Stopping a Card Game and Pricing a Coin-Flip Call

# What are the answers to these questions on card deck and option pricing?


# What are the answers to these questions on card deck and option pricing?












here are 3 questions I have some trouble dealing with. Your help will be greatly appreciated!

1 - We have a deck card: 26 red, 26 black. we play a game: you draw a card from the deck without putting it back. If it is red I will pay you 1£. If black you will pay me 1£. You can stop whenever you want. what is a fair value of this game, assuming risk-neutral? what if there were infinite red/black cards?

2- assume rates are 0. There is a call option written on coin flips, that is the payoff of the security is the number of heads that comes up after a number of flips. Strike price is 2. Value this option for 4 coin flips. What is its delta?

3 - Given the BS implied volatility are the same for a bunch of calls with different strikes, other things being equal, how could one make an arbitrage if we just know the underlying volatility follows a stochastic process?

my guess (please correct if you find it wrong):

1 - stopping time has all its importance here. Lets say I pick up a black card: I will get 1£ but the proba of getting a red on the next round is higher. I tried to do something like: P(getting red at the nth attempt/s red cards have been picked up) with s<=n compared to P(getting red at the nth attempt/s-1 red cards picked). But this was not really working...

2 - when I flip 4 coins there are 16 possible paths. My understanding is that the strike is the number of flips. Then I break down the payoff: - if n_heads=0 then payoff = 0 wp 1/16 - if n_heads=1 then payoff = 0 wp 4/16 - if n_heads=2 then payoff = 0 wp 6/16 - if n_heads=3 then payoff = 1 wp 4/16 - if n_heads=4 then payoff = 2 wp 1/16

so the value of my call will be 4/16 + 2/16 = 6/16 = 3/8

For the delta: - if n_heads=0 then payoff = 0 wp 1/16 => delta = 0 (OTM call) - if n_heads=1 then payoff = 0 wp 4/16 => delta = 0 (OTM call) - if n_heads=2 then payoff = 0 wp 6/16 => delta = 0 (OTM call) - if n_heads=3 then payoff = 1 wp 4/16 => delta = 1 (ITM call) - if n_heads=4 then payoff = 2 wp 1/16 => delta = 1 (ITM call)

=> delta = 4/16*1 + 1/16*1=5/16 so I will short 5/16 of stocks

3 - if all implied vol are equal for a bunch of calls, low trike calls are underpriced compared to high strike vols. So i will go long low strike calls and short high strike calls for the same maturity (basically being long call spreads). but is there a way to make it more formal?

## Answer by Quantuple (score 1)

https://quant.stackexchange.com/a/25672

- This question is extremely interesting and not that straightforward. See answer here. From a financial perspective this is very much like pricing an American call (stopping rule = intrinsic value from exercice (i.e. current cash earned) > continuation value (i.e. what you can expect to gain). Note that you can never win more than 13 nor lose (at worst you play to the end and finish with 0 since there is the same number of red/black cards in the deck).

- Its not 4 coins, but 4 flips. So its a simple binomial distribution on the number of heads. The strike is also a number of heads: you only win something when you end up with strictly more than 2 heads over the 4 flips (hence 3 or 4 heads over 4 flips).

- I would use the known stochastic vol process to price and delta hedge the option. Call $V_a $ the price using that actual volatility and $V_i $ the implied market price. If $V_a > V_i $ (else do the opposite of what follows), buy the option at its implied price $V_i$ and synthesise a long option position using a replicating portfolio (stocks and cash), using the actual volatility for the $\Delta $ computation. Dynamically rebalancing up to expiry will leave you the difference between the actual option price and the market price as a terminal wealth.