Early Exercise Value in a Coin-Toss Stopping Game
Summary
The document poses a fair coin-toss game with up to 100 flips. At each point, the player may stop and receive the fraction of flips that were heads; the question is how to price the game when stopping is allowed. It contrasts this with the no-early-exercise case, whose expected payout is presented as one half, and notes that the ability to stop can increase the player's opportunity.
A binomial tree is suggested as a way to reason about possible outcomes and early stopping, with examples of favorable branches and their payouts. However, the document contains only the interview question and the asker's partial reasoning; it does not provide the optimal stopping solution, a fair entry price, or a complete backward-induction calculation. Its branch discussion alone does not establish the game value, since pricing requires evaluating the expected value under an optimal stopping policy across the full tree.
Key ideas
- The game pays the current fraction of heads whenever the player chooses to stop.
- Without early exercise, the document treats the expected payout as one half.
- Early stopping creates an optimal stopping problem that can be represented with a binomial tree.
- The prompt does not provide a solution or establish the minimum price to charge.
Tags
Full text
# Early Exercise Options and Coin Flipping
# Early Exercise Options and Coin Flipping
This problem was presented in an interview, and I know I got it roughly correct. But I am still not entirely understanding the early exercise component of it:
Say I am advertising a game where I would flip a coin up to 100 times. The payoff at any time $t$ is $P_t=\frac{\# of heads}{\# of flips}$. If I allow you to walk away with your earnings at any time, how much should I charge for the game?
I understand logically how without early exercise, the option should be priced as $\frac{1}{2}$, since for any large number of tosses the Payout should be $P_t=\frac{\frac{1}{2}*\infty}{\infty}=\frac{1}{2}$ from the law of large numbers, expected value etc etc. The interviewer then proceeded to show me how a binomial tree could be used to price the option:
The maximum upside for the top half of the tree is 1, with the probability of that occurring at $\frac{1}{2}$ and the maximum upside of the lower half of the tree being in the above tree $\frac{1}{2}$. Extending this out to 100 flips results in payout 1 in the top half again, but the bottom half could potentially result in $\frac{99}{100}$ if you flipped tails first and then flipped heads 100 times.
I know there are only one or two steps left to finding the minimum price that me, as the game maker, should charge to play this game. I know it should be higher than $\frac{1}{2}$, because I am giving the early exercise option and this will only give me more downside while giving the user more upside. The minimum cost means I should be pricing it on the potential upsides of the two branches. Can somebody bring this home for me? Would help me understand for the next time a question like this is asked.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.