Replicating a Series-Winning Digital Payoff with Sequential Bets

# Mark Joshi, Quant Interview Question problem 2.34; replicating a digital option on a 4-step symmetric binomial tree


# Mark Joshi, Quant Interview Question problem 2.34; replicating a digital option on a 4-step symmetric binomial tree












Question:

> Team $A$ and team $B$, in a series of $7$ games, whoever wins $4$ games first wins. You want to bet $100$ that your team wins the series, in which case you receive $200$, or $0$ if they lose. However the broker only allows bets on individual games. You can bet $X$ on any individual game that day before it occurs to receive $2X$ if it wins and $0$ if it loses. How do you achieve the desired pay-out? In particular, what do you bet on the first match?

Thoughts:

My initial thought was breaking this problem up in terms of combinatorics and probability by asking questions like: how many possible combinations are there for one of the two particular teams to win? What is the probability that $A$ wins?, what is the probability $A$ wins given $B$ wins the first game?, etc...

I was a bit stumped by this question so turning to the solution the author suggests that well this is just replicating a $4$-step symmetric binomial tree. Continuing on I could not really follow his solution, I was wondering if there were other ways of answering this problem. Any suggestions or guidance are greatly appreciated.

## Answer by will (score 6, accepted)

https://quant.stackexchange.com/a/36257

The answer above is only confusing because it is missing the the bet amounts.

You have a series of events, you are only allowed to bet on single events.

You want to construct something such that the payoff is dependent on the various outcomes:

Now we need to fill in the blanks - the current winnings, and the bet at each point. for each node $n_{ij}$, the current winnings must be the value at the previous nodes $\pm$ the bet (depending on if it were a win or loss).

The first layers are pretty trivial:

if i have X, and betting Y leaves me with 0 if i lose, then X must equal Y. and if winning leaves me with 2Y, then X = half the winnings. So the final node has me with a balance of 100, and betting 100. I do the same for all the nodes where i can infer the same thing:

Note that you can only fill in nodes where you know both possible outcomes. Fortunately these always exist. Now you fill in the middle node where the score would be 2-2, and then all of those diagonals:

And you then arrive at the same answer as above.

I'll have a think about it tomorrow on how to do it with just 4 nodes though.

## Answer by Lliane (score 2)

https://quant.stackexchange.com/a/36209

You can solve it with a 7-step binomial tree, maybe there is a shortcut to get to 4 ?

Start with the green nodes which is your final gain/loss, the previous gain/loss is just the mean of those two values, the previous bet is simply half the difference between those two child values (I haven't written those).

First bet is a 31.25 bet, then 31.25 again if you lose, 15.625 if you win.