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Here is the link to my simulation: (The data is based on using the Martingale betting system in European Roulette) https://docs.google.com/spreadsheets/d/1GH48faKeK5clonmYO6aySzchGeRhp7nx8Kc0cA0UFVA/edit?usp=sharing

As you can see, there is a linear growth rate with a few dips here and there, but overall, it seems to be profitable.

So back to my question, why don't rich people just make a simple program, hook it up to some online casino and make money?

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    If you run it long enough one of the dips will go down below zero, and then your billionaire is bankrupt.2017-01-18
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    Because rich people have more profitable and less risky ways of making money. Like stock, investments, real estate etc.2017-01-18

2 Answers 2

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Two main reasons:

  1. It is actually unprofitable in the very long run, no matter how large one's finite wealth is. In your example, the player is one loss away from going bust as early as round 17 (edit: well, seems to me the sheet rerolls the results on page load or something. Bummer). I would've guessed a bust to have happened in your experiment too, but the periodicity of pseudorandom number generators might play a part here.

  2. There is a big practical hindrance - casinos don't tend to allow arbitrarily large bets. If the betting cap for a single spin of the wheel is set, for example, at ten thousand dollars, that will nullify the billionaire's ability to take advantage of the Martingale strategy despite their virtually limitless wealth.

(A note about simulation methodology: given an infinite list of random win/lose results, we can always retroactively set the player's bankroll high enough, or cut the length of the sample, to make the Martingale appear profitable. This can happen easily by accident but is not an indicator of the strategy actually working while it will seem deceivingly promising)

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Let me give an answer in the context of Martingale Theory.

First, let us translate the setting into probabilistic notation. Let $Y_1,Y_2,\dots$ de iid variables with $$p=P(Y_1=1)=1-P(Y_1=-1),$$ where $p\in (0,1/2)$. Define the so-called martingale betting strategy: $$\phi(y_1,\dots,y_{n-1})=\begin{cases}2\phi_{n-1}, & y_{n-1}=-1\\1&y_{n-1}=1 \end{cases}$$ It's quite straightforward to show that in this setting, the capital at time $n$, defined by $$X_n=\sum_{k=1}^nY_k\phi_k(Y_1,\dots,Y_{k-1}),$$ is a (strict) supermartingale, so that the expected capital is (strictly) decreasing through time.

One may then wonder why is this system famous, or why one may even consider to play with this strategy. The answer lies in the fact that one may decide to look at the process only at specific times and this new "subsequence" of the initial capital process will be a submartingale! More precisely, if one decides to cash-out only when one has just won, i.e., at times $$\tau_1=\inf\{n\in\mathbb{N}:Y_n=1\}\:\:\:\tau_{k+1}=\inf\{n>\tau_k:Y_n=1\},$$ then the new capital process defined by $$Y_k=X_{\tau_k}$$ is a (strict) submartingale. In fact, $Y_k=k,\:\forall k$. (This explains -to some degree - the linearity of the top of your graph).

So now we have established that it is a feasible idea to use this strategy and then just cash-out at a winning time. The reasons for why this strategy can't be carried out in practice are:

  1. Casinos have an upper bound of how much money you can owe them, and if $(X_n)$ is replaced by a modified version of it which is bounded from below: $$X_n'=\min\{X_n,K\}$$ for some real number $K$, then actually $(X'_n)$ is uniformly integrable, which implies by Optional Sampling that for any non-decreasing sequence of a.s. finite stopping times $\sigma_1\leq \sigma_2\cdots$ the process cashing out at those instances, defined by $$Y'_n=X'_n$$ is still a supermartingale.

  2. Even if there was no money barrier, if $p$ is small, as it usually is, then the (geometric) times between successes have big mean and huge variance, such that a player may need to spend a ridiculous amount of time at the casino before he can leave at a favourable point.

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    @bof well-spotted typo. Thanks!2017-01-18