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In which by using modern technology and literary devices, various details are given for systematic speculation in what Julius might have called the ludi fortium and in which are discussed certain apparent paradoxes in the hope of furthering the cause of truth, justice, and the American pocketbook and losing your shirt in a rational manner.

Everyone knows that Arnold Rothstein (1882 - 1928), known to his many acquaintances as "The Brain", was famous for his "bankrolling". That is, if you needed some dough for some - ah - "enterprise", you'd go up to Arnold and explain your needs. He'd reach into his pocket and pull out a thick roll of bank notes and hand it over. Of course, you had to pay Arnold back and at interest¹, but that was the way Arnold did business.

So suppose we have the following - and purely literary - scenario. Arnold was going about his business - as we see above with his friends Alvin "Titanic" Thompson and George "Hump" McManus - when he ran into an old acquaintance named Gamblin' Joe Blow.

You see, Gamblin' Joe considered himself something of a mathematical genius, and he was able to construct a copy of Charles Babbage's "difference engine". With the "engine", Joe came up with a strategy for betting in various games of chance.

Charles
He invented the engine.

On the first bet, Joe said, you you'd bet the minimum allowed (all gambling tables have a minimum). Say it was a $1 bet on red at the roulette wheel. Then if you won you'd pocket your winnings and again bet the minimum.

But if you lost, you double the previous bet. That way if you lost your first bet, the next bet would be $2. If you then won, with the $1 lost on the previous bet and your $2 win, you would be ahead by $1. You'd then go back and bet the minimum. You would continue with this betting sequence, playing the minimum after a win and doubling your bet after a loss.

It was impossible, Joe claimed with a smile, for him to lose. By doubling your last bet each time your lose, once you won then you would wipe all previous losses and increase your bankroll by $1.

Since you're betting red, Gamblin' Joe nodded - remember he was a mathematical whiz - you'll win 47% of the time. Therefore you'll have a net gain of $1 every 2.1 spins of the wheel. And as seeing is believing, Joe provided a graph of the expected wins.

Expected Wins of
Joe's "Double Up" Method

So at the start of a game, your bets might run like:

Sure enough, with each win your bankroll increases by $1.

Of course, Joe pointed out that there were streaks of bad luck. But even if you lost 6 times in a row that would only be $63. So if Arnold could supply the "ready" to cover such losses, once Joe won the next spin Arnold could get his dough back and Joe would show a profit.

Once more to emphasize his argument Joe used the difference engine with what are known as resampling or "bootstrap" statistics to plot out a - quote - "typical run" - unquote. But even with a string of losses, Joe kept making money.

Joe's Bootstrap Graph

So what he'd need, Joe said, was enough of a bankroll to cover any early losing streaks. Then after a while Joe would have earned enough so he would have a cushion against future losses. If Arnold could hand over a good stake, say a grand, Joe would be able to win - as the mathematicians say - in an increasing monotonic linear manner.

"Nihil Sub Sole Novum"
 - Arnold Rothstein Ecclesiastes 1:9

Of course, with his own experience Arnold immediately recognized Joe's - quote - "amazing betting method" - unquote - was nothing new. It was, in fact, simply the well-known Martingale System. The Martingale System is one of the oldest "progressive" gambling systems and every gambler knew about it. But Arnold liked Joe's moxie and handed him the grand.

In any case fortified with his system (and Arnold's grand) Joe hied over to Bugsy and Virginia's Gambling Casino. He quickly found a table with a $1 minimum and began to play.² Naturally he kept tabs on his winnings.

Bugsy and Virginia.

Although Joe's increase was modest, for the first hour and a half he saw a fairly steady increase.

So things still seem to be working.

"The world of reality has its limits, the world of imagination is limitless"
 - Jean-Jacques Rousseau

But suddenly Joe found himself loudly and teemingly expressing surprise at what happened next. He began loosing - a lot.

In fact, he kept on losing until his brankroll dropped to $900.

Now he was $100 short of his initial bankroll! That wasn't according to plan.

Well, since he lost on his last bet all he could do was to double up. As his last bet was $64 he now laid down $128.

But then Joe comes face to face with the reality of casino gambling:

Spin	Bet ($)	Outcome	Spin Win	Bankroll ($)	Comments
67	128	-	-	900	ERROR! ERROR!

Instead of covering the bet, the dealer turned to Joe more in sorrow than in anger.

"Sorry, buddy," he said. "You can only bet $100. That's the limit at this table."

Yes, all casino bets have not just minimum bets but betting MAXIMUMS.³

So here's Joe's dilemma. If he bets the $100 and wins, he'll still be $28 short of recouping his losses. So the idea that on each win you'll always increase the previous high bankroll by $1 isn't correct.

Well, at least if Joe wins, he'll be back with his original $1000 bankroll. So he lays $100 on red.

But Joe's (bad) luck continues:

At this point, Joe pauses - not for refreshment - but to plot out his wins and losses.

Joe's First Session
(Continuing)

Hm. After 70 spins - two hours at the table - far from seeing his steadily monotonic linear increase in winnings, Gamblin' Joe has lost a bundle - half his (or rather Arnold's) bankroll.

But at least Joe still has money and so he keeps betting and - thankfully - on the next spin he wins.

Spin	Bet ($)	Outcome	Spin Win	Bankroll ($)
71	100	WIN	100	600

But now he has yet another dilemma. There's no way a single bet will recoup his losses. So there are two ways to proceed. He can keep betting the maximum until he recoups his losses. But this way in five rolls he might lose everything.

Or he can start the betting sequence again starting out with his $600 bankroll. That is, he'll first bet $1 and then when he wins he'll again bet $1 and double up if he looses. Prudence dictated this more economical option and thankfully Joe's bankroll began to slowly climb back up.

Joe's New Session
(Starting Afresh)

So after another six or so hours at the wheel, Joe's bankroll has upped to over $670.

Perhaps with just a couple of more days at the wheel, Joe will be back at $1000.

And indeed Joe's luck continues. His bad luck, that is.

Now what does he do? He can't even bet the maximum - $100. And if he bets all he has left and wins he'll only have $90. And if he looses he'll be wiped out.

And sure enough the last spin is:

Joe now looks at his progress ...

Joe's Final Session

... and checks when the next boat to Rio sails.

"The roulette table pays nobody except him that keeps it."
 - George Bernard Shaw, Maxims for Revolutionists

Curmudgeons, of course, will point out all this brouhaha is merely a fictitious and contrived example aimed at warning bettors away from gambling systems. After all, a single example does not general principals dictate.

Instead what if Joe had a lot of partners all loaded up with $1000 bankrolls? And let the examples be dictated by the true random spins according to the odds of roulette. Only then can we see how such a betting system really fares.

That is a reasonable request and can be realized using the modern equivalent of Charles's difference engine. So we'll have Joe team up with Bob, Freda, Henry, Mia, Willow, George, Harry, Olivia, Florence, Joseph, Leo, Archie, Oscar, Isla, Ivy, Arthur, Noah, Lily, Ava, Oliver, Alexander, Amelia, Martin, Mary, Edward, Anne, Catherine, and Arthur. Fortified with their bankrolls everyone begins to play:

And voilà!

A Little Help from Joe's Friends

Note that here every single one of Joe's friends - without exception - also went broke. And there was no fiddling with the sequences here. The play of everyone was what would be seen by true random spins of the wheel. Repeat: EVERYONE went broke playing the Martingale System.

So despite Joe's early analysis, none of the players - that's NONE! KEINS! NICHTS! - had a monotonic linear increase in winnings. Although some of the players did have an initial increase in their bankroll none so much as doubled their money. Instead all the winnings showed an overall downward trend until the players lost everything.

But it's not actually the downward trend that raises eyebrows. It's the amount of the downward trend which is - as Artie Johnson might have said - verrryyyyyyy interesting.

Further elaboration will require a slight digression. Let's look what happens if Joe and his friends simply play only $1 a spin and keep track of their bankrolls. Playing at casino odds and payoff, it's clear that the casino will gradually whittle their money down.

Joe and His Friends
Playing the Minimum

Now if you do what's called a least-squares linear regression on the points, the line has a slope of -$0.0526 per spin. In other words, the players are losing 5.26¢ on each turn.

Joe's and His Friends
(Regression Fitting)

Students of the mathematics of gambling will immediately notice that the downward slope of 5.26¢ is the same as what gamblers call roulette's house percentage. Also called the house advantage, the rake, the juice, the cagnotte, or the vigorish ("vig" for short, a great Scrabble word), the house percentage is the average rate of loss for each bet.

The vigorish arises because the casinos pay the bets off at less than correct odds. In other words, when you bet $1 on red and win, the casino pays you $1. But they should pay you $1.11.⁴

                               Player's Odds
       Payout($) =         × Player's Bet
                               House Odds

Now if you take the regression line and overlay it on the graph of Joe and his friends playing their system, it actually fits the data, maybe not exactly but still pretty nearly as Isaac Newton would have said. So in addition to the fact that you can't always double up, the fact that the house doesn't pay the player what it should seems to be another contributing factor in their losses.

Joe's and His Friends
System Playing

OK. The empirical evidence seems to show that Joe and his friends will eventually lose. But then how was Joe able to theoretically prove to Arnold that he should win? We've always wanted to know that.

No doubt you have, as Captain Mephisto said to Sidney Brand. It's very simple really.

It's very simple.

The graph that Joe showed to Arnold was based on the idea that no how many losing spins he had in a row he had, he could always double up and so recoup his losses. But because of the betting limits, you can't always double up. So Joe's graph does not represent the real world.

And although it's not immediately obvious, Joe's theory also requires he has an unlimited bankroll. With unlimited dough, no matter how long a losing streak he has, he would always be able to bet enough to recoup his losses. But in the real world where everyone has a finite amount in the kitty - no matter how large - Joe will eventually have to bet so much that he will loose it all.

Yes, yes. We know that theoretically Joe can loose so many bets in a row that no matter how big his bankroll he'll get wiped out. But we want a realistic assessment. Not some preachy sermon on the hazards of gambling. So just how likely is it really that Joe will have those long sequence of losses in a row?

One thing about randomness is it seems to have a lot of unexpected order. What shouldn't happen seems to happen more often than you might think. So let's look at another betting session that is based on Joe's theory - and a session that actually does come about when performing computer - that is, the "difference engine" - simulations of playing the wheel:

As a picture is worth a thousand words, we can summarize this session in a nice neat graph.

Unlimited Bankroll

So yes, in our example Joe could recoup his losses. But he would have had to go to Arnold and borrow over half a million smackers to do so. Needless to say Joe didn't show this graph to Arnold.

Again the curmudgeons object. It's all well saying you see a long string of losses more than you expect. But c'mon! In this - quote - "example" - unquote - Joe loses 19 times in a row, for crying out loud! How likely is that?

Now odds makers tells us this will happen only about once in every two hundred thousand spins of the wheel, for Pete's sake! What we need to know, though, is just how many successive losses can Joe expect in a typical session at the wheel and how much can he expect to lose?

Well, to answer this question we need to first calculate the expectation value - that is, the average number - of consecutive losses. That is, what will be the average number of sequential black and green numbers (0 and 00) with no red numbers?

First of all, the probability that any spin will be a losing spin of length 1 is simply ²⁰/₃₈. That's just the probability that you will lose a single bet. This probability is simply the sum of the 18 black numbers and the two green numbers divided by the total numbers appearing the wheel.

So the odds of Spin #1 being a run of losses of length 1 is:

... or 52.6%.

OK. What about getting two losses in a row? Well, that's the probability of getting one loss in a row followed by another loss. So you multiply the probability of one loss by ²⁰/₃₈.

And for other successive losses we keep up the process:

So in general the probability that you'll have a run of losses of length i is:

Now we need to cite another and well known rule of mathematics. And that's for any statistical property (or random variable), X, the expectation value of the property, X, is the sum of the product of all possible values of the property at value i, X_i, times the probability of that X_i will occur, P_Xi.

Now note that in our case, there is no limit to the number of spins, (i). That is, all possible values is infinity (∞). So we multiply the length of the run with the probability of that length and sum them up and we get the expectation value of the length of the losing runs - which we call L - is:

... which is a little more comprehensible if it's expanded as:

Now at this point we can pause and look at some specific examples of the sums.

Hm. Most interesting. It seems that as you spin the wheel more and more, the expectation values of sequential losses, L, hits a constant number. That is, the summation converges. So the average number losses before you get a win is a rather small value.

OK. But these are merely specific examples of average sequential losses for a finite number of spins. Is it possible that if we continue with more spins eventually the numbers do get larger and larger? Is it possible that the average number of consecutive losses is eventually infinite?

So we really DO need to calculate the sum over all possible values.

True it's rather tedious - and actually impossible - to individually sum up an infinite number of terms. Instead we introduce the notion of the partial sum of a series.

A partial sum is a formula that gives us the total sum of a given number of terms in one nice neat formula. For instance, suppose you have the summation:

... which is:

S_n = 1 + 2 + 3 + 4 + ... + n

Clearly summing up a lot of numbers, say the first 65,537 positive integers, would be a rather tedious task. But the formula for this partial sum is simply:

So the sum of the number from 1 to 65,537 is:

Now if we have the summation:

... which can be expanded as:

This, in turn, is similar to a famous series:

... which with a little middle school math can be shown to have the partial sum:

Now notice that if we let ²⁰/₃₉ = a = x then our original series is:

... which is 1 less than the series where we have the partial sum. Conveniently the partial sum we just calculated has the number 1 at the end. So we just lop that off and the partial sum of our series is:

Remember that we want to calculate the sum of all possible terms, and in our case that's an infinite number of terms. And to calculate the sum of the infinite number of terms we must take the limit of the partial sum as n goes to infinity.

Now delving into a textbook of calculus we learn that:

And that:

... which fits our series where x = a =²⁰/₃₈

So with minimalistic algebra we have:

Putting the numbers in the formula for the partial sum, we have:

... and ...

... which is:

... and hey, presto!, we know that:

... which fits the numbers in the previous table.

So the average number of times you'll lose in a row is between 2 and 3 spins of the wheel. It's left as an exercise for the reader to show that the average number of wins in a row is ³⁴¹/₂₀₀ = 1.71.

Well, say the optimists, with only 2 to 3 losses expected in a row, doesn't that mean we can expect to lose less than $10 before we win? So our bankroll of $1000 oughta be plenty to tide us over.

Sounds good.

Except that Joe and his friends all lost their shirts.

"I understand"
 - Manhunt of Mystery Island, Republic, 1945

OK. We've seen the average number of losses in a row is between 2 and 3 spins. That's the expectation value of the sequential losses.

But what is the actual expectation value of the amount of money lost due to sequential losses? Didn't we guess that it shouldn't be much more than $10?

Once more we'll use the partial sums of losses. Then we'll look at the the moolah risked before we win the next dollar.

Remember the probability of a given number of losses is:

Also remember that you double the amount you bet when you lose. So the doubled bet times the probability of the number of losses:

A little more algebra gives us:

So the summation of the first n spins is:

... and this is:

Hm. This looks familiar. It is in fact our old friend.:

... where x = ²⁰/₁₉.

So it should converge nice and neatly to a small number, no?

Well, we must be honest and say:

NO!

Eins mehr let's make a table of the data. Remember we don't use the formula for the infinite summation but we have to use the formula for the finite partial sums which is:

... where x = ²⁰/₁₉.

So putting the right numbers in the right place we end up with:

Sacrebleu! (to quote Zola). How can such a formula which earlier produced a nice small and convergent number keep going up? Ce n'est pas possible!

Well, it is pose-ee-blay. Remember that our partial sum converged only if the limits of (n + 1)aⁿ and naⁿ both converge to 0. But that happens only if a < 1.

But when calculating the money lost, a = ²⁰/₁₉. This means a = 1.053 - which is bigger than 1. So as i gets bigger (which sounds like horrible grammar) this new summation grows without limit. That is the summation "blows up" as the mathematicians say.

Remember that the intuitive feeling was that since the since the average number of losses between wins is only around 2.34, we wouldn't even lose as much as $10 before we recoup our losses. But according to the numbers we're going to lose a lot.

How much? Well, take the limit of the formula for partial sums where n go to infinity.

And without going into tedious details we can prove:

... which means that for a sequence of losses - and pardon us if we shout -

THE

AVERAGE EXPECTED

LOSS FOR THE

MARTINGALE SYSTEM

IS

INFINITE!

"If you shall fright me into labor and concentration, I shall win my game."
 - Ralph Waldo Emerson, The Correspondence of Thomas Carlyle and Ralph Waldo Emerson.

OK. We can loose it all. But how much can you win?

Well, that's easy. As long as you have your unlimited bankroll and can always double your bet, no matter how often you loose, once you do win you'll recoup your losses and come out $1 ahead. So whenever you win you win a dollar. So the expectation value of winning following a sequence of losses is $1.⁵

So as Flakey Foont asked Mr. Natural, what does it all mean? Well, not Mr. Natural's famous reply. It does mean that with using our betting strategy, then

THE

EXPECTED

MONETARY

RISK

FOR THE

MARTINGALE SYSTEM

TO WIN

$1

IS

INFINITE.

And In Conclusion
 - William Shakespeare

Now all together:

   A

NECESSARY

CONDITION

FOR THE

MARTINGALE SYSTEM
TO WORK

IS AN

UNLIMITED
BANK ACCOUNT.

... and ...

   A

NECESSARY

CONDITION

FOR THE

MARTINGALE SYSTEM
TO WORK

IS

NO BETTING LIMITS.

But you say even if - as the Spartans told Phillip - Joe can't sit at the table with an infinite amount of money to cover the losses he can have someone slip him an extra grand if his initial bankroll goes bust. So at least in principle he can keep on playing.⁶

And yes, what we see is interesting. Suppose we have to live with the betting limit but even if we don't exactly have an unlimited or infinite bankroll, we have friend who will alway lend us another grand if we go bust.

We then we start to bet. We also keep tabs, not just of our bankroll which is being replenished if necessary, but our total winnings. That is, how much dough we have taken from the casino.

And what we find is:

Joe's Replenished Bankroll
Betting Limits: $1 to $100

Hm. The amount we "win" is a negative number - which mean we're losing - and our "winnings" keep going down - which means our losses get bigger! So even if Joe has the potentially unlimited bankroll where he can go to replenish his losses, he doesn't really win. As he keeps playing his losses - and his debt to his benefactor - get bigger and bigger.

A Most Amusing Paradox
 - Gilbert and Sullivan (The Pirates of Penzance)

In the end the Martingale system comes off as something of a paradox. In an imaginary world with no betting limits and an infinite bankroll - you can always turn a profit even if the payout is at casino odds. But in the real world, the combination of having a finite amount of money and betting limits - not to mention the House's vigorish - means you'll eventually go broke.

But it's never good to leave on a negative note. We can always find a few bon mots suitable for the topic.

... and finally there's:

References and Further Reading

The Mathematics of Games and Gambling, Edward Packel, The Mathematics Association of America, 1981.

Understanding Probability: Chance Rules in Everyday Life, Henk Tijms, Cambridge University Press, 2004, (Third Edition, 2012).

Scarne's New Complete Guide to Gambling, John Scarne, Simon and Schuster, First Edition (1961), Second Edition (1974).

The Big Bankroll, Leo Katcher, Harper Brothers, 1958.

The Unsinkable Titanic Thompson, Carlton Stowers, Eakin Press, 1982.

The Odds Against Me, John Scarne, Doubleday, 1966.

The Weekend Gambler's Handbook,, Major Riddle, Random House, New York, 1963.

An Introduction to the Bootstrap, Bradley Efron and R.J. Tibshirani, Chapman and Hall/CRC Monographs on Statistics and Applied Probability, 1993.

Emile or On Education, Jean-Jacques Rousseau, Amsterdam Néaulme, 1762.

Maxims for Revolutionists, George Bernard Shaw, Archibald Constable and Company, Ltd., 1903.

The Correspondence of Thomas Carlyle and Ralph Waldo Emerson, 1834-1872, Thomas Carlyle and Ralph Waldo Emerson, Houghton-Mifflin, 1883 - 1884.

"When the U.S. Air Force Discovered the Flaw of Averages", Todd Rose, Toronto Star, January 16, 2016.

De Garrulitate, Plutarch, Perseus, Tufts University.

Diamonds Are Forever, Ian Fleming, Jonathan Cape, 1959.

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