If you found something that was a true 50/50 bet, winning and losing had an equal chance of occurring and each result was independent of the last, would there be a way to manipulate this to guarantee a profit with a certain betting strategy, or theoretically would you always just end up breaking even in the long run?

No, you could not guarantee a profit. You could guarantee a loss, but not a profit.

If you had a (a) specific target profit in mind, and (b) had a large enough bankroll, you could create a system to make that amount *almost* guaranteed at any level of "almost" you want, 90%, 99%, 99.99%, 99.99999999%.

But not guaranteed.

I've used the Labouchere system currently to turn 500 units into approx 2500, after every line banking the profits, so i could only ever lose a maximum of 500 units.

However I am sure this is not the optimal strategy so was looking for a bit of advice really on the best way to go about the betting.

How can you guarantee a loss?

It depends on how you define "guarantee", I guess.

By playing long enough. No matter how you're betting, your chance of going broke will approach 100% the longer you play. Yes, it's unfair that it's like that with a 50/50 proposition, but that's how math works. You're guaranteed to reach 0 at some point. If you could keep playing with 0 or negative dollars, you'd also be guaranteed to reach +trillions at some point, but unfortunately once we reach 0 we can no longer play.

The larger your average bet relative to your bankroll, the less time you need to play to have a high chance of going broke.

Given what I said above, the optimum is to not play at all. Or if you must gamble, set a limit and don't play long. If you get ahead, quit.

The logic that the optimal strategy is to not play at all because the probability of going broke approaches 1 as you play longer and longer is not sound. The $EV of any combination of wagers is obviously still 0, causing the player to be indifferent between all possible gambling strategies, including not playing at all and playing until you're broke or dead (assuming risk neutral player/constant marginal utility of money/...).

I actually agree, because it all depends what OP means by "optimum", when all roads lead to busto. Had he asked about the optimum way to play a -EV game, I'd have given the same answer and been technically wrong for the same reason.

EV is so 2005. It's time to move on to EG.
An assumption which describes ~0 people on earth.

Where and what do you bet on that is exactly 50/50?

This is why "guarantee" should be defined. One could take it to mean certainty. 100% probability is not certainty.

I'm done nit-picking.

People are answering a different question than he is asking. They are answering the question "With a finite bankroll is there a way I can guarantee that at some point I will be ahead?" He wants to know if there is a way to manipulate his bets to give him a positive EV.

OP, if that's the case then the answer ofc is no.

1st bet has 0 EV.
2nd bet has 0 EV.

EV of those 2 bets = E(1st bet) + E(2nd bet) = 0 + 0 = 0
Same for any number of bets.

It's funny you'll laugh, but minus the risk or a powercut or DDOS etc, but it's on an online game, you stake virtual currency, with 50/50 odds every time, but the virtual currency has a real life value when you sell it.

Exactly, or to know the strategy that will give me the smallest chance of getting wiped out and having 0 bankroll. I understand that in the long run if I played infinitely I would always have the run of doom.

PS. Love the books David, Theory of Poker was probably my favourite Poker book I have read. *Ass kissing over*

Often both questions are asked in this forum, and people often mix up the two things, even smart people. I think it is always important to clarify that it you can be virtually certain of a profit in most games, but that does not equate to +EV. Illustrating the difference is often an education.

I've posted before on the math around maximising the likelyhood of specific profit amounts, as this is a very interesting topic imo. This is especially interesting I think because people dismiss -EV gaming as just "guaranteed zero dollars ruin every time burning money etc omg obv" when the fact that it is not guaranteed every time is the reason the gambling industry exists.

The vast majority of people who walk into a gambling hall are taking a -EV proposition, but will have experiences that will vary from big wins, to big losses, at differing frequency and cadence depending on their choices.

Minbet every time.

If you bet 1% of your current stack you'd never go bankrupt even on a bet that you won 1% of the time.

If you bet 100% of your current stack you'd always go bankrupt, even on a bet that you won 99% of the time.

I guess the question is, do you have to bet? If you do, there are strategies that can help you achieve particular distributions of outcomes - but you can't change the overall EV of the bet. If you're going to make a LOT of bets then betting strategies matter even less.

Yes, if you're allowed to bet infinitesimal amounts, which I highly doubt. He's talking about a computer game, which probably can't exceed 64-bit precision.

Even if you could bet infinitesimal amounts, you'd get infinitely close to 0 with -EV bets. Is that any different than broke? (But with 0 EV I'd have to think about it.)

If you bet a fixed fraction of your bankroll every time, your median bankroll still approaches 0.

Good answer.

Your BR approaches 0, but I'm wondering if you can guarantee a profit by quitting at a certain target (when arbitrarily small bets are allowed).

Since you can play forever without reaching 0, you're guaranteed to be up a million flips at some point, but the longer that takes, the less that will matter because losses count for more than wins. You might be up a million flips but down money!

Suppose you start with $100 and want to win $1 net profit and quit. If you bet a fixed 1% of your bankroll, what's the chance you'll ever win that dollar? It's higher than 50%, but is it 100%?

I'll try to calculate it, or at least visually look for a limit, in my next post.

Yes, at least in the P converges to 100% sense. It's just a 1-d fair random walk with the axis renumbered. That doesn't make the game +EV without a dumb definition of EV though.

Actually, this isn't right. It is a fair 1-d walk but the axis is renumbered continuously if you treat it that way. You'll get arbitrarily far ahead of 0 in terms of wins-losses but you have to do it in time, and that's not guaranteed AFAICT.

Unless I'm missing something trivial, that chance cannot be 100%. Say that you have a strategy that consists in winning 1$ or go broke (-100$). Since there are just two outcomes, calling with p the probability of winning 1$, you must have:

p*(1$)-(1-p)*100$=0 -> p=100/101

since combining 0 EV bets will result in 0 EV for all the bets your strategy dictates.
Of course, if your bankroll is allowed to grow without bounds, the probability of losing it will approach to zero.

Haha I'm glad I didn't attempt it yet (I was about to now), because pokergods only know how much time I'd have wasted applying the complicated approach I had in mind. Thanks as always. When I saw your name as the last poster, I knew before I even clicked the thread that you solved my question. You and TomCowley are like lurking BruceZ's

David, if you're lurking, now is a good time to give your lecture about using simple reasoning whenever possible.

This is one of the greatest compliment I ever got. Thank you sir, I really really appreciate, although I don't even remotely play in the same league of the guys you mentioned.