I just had a conversation with my dad, who is well versed in maths in general. He understands that you can't turn a negative expectation game into a positive expectation game using betting systems. However, he claims that martingale (e.g.) will work on a 0 expectation game e.g. flipping coins.

My claim is that you cannot change the expectation of a game at all by using betting systems of any kind - i.e. if the game has an expectation of +5, 0 or -5, that is invariant under whatever bets you choose to make on any given trial of the game. The more I think about this, the more I realise that I am just taking it on trust and have no idea how it's proved.

Can anyone direct me to where I can find a proof of this (or correct me if I'm wrong)?

For any finite bankroll your intuition is correct and you can simply sum the series and it will have an EV of 0. For a large enough bank roll you may win 1 bet 99% of the time, but the EV is still 0 because you lose everything 1% of the time. If you don't restrict yourself to finite bank rolls and number of wagers, you can argue that the game can have positive EV, but that's sort of philosophical more than practical. Here is a famous example on those lines.

https://en.wikipedia.org/wiki/St._Petersburg_paradox

So I would say you're both possibly correct, depending on what constraints the game has.

I actually used that as an example in our conversation, and also the coin flipping thing where you should theoretically pay infinite money but for practical purposes your ev is about 20 bucks (can't remember right now specifically what that one is called).

I did start talking about how reciprocals of powers of 2 sum to a finite value and all that, but I don't think that really covers it. I want to show that for any betting system you can't change the EV of the game.

As a toy example, let's say we flip coins and I come up with a system that after I lose I bet x and after I win I bet y. I want to show that after N flips, my EV is still 0 in this game.

When I said "he's a bit handy at the maths", I might have understated a bit. He has a PhD in structural engineering, and he used to compete in math olympiads in the 70s. I still think he's wrong on this, though. You can sort of see why I might be second guessing myself.

Structural engineers typically know nothing about probability. They just memorize formulas. The most dangerous gamblers are ones who feel themselves appropriately educated when they aren't.

Oh lol, I thought that was the two envelopes thing, that is the coin flipping thing haha. Sorry, I was in the middle of a bunch of stuff and didn't click the link!

Well, he knows enough to say "prove it", and if I send him a proof that doesn't make sense he will know its wrong. I mean, it seems to me that for N bets in independent trials your EV should just be the sum of the EVs of the individual bets, and that's that. Am I overthinking it?

Look up the definition of EV. Show that the EV of N flips is the sum of EVs of the N individual flips. Show that the EV of each individual flip is zero, no matter how you scale the bets.

Each of these should be trivial.

Post this in the probability forum. They love 100% of martingale question threads, especially those that involve infinite bankrolls or 0 EV choices. You are not overthinking it. Casinos were built with the help of your buddy and his magical system. Feel free to sprinkle in some Monty Hall scenarios as well, as those add a festive element to the topic.

Does he have any doubts about the finite bankroll case? I think that should be pretty obvious to him that the EV is 0. For the more general case, you have to sort of define what "works" means. Losing can be a zero probability event, (1/2)^n as n---> inf, but I don't think that is sufficient to say the game has positive EV, when each term has 0 EV.

Oops, thought I had posted this in Probability, don't know WTF happened there. Mods, feel free to move it there if you want!

It's because your bet is being covered. It's not a game where you play then they play. So you always get the chance to win one bet and if you can play martingale forever, until you win then eventually you will.


If you then had to cover the bet the odds would be even and zero ev.

You're just a politics junky. Admitting you have an addiction is the first step to recovery.

I would turn it around on him. I'm sure it is evident to him that each specific toss of a coin (or whatever) has an EV of 0. It is a fairly standard mathematical practice to calculate the expectation of a series of independent events by summing the individual expectations, and since each one is 0, the sum, no matter how long the series, must be 0.

He is the one making the claim that this isn't true. The burden of proof is on him, not you. Your only claim is that 0 + 0 + 0 ... to infinity =0. You can demonstrate it pretty easily by showing that 0 + 0 + 0 =0 and that if you do it a bunch more times it just doesn't change. You could call it a proof, but he obviously thinks there is something wrong with this, so he needs to be the one to prove it sums to more than 0.

If you do want a simple logical approach, try this.

If betting heads every time will yield a positive expectation, then betting tails every time must yield a negative expectation. But clearly choosing heads or tails is arbitrary, and if you said you would bet tails every time then (according to him) that would be +EV. Now, imagine if you and your dad bet on the same flips - but he bet heads every time and you bet tails every time - if he is correct then you would BOTH end up winning, no matter how long the game goes. Maybe he will see that this makes no sense. Maybe not.

I'm not a math guy at all but AFAIK that's called riding the variance which may work in case he'd start with positive variance (that's totally up to chance / luck if you will) and quits for good once he is up a reasonable but small amount not too far from his mean expectation than never plays ever again.

This "system" if you can even call it that also not works because humans usually don't stop until they haven't turned that small win into a huge loss. Actually starting out with positive variance is more like fuel to the fire in the case of these genius winning systems

Ok, this is ****ing annoying. Just spoke to him again, and I think he does this on purpose because he knows that I only call him when I'm drunk and can't answer this **** on the spot.

His new one is "I didn't say martingale. I said, I do two parallel bets. One is on tails which is martingale. The other one is on heads, which is a flat unit. So, he places the following bets on both heads and tails, until tails drops: [1,1], [1,2], [1,4], [1,8], [1,16] etc.

I started talking about supertasks and all that stuff, and he is just like "prove that my system doesn't work with no table limits". I ****ing can't. I sort of stumbled and said "well, you have two series that diverge and you're intermingling terms, and you can't do that", and he's like "what series that diverge? show me that what series diverge". **** man, I know he's doing this to test me, and while I know intuitively that the answer is that there are divergent series here, I can't show it.

I like the sum of zeroes explanation the best, and nobody disagrees that the EV of any individual trial is zero. But he's like "well, when I win, I win more than I lost, because of the flat bets". I tried to explain this like one of those supertask things, like when you put 10 balls in a basket and take 1 out and you're still left with them all at the end (or not, depending on which ones you take out, and the axiom of choice and stuff, and that's where I got myself all confused).

Love the no table limits part... Welcome to the theoretical millionaires club

Can't see how this could be of any help. Solutions like those were attempted by gambling experts centuries ago with no avail.
As long as the production remains random independent, no betting system could invert the initial EV.

I called him back today and had him work through an EV calculation if he bets, e.g. 1 on heads and 4 on tails on a single trial to show that the EV was in fact 0. Then I used the sum of zeros thing and he begrudgingly agreed.

I think the issue is that he wasn't going "across", he was going "down", and when you have two infinite series that sum to zero together, but diverge individually, you can't just intermingle terms. If you start doing that, you can just make e.g. the harmonic series add to arbitrarily close to e.g. pi or e or 10 or log 50 or make 1-1+1-1 add to 10 million (+/- "something" at the end)

Anyway, pretty sure he knew the answer, I think this was all a big ruse to get me to call him when I was sober enough to explain it lol.

Also, I had to explain that loose words like "eventually" were doing a lot of work in his argument, and were in fact introducing infinity/limits into the equation. He couldn't argue with me for any finite n, so it came down to "but eventually..."

I understand.
In the past one roulette expert suggested that after a 3 or superior sigma deviation happening within a short sample of even chances, the future probability to get the silent chance showing up by clustered series of any lenght was higher than 50% thus giving the player an edge (capable to even erase the zero impact).
He adviced to seek for just one unit of profit after any occasion like that by flat betting, then waiting again for another deviated event.

Theorically this procedure doesn't seem to be so bad (the worst scenario is always to get an EV=0), actually long term trials have shown that no matter which point we decide to bet and as long as we have no means to dispute the randomness production, this method is perfectly comparable to a random betting.

Quite likely, if we take as a starting betting point a 5 or superior sigma deviation that happened within a relatively short term sample and of course considering a fair game, it's plausible to expect the silent chance to partially catch up sooner or later. At least for just one recovering step and after considering some pattern situations.
Is this a math eresy?
Yes it is.

I mean, my whole argument was that you can't change EV of a given game/bet through any strategy that involves modifying the amount wagered in a given trial of said game. This is well known in these 'ere parts, but when someone who is fairly well versed in maths turned around and said "prove it", I stumbled a bit.

By the way, if a real-life roulette wheel rolls in a specific quadrant 10 times in a row, it is actually more likely to roll in the same quadrant the 11th time than dictated by pure chance for an idealised wheel.

Yep, but only if you have reasons to think some unrandomness is acting along the way.
No unrandomness = no party! :-)

Also, sorry, but lol @ "roulette expert" using betting systems. As far as I know, the only ways to beat a wheel with standard payouts and no comps are looking for bias and clocking the ball.