# 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.

• This is basically like American option pricing. You work backwards and at every node you check which is larger: the expected pay off if you keep on playing or the pay off now. If you work this out for your two step tree, this gives you 5/8. – Raskolnikov Nov 11 '17 at 15:45
• Did anyone ever fully explain this? I feel like the answer should be .75, since 50% odds of getting 1 and 50% odds of approaching .50 in payout - so .5+.25 = .75 Am i wrong? – BILLYBOB May 3 '19 at 23:14
• I did backward induction on a n-step tree, got the following values : 0.743(n=10), 0.761(n=18), 0.778(n=50). It seems to be converging slightly above 0.75, if I did it right. – dm63 May 4 '19 at 9:54