# Fair games and the martingale strategy III

Gambler’s Ruin

Continuing directly from the previous post, the nicest example of the optional stopping theorem we developed there is to example a simple random walk constrained between two values, say 0 and N. This represents an idealised gambling situation, where the gambler stops playing either when they reach some pre-agreed profit, or when they go bankrupt. We assume that we start at level k, for k = 1,2,…,N-1.

Naturally, we want to know the probabilities of winning (ie getting to N) and losing (ie going bankrupt). We could set this up by conditioning on the first step. Let $p_k$ be the probability of winning starting from level k. Then we must have

$p_k= \frac12 p_{k+1}+\frac12 p_{k-1},\quad k=1,\ldots,N-1,$ (*)

with the obvious boundary conditions $p_0=0, p_N=1$. In an ideal world, we just know how to solve second order difference equations like (*). Well, actually it isn’t too hard, because we can see from (*) directly that

$p_{k+1}-p_k = p_k-p_{k-1},$

and so $p_k$ is a linear function of k, and so $p_k = k/N$ follows pretty much immediately.

But, we can also use OST profitably. Let T be the time at which we first hit 0 or N. It’s intuitively clear that this should have finite expectation, since the problems you might encounter with just the hitting time of a single level shouldn’t apply. Or you can consider the expected number of steps before you see N ups or downs in a row, which certainly provides an upper bound on T. This random number of steps is sort of geometric (at least, can be upper bounded by a geometric RV) and so has finite expectation. So can apply OST to X at T, and we have

$\mathbb{E}[X_T] = N\cdot \mathbb{P}(X_T=N) + 0 \cdot \mathbb{P}(X_T=0) = \mathbb{E}[X_0]=k,$

from which we also derive $p_k=k/N$.

The reason we talk about gambler’s ruin is by considering the limit $N\rightarrow\infty$ with k fixed. After a moment’s thought, it’s clear we can’t really talk about stopping the process when we hit infinity, since that won’t happen at any finite time. But we can ask what’s the probability that we eventually hit zero. Then, if we imagine a barrier at level N, the probability that we hit 0 at some point is bounded below by the probability that we hit 0 before we hit level N (given that we know we hit either zero or level N with probability one), and this is $\frac{N-k}{N}$, and by choosing N large enough, we can make this as close to 1 as we want. So the only consistent option is that the probability of hitting 0 at some point is one. Hence gambler’s ruin. With probability one, ruin will occur. There’s probably a moral lesson hiding there not especially subtly.

So the deal here seems to be that if you just care about your average, it doesn’t matter how to choose to play a sequence of fair games. But what if you care about something other than your average? In any real setting, we maybe care about slightly more than this. Suppose I offer you a bet on a coin toss: you get £3 if it comes up heads, and I get £1 if it comes up tails. Sounds like a good bet, since on average you gain a pound. But what about if you get £10,003 if it comes up heads and I get £10,001 if it comes up tails? I’m guessing you’re probably not quite so keen now.

But if you were an international bank, you might have fewer reservations about the second option. My intention is not to discuss whether our valuation of money is linear here, but merely to offer motivation for the financial option I’m about to propose. The point is that we are generally risk-averse (well, most of us, most of the time) and so we are scared of possible large losses, even when there is the possibility of large profits to balance it out.

Let’s assume we have our simple random walk, and for definiteness let’s say it starts at £1. Suppose (eg as a very niche birthday present) we have the following opportunity: at any point between now and time t=5, we have the right to buy one unit of the stock for £2.

We want to work out how much this opportunity, which from now on I’m going to call an option, is worth on average. Note that now it does seem that when we choose to cash in the option will have an effect on our return, and so we will have to include this in the analysis.

Note that, once we’ve bought a unit of the stock, we have an asset which is following a simple random walk (ie sequential fair games) and so from this point on its expected value remains unchanged. So in terms of expectation, we might as well sell the stock at the same moment we buy it. So if we cash in the option when the stock is currently worth £X, we will on average have a return of £(X-2). This means that we’ll only ever consider exercising our option if the current value of the stock is greater than £2. This narrows down our strategy slightly.

This sort of option minimises the risk of a large loss, since the worst thing that happens is that you never choose to exercise your option. So if you actually paid for the right to have this option, that cost is the largest amount you can lose. In the trading world, this type of opportunity is called an American option.

The trick here is to work backwards in time, thinking about strategies. If at time t=4, the stock is worth £1, then the best that can happen is that it’s worth £2 at time t=5, and this still gains you no wealth overall. Similarly if it’s worth £0 at time t=3. So we’ve identified a region where, if the stock value enters this region, we might as well rip up our contract, because we definitely aren’t going to gain anything. Remember now that we’ve also said you won’t ever cash in if the stock’s value is at most £2, because you don’t gain anything on average.

Now suppose that the stock has value £3 at time t=4. There’s no danger of it ever getting back below £2 during the lifetime of the option, so from now on your potential return is following the trajectory of a simple random walk, ie a fair game. So on average, it makes no difference whether you cash in now, or wait until t=5, or some combination of the two. The same argument holds if the stock has value £4 at time t=3 or time t=4, and so we can identify a region where you might as well cash in.

What about the final region? If the stock value is greater than £2, but not yet in the definitely-cash-in area, what should you do? Well, if you think about it, the value of the stock is a fair game. But your return should be better than that, because the stock price doesn’t take account of the fact that you wouldn’t buy in (and make a loss overall) if the value drops below £2. So at this stage, your future options are better than playing a fair game, and so it doesn’t make sense (in terms of maximising your *average*) to cash in.

Now we can actually work backwards in time to establish how much any starting value is worth under this optimal strategy. We can fill in the values in the ‘doomed’ area (ie all zeros) and on the ‘cash in now’ area (ie current value minus 2), and construct backwards using the fact that we have a random walk.

The final answer ends up being 7/16 if the stock had value £1 at time 0. Note that the main point here is that working out the qualitative form of the strategy was the non-trivial part. Once we’d done that, everything was fairly straightforward. I claim that this was a reasonably fun adjustment to the original problem, but have minimal idea whether pricing options is in general an interesting thing to do.

Anyway, I hope that provided an interesting overview to some of the topics of interest within the question of how to choose strategies for games based on random processes.