I'm starting this thread because I worry that someone may misconstrue something that heehaww pointed out elsewhere and perhaps go on to do something foolish. Also what I am about is counter intuitive and may start an interesting discussion.

The martingale system for even money bets means you start off betting, say a dollar, and keep doubling up after each loss and go back to a dollar after a win. When you do win, your bankroll has increased by one dollar compared to when you started the sequence.

If you are betting on something that is less than 50% and have a limited bankroll you will eventually hit a streak so bad that it will wipe out your staring bankroll plus all your accumulated dollar wins. (That would not necessarily happen if you were over 50% because your accumulation pace outruns bad luck chances.)

But if you have an unlimited bankroll each one dollar increase must eventually happen. Even if the bet is well below 50%. (As long as it remains the same odds throughout.) So, as heehaww pointed out, you are essentially 100% to reach any goal you set for yourself. With an unlimited bankroll you will eventually be a trillion dollars ahead betting on 49% shots.

But what about the principle that a group of bets that are all negative EV has a negative EV? This seems to violate that.

But it doesn't. Here is why:

If George has an unlimited bankroll and embarks on the martingale system for 49% shots and you interrupt him and ask him how he is doing overall at that precise moment, whether it be after one flip or a googol flips, his possible answers average out to a negative number. Do you see why?

I'm going to elaborate a little. Suppose I decide to check on you after your millionth flip. At that point you will most likely be up about $490,000. Every one of your winning flips moves your bankroll up one dollar, (compared to where it was after your previous winning flip).

If your millionth flip was a winner, you are at a new peak as to how much you are ahead.

But if your millionth flip was a loser, you will be below your peak. Well below if you are in the midst of several consecutive losses. It is even theoretically possible that the losing streak has temporarily wiped out your profit to that point. This of course doesn't bother you because when the losing streak ends you will have overcome the terrible loss and be at another new peak. But if you knew I was lurking nearby and might interrupt things you might feel differently.

Isn’t this what Phil Laak means when he answers:

I’m up...

Or

I’m stuck...

Or

I’m upstuck...

I should add one more thing. The inevitability that somewhere in your adventure you hit a streak that wipes out your accumulated gains does NOT apply when you are betting on 51% shots rather than 49% shots. A losing streak of any size will eventually happen but even with an infinite number of flips it is not certain that any of those losing streaks wipes out you win up to that point and your overall EV martingaling good bets is positive, just as you would expect.

Nowhere has unlimited table limits so the point is moot.

On infinity tries does it even matter what the odds are?

Like martingaling on something like a 0.00..1% chance of a win per flip (however many 0s you wanna add as long as it's not infinitely small.) Would you not reach the same "being up a trillion" at some point if you martingale flip infinity, and if not what is the impossibility threshold for the odds.

There is no threshold. The OP was not written for gamblers but rather for the ten percent of the population who become concerned that something is going on that seems to break a law of logic. Or to put it differently, 80% of those who have had, or have a chance to make a major impact on the world.

Of course, if your bankroll is infinite and your starting tier 1 wager is finite, your return on your capital must be 0%, no matter how many you win or lose.

But yes, Martingale is a terrible way to throw your life away.

Why the Martingale strategy seems to work
There is a very nice logical explanation why Martingale should work. If you keep doubling your initial bet, the roulette ball must sooner or later land on a number which grants you a win. Right?

Yes, that’s true. With the number of attempts approaching infinity, the probability of overall failure approaches 0. That means that with infinite resources and no limitations from the casino, you will eventually manage to win and finish a cycle with net winnings of $1.

With infinite resources and no limitations from the casino, the entire system would work. However, as you have probably already guessed, these two conditions are never met in the real world.

source: Roulette scams strategies

So i just saw this thread - and I tried to formalize it but again seem to be coming up with 2 diverging results
Start off betting B on a game with prob of winning p which returns 2x the bet (ie get your money back + same again)

Each time you lose you double the bet

Eventually the amount you win (after n iterations if the starting bet is B) is B.2^n (as you have started betting B.2^0 and most recently bet B.2^(n-1)). Your total of bets placed is B.Sum{k=0 to n-1}2^k = B.(2^n -1) à ie you net return a profit of B

And then start over and keep repeating this strategy ad infinitum



So I thought about the expectations of a single cycle. Assume p = probability of winning a game and q = 1-p losing + each game is independent.

Now expected value of the strategy for a single cycle (play til winning) is Expected Profits – Expected Wager amounts

[Couldn't get spacing to work here so i've just color coded stuff instead]

Period Wagers Profit

1 B.2^0 [with certainty] B.2^1.p

2 B.2^1.q B.2^2.pq

3 B.2^2.q^2 B.2^3.pq^2 etc

…

n B.2^(n-1).q^(n-1) B.2^n.pq^(n-1)



Sum of wagers = B.Sum(k=0 to n-1)(2q)^k = B((2q)^n -1)/(2q-1)

Sum of profits = 2B.p.Sum(k=0 to n-1)(2q)^k = 2B.p.((2q)^n-1)/(2q-1)



Difference = B.((2q)^n-1)(2p-1)/(2q-1)



Now if q = p – 0.5 => expected value isn’t defined but looks like it would limit to 0

If q > 0.5 (ie probability of winning 1-q or p is less than 50%) expected value gets increasingly negative (goes to –ve infinity)

If q < 0.5 (probability of winning is greater than 50%) then the limit converges to B

*** However ***
Probability of winning at least once in n bets is 1 - probability of losing each bet = 1 -q^n
since q < 1 (ie there is some probability of winning however small) then 1-q^n --> 1 you will always win on a long enough time horizon

Sure enough when you construct this in computer sim every return = B and is independent of value of p or q

So now I'm not convinced of the maths above....and think it has something to do with the fact that the sequence stops on a condition (so the standard sum n from 0 to infinity unconditionally approach might not be right).

any thoughts?

If you have an "infinite bankroll" then it seems pointless doing this anyway as you'd be no better off after you win...

Juk

Yes.....but then make the same argument with
- no bankroll
- but infinite credit for zero interest payments

then you would be interested.

In any case - i spoke to a financial quant who said the idea of expectations in a strategy like this doesn't make sense (the strategy can't be considered arbitrage free because it depends on infinite access to money ........... apparently in mathematical nomenclature this is considered 'non-admissable').

My guess is that is the same paradox at work in the St Petersburg problem
- keep flipping a coin (fair)
- game stops when it comes up heads for first time
- you get paid 2^(n-1) where n is number of flips
- prob of it being heads for first time at n is 2^-n
- so expected value = sum{i=1 to infinity}(2^(n-1) * 2*-n) = sum{i=1 to infinity} (1/2) = infinity

i tried it on 100million sims on only got expected value of 15-20

its also strange as the value grows as prob of coming up heads goes down but there is a discontinuity it doesn't tend toward infinity as pr of heads -> 0 because at 0 the 'value' of the structure has to be 0 [can't ever payout]

With q>p, the EV becomes more negative the more credit you have, except there's a discontinuity "at" infinity, where the EV of a sequence is equal to +(initial bet) because with infinite credit, you have a 100% chance of winning. The infinitesimal chance of eternally losing has zero measure. If you want to delve into the weeds, read about measure theory.

This is a really interesting thread and I think exposes a common problem of the use of the number infinite. I think it's perfectly fair to think of the concept of an infinite bankroll, but where the disconnect comes from is not also being able to conceptualize the likewise possibility of being able to lose an INFINITE number of times in a row. With a fair definition of infinite applied to both bankroll and statistical probability should give a clear indication of how ROI is indeed still negative. I think the main problem with OP's insertion is that if we are accepting the possibility of an infinite bankroll at face value while falsely rejecting the equal possibility an infinite downswing in the same breath. It's easy to reject the second because in our ordinary lives we deal with real numbers and see there's always an end to things, but then we thus shouldn't be so quick to accept the concept of an infinite bankroll, as this also of course can't be comprehended in the real world.

In fact, it's ironically the same inherent flaw with the original martingale problem. For example, I have a million-dollar bankroll and as a mere mortal, it seems a no brainer to place a starting bet of $1 with this strategy, because the concept of losing 20 flips in a row to lose my entire bankroll is so foreign in everyday life that my brain can't possibly imagine that it could actually happen the first time I go out and decide to take flips until I win back $1.

Well, if I somehow get my hands on an infinite bankroll, the mere amount risked is still statistically a NEGATIVE ROI investment because there is indeed a non-zero chance of losing an infinite amount of flips in a row. Any rejection towards this thought process is incorrectly pulling the sum of infinite out of the conceptional realm yet incorrectly keeping the idea of losing an infinite amount of times in a row back to real-world assumptions. This is incorrect-- if we can accept the concept of an infinite bankroll, we also must accept the concept of losing an infinite amount of flips.

What's the chance of losing a billion flips in a row? Very low but non-zero
What's the chance of losing an infinite number of flips in a row? Ridiculously absurd, but with an infinite # of trials, still non-zero.

I am just a random kid so feel free to correct any mistakes I made. I thank you for these threads, thinking about this stuff is making me want to major in statistics.

(1/2)^∞ = 1 / 2^∞
2^∞ = ∞
1 / ∞ = 0

What you've stumbled upon are the probabilistic concepts of "almost never" and "almost surely", of which repeated coinflips are a textbook example. The set of infinite sequences involving no tails is non-empty, hence the need to say "almost", but the probability of a sequence from that set is still zero.

If one attempts to assign any real nonzero probability to each individual sequence, the total probability will exceed 1. No can do.

However, in the nonstandard approach to analysis (where infinitesimal "hyperreal" numbers are used), it may be possible to have infinitesimal probabilities. A free and good-looking paper from 2016 makes the case: https://academic.oup.com/bjps/article/69/2/509/2669779

What might we do with that information? I don't know. To me, saying, "My chance of flipping infinite heads is infinity to 1 against, or epsilon" does not sound any more optimistic than "zero". Both convey the same thing: call it zero, call it epsilon, call it Santa Claus, either way it simply won't happen.

Here is another reason why trying to think about what happens when our bankroll is infinite breaks our intuition:

When we have a finite bankroll, keeping track of how much we have won or lost is equivalent to knowing our starting bankroll and our remaining bankroll. With an infinite bankroll, it is not. When we are playing for finite bets with an infinite bankroll, our bankroll remains infinite no matter how much we have won or lost or no matter how much we bet. This reduces finding the EV of a Martingale to a paradoxical futility.

No such thing as infinity

Imagine trying to increase your bankroll when it's already infinite.

Bet entire infinity on one flip.

win the flip (important).

profit

Actually it’s not important, and you don’t profit. Infinite sets always admit a disjoint union of infinite subsets, so you can lose a flip for infinite money and keep your bankroll the same size. Also, if you win the flip, your bankroll stays the same size.

Agreed. It’s just a variant on the Hilbert Hotel problem. For thoose unfamiliar, you go to check into a hotel with a (countable) infinite number of rooms. The clerk says “Sorry, we have no vacancy). You correctly argue that this is untrue since the guest in room 1 can be moved to room 2. The guest in room 2 can move to room 3, and in general for each N the guest in room N moves to room N+1, and hence the clerk can open up room 1 for you. The same works if you come with a group of P people, and continues to be true even if P is itself infinite.

The point is that the union of any finite or countably infinite number of countably infinite sets is itself countably infinite. There is no way to increase the cardinality of a countably infinite set (such as the set of chips or bills comprising your bankroll) by adding to it another finite or countably infinite set (such as the set of chips you win). You could also lose an infinite amount, for example by numbering each chip in your infinite bankroll and betting only the even numbered ones, and still keep an infinite bankroll.

If the infinite set of dollars is countable, then how does one counter the following argument: what if you start with a $2 bet, double it after each win (anti-martingale) and achieve an infinite streak of wins? You'd have 2^∞ dollars, and since the infinite streak of wins is countable, wouldn't that be the same as 2^ℵ0 ? And if so, that would mean that you've increased the cardinality of your wealth and it is now uncountably infinite (given that the cardinality of the powerset equals |ℝ|).

I'm pretty sure it's a flawed argument, and the error must be in equating "numerical ∞" to ℵ0, but why aren't they the same here?

(I realize that the probability of an infinite streak is zero, even with infinite start-overs, but humor me anyway.)

That’s not where the flaw is. The flaw is that you’re trying to apply a limit computation rule that only applies to finite variables to an infinite variable.

Even humoring you about the infinite streak, you’re taking a limit of finite numbers. That limit must be either finite or countable. In this case it is countable.

e^x-e^x=0

that's how I see it

You are proposing a sum of a countably infinite number of finite terms. Each term in your sum increases without bound, but each term is still finite. The sum of an infinite number of finite terms can either be finite or countably infinite; it can never be uncountable.

The mistake you make is equating the sum to 2^(aleph null). We have to be careful with infinite cardinals, you can’t just substitute them into formulas as if they were finite numbers. In this case, your sum does not contain a term equal to 2^(infinity), so the sum is countable. The sum 2^1+ 2^2 + ... indeed is itself equal to aleph null.

(Apologies, but I’m posting on a mobile and can’t render the mathematical symbols; hopefully everything is clear enough)