Topic: No House Edge -- How Will It End? - page 3. (Read 786 times)

A player that has sufficient bankroll cannot go bust mate, because the RTP will always return to the theoretical RTP of 100%. Yes, there will be variance; however, it will return to the theoretical after xxx bets. Meaning if a player wagers $1 he will get $1 back.

Anyways, sure one can get bust if the bankroll is not sufficient. *I use the term "sufficient" because I believe we don't need to have an infinite bankroll in order to see the "stability."

Well, you can't speak for everyone. Moreover, you can't even assume that everyone is going to agree with this statement in the first place


I gambled, like a lot. Lost as much, due to a failure to see and understand how things work in practice and for trusting other people a way too much. Remember that climax moment from Interstellar (probably worth watching the whole movie)?


Does coin tossing count?

If it counts, then you can start out with the example presented in the OP. A simple setup, known for literally millennia. You have 100 dollars and I have 100 dollars, the base bet is 1 dollar. Someone has to bust and relatively fast at that. Pretty much a "street game" if you ask me, and nowhere near a "zero-sum game" (as you mean it), Certainly worth trying it out with your friends, to feel it with your skin in the game, so to speak


To begin with, no sane person will gamble on the street. Yes, I was basically insane back in the day to gamble on the street if this is what you want to hear. But I was young, stupid and stubborn as youths can sometimes be. You can't leave while in profit unless the other side agrees or you have agreed beforehand upon the time when it is allowed to take dough and run. Indeed, if you could force your way out you would, but no one would be playing with you again

I think everyone will agree. And then ? What is your conclusion for real world ?


I never gambled on the street, no, sorry And while I agree this is being "practical and realistic", it's unrealistic again then. You are comparing a realistic scenario with unrealistic assumptions and rules.

Please give an example how such a "street game" would look for you with all the parameters and what you would expect the result to be.

Lets say you have 1,000 USD bank roll and there are 200 players with 1 USD bankroll each. Now you toss a coin with each one of them, stake 1 USD. What do you expect your profit to be ? In first round you would eliminate 100 players. 100 players remain with bankroll of 2 USD each; your bankroll is still 1,000. Then in second round, third round etc. you would keep eliminating players, until there is only one player left with bankroll of 200 and you still have your 1,000 bankrolll. You will eliminate that player too sooner or later. And then we are here again:


Is all this a realistic scenario ? No. Players that are in profit will leave the game before it ends. Now you will say: but, street rules. Yes, street rules, but no sane person will play against you.

It's merely a reliable way to test your different tactics and strategies in a game. For the house, it wouldn't be so ideal, though, because if you have no edge, there's no additional chance for the house to win, no profit for them. For players, it's ideal, but there wouldn't be anything good coming out of it for the house, so that isn't happening. It will end only in the player in profit and the house losing the "chance" or vice versa.

There's not infinite gameplay, there would be a time where one side is having more "winning" than the other, so whichever comes first, that's the game-ender. It's bound to luck, I guess.

Now let's wait for other folks to come to terms with this


If we thoroughly follow this logic through, then you shouldn't be tossing coins in the first place

How come? Cause if you assume proper money management, you should assume it for both sides, right? Then the whole endeavor becomes an exercise in futility whether your goal is to win or not to bust as there are only two outcomes possible, a total win or a complete loss. In practice, though, you typically can't stake less than the base amount previously agreed upon, and you can't leave when in profit either (street gambling rules). So someone has to lose and accept it


Ever gambled on the street? This topic has everything to do with how things are going to pan out there. And you are in for a big surprise if you expect "a zero-sum game" there. It is also called "learning it the hard way". Can anything be more practical and realistic?

Yes, this is a correct statement. And now ?

It's against proper money management though, since when you are down to 1 USD, you shouldn't stake that 1 USD on one bet/toss And if the (finite) bankrolls are big enough, assuming 1USD stakes, this will take like forever - at least longer than this world will exist

And with "my" case I can't go broke, which is more realistic imo - the staking with % of your bankroll that is. Always staking the same amount regardless of your bankroll is not what any gambler should do.

I admit, that I have trouble to understand what you are trying to achieve with this and the other thread. This is so far from reality, that it's just theoretical discussion, which is of course fine and interesting too. But it seems to me, that you are trying to prove something, what will have no added value to you, since it's not realistic to happen in real world.

That's an extension for the case of different bankrolls

And that's also the reason why an extremely wealthy casino (or way wealthier than all its players combined) doesn't need a house edge at all to stay profitable. But that's a point of discussion in this thread. The present topic is a sort of spin-off of that thread to deal with a certain misconception that was revealed there


As long as your bankroll is finite, the staked amount is not zero, and remains the same (e.g. 1 dollar), on an infinitely long timeframe you will either bust or win provided the same terms are honored by your opponent too


This is a separate case which needs to be dealt with independently (read, it is a different case)

Ok, so I have a finite bankroll of 100,000,000,000,000 USD and I stake 0.00000000000000000000000000000000000000000001% of it with every bet. My Stake is divisible into any imagineable amount - it can get as low as I want or as I need it to be, if I get very unlucky and lose 100000000000000000000000000000000 bets in a row. How can I bust ? 

Even if I have only 0.01 USD left and I always stake 50%, the next bet would be 0.005, next bet 0.0025 etc.. I can't go broke, it's not possible.

When playing on a 50/50 odds PvP, the player with the most potential attempts will be least likely to bust.

For example, a player with $10,000 going up against somebody with $10 on a 50/50 odds table, the latter is most likely to first to bust—simply because he only gets 10 attempts vs 10,000 for the other guy.

If you scale this to infinity, both players will eventually bust.

As this topic has shown, there are many people who erroneously assume that on a 50-50 win chance and no house edge in a game of chance, there'll be no clear winner on an infinite timescale, i.e. until someone busts with no time limits imposed (a kind of "a zero-sum game"). In fact, I was rather surprised with this view because it is pretty simple to prove the opposite, which I'm going to do below

For simplicity's sake, let's consider a simple game of coin tossing. Two players are staking 1 dollar at each toss of a coin. It should be obvious without any further explanation that if they have only 1 dollar, one of them is going to bust straight away. If they have 10 dollars each, it will take a little bit longer but one player will still bust in the end, and it was estimated that it is going to happen under just 200 tosses on average

It doesn't take a lot of brain power to see the overall dynamic, and draw a reasonable conclusion that it doesn't really matter what fraction of the bankroll the two players start off with since on a long enough timescale one of them is still set to bust. It could be further generalized that on an infinite timeframe gambling on a 50% win chance with a finite bankroll inevitably ends in a bust of one of the players. This is what I would call a statistical certainty

And this is in stark contrast to what I've seen people claim. So what is your opinion?

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To avoid ambiguity and misunderstanding, I changed the part "it doesn't really matter what fraction of the bankroll the two players stake at each toss" to "it doesn't really matter what fraction of the bankroll the two players start off with"