play where risk matters?
### What a Strong Answer Covers
- A precise formulation of the state, the stop-or-continue decision and the boundary cases
- A recursion that compares stopping with continuing, and why it gives the optimal strategy
- Small cases computed by hand that check the recursion
- The value of the full game, computed or credibly estimated, and how it becomes a bid that reflects risk preference
- The shape of the optimal stopping rule, including when continuing is clearly right
### Follow-up Questions
- Give a simple strategy with a positive expected value and compute that value exactly without the full recursion.
- Show that it is never optimal to stop while at least as many red cards as black cards remain.
- Under the optimal strategy, can you finish with negative winnings?
- How many states does the computation need for a deck with n red and n black cards, and how long does it take?
Overview: A quantitative interview puzzle: cards from a shuffled deck of 26 red and 26 black are turned one at a time, each red wins a dollar and each black loses one, and you may stop whenever you like. It asks how much you would pay to play, testing optimal stopping, backward induction, small-case checks and turning an expected value into a price.
Read the full Soros Quantitative Researcher interview experience this question came from