No House Edge -- How Will It End?

[deisik]

Sorry, if my comments offend you, that is not my intention, don't take it personal

Well, if you think that they are offending me, then let me tell you that they are not

From a theoretical point of view, one player will bust in the end. This could be the player with the 200 USD, but also the player with the 1000 USD. It's more likely to be the player with the 200 USD though, because to put it very simple: It's more likely to occur a 200-winning-streak with 50/50, than to occur a 1000-winning streak with 50/50. But both are possible

We have already established that

But it makes a by far more interesting case when the bankrolls are the same and huge (but finite) in respect to the amount wagered at each toss (if we are talking about a coin flipping), following the conditions set forth in the OP and worked out further in the thread. It is a seemingly borderline and indeterminate case but since you can't have it both ways and there can be only one definitive answer or conclusion (either bust and win or no-win aka a "zero-sum game" by your invention), that makes it ever more intriguing to find the correct solution

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[FontSeli]

In my time at the University, I studied probability theory.
My teacher had a joke example that I remembered: What is the probability of meeting a dinosaur in the city? 50-50 - either you meet him or you don't.
If this is transferred to gambling, then everything will depend heavily on luck. Even at a long distance, a guy with one dollar can beat the one with 10 dollars. Of course, if the ratio of starting capital will differ by more than ten times, then the final win will be for those who initially had more money.

[BITCOIN4X]

If this is transferred to gambling, then everything will depend heavily on luck. Even at a long distance, a guy with one dollar can beat the one with 10 dollars.

Right, this usually happens in games that rely on luck. There is an opportunity for those who have less capital to win on this game system.

Of course, if the ratio of starting capital will differ by more than ten times, then the final win will be for those who initially had more money.

Having more capital can increase your chances of winning bets on games that depend on luck, so players often use a strategy called a double bet or martingale. I agree that regardless of the type of game in gambling, we always hope that luck will bring us to victory. Our strategy, experience, and knowledge in gambling are just supporting factors to create chances of victory.

[bbc.reporter]

The infinite toss theory assumes that you also have infinte bankroll for the very long timescale to occur and for the wins and losses to always become 50-50

That's not the case here

It is indeed about a very finite and, moreover, equal bankrolls. So the question is that whether one of the players is going to bust or it will be a "zero-sum game", i.e. no one winning or losing anything (give or take) provided the player's bankrolls are big enough to allow for a very large number of tosses

We have two polar opposite outcomes which exclude each other, and I'm utterly curious to find out which one is actually going to play out in real life (as it may have very important and practical implications in other domains). So far it looks more like one of the gamblers will invariably bust eventually

I reckon there might be another misunderstanding. It is a zero sum game. A zero sum game is a player's wins is balanced by another player's loses to the winning player.

In game theory and economic theory, a zero-sum game is a mathematical representation of a situation in which each participant's gain or loss of utility is exactly balanced by the losses or gains of the utility of the other participants.

Source https://en.m.wikipedia.org/wiki/Game_theory

[Darker45]

It doesn't mean that since the chances of winning is 50:50, the result is going to be equal between players. In 10 tosses of a coin, it could happen that the result is 0 heads and 10 tails.

Therefore, on an infinite timescale with finite bankrolls, one will always prevail over the other. And with absolute certainty.

What's the people's claim, by the way, that this truth is in contrast to?

[Debonaire217]

In an infinite time scale, most of the people will not really prolong the game that much, in a 50/50 chance game between to person, there's likely an instance that one will stop playing either because they win or because they see their funds losing streak. Still, falling to your conjecture that one will prevail over the other, though, there's still a chance that both will stop playing if they retrieve their funds no one wins or lose, though the chance of this happening is quite low for me because as a gambler, you are playing to test your luck, you should accept your faith either to win or lose.

[deisik]

and it is not like everyone agrees with the conclusion drawn in the OP

Everyone doesnt have to agree with the conclusion but it is one of the conclusion that would happen

There's definitely a difference between being a blind contrarian (mostly for the sake of just expressing your disagreement) and disagreeing on some factual or logical grounds

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

So let me draw these conclusions . Assuming that there is no house edge or in the case of two people tossing coin game with exactly 50% chance for each player, Also finite bankroll and infinite time to play

There would be several conclusion after X number of game
1. Each player's bankroll would remain the same
2. One player will have higher bankroll and another with lower bankroll than what they started with
3. One would be the winner while one bust

These are not conclusions, these are possible outcomes

It isnt 100% certainty that one will bust while one win because at exactly 50% chance to win with no edge, no one could tell how things will be after several number of games

We are talking statistics here

It doesn't matter what the outcome will be in a few games, generally speaking. What matters, though, is the direction where things are going. Besides, after just one bet, there'll already be an edge in the form of a bigger balance because one player necessarily loses and the other necessarily wins. And technically, if they wager all, there is 100% certainty that one of the players will bust. Since bankrolls are finite, this can be the case on an infinitely long timescale as well

[kryptqnick]

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 stake at each toss 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

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.

The problem is that the infinite divisibility is not the case. There's always a minimum bet, and eventually you'll reach it and won't be able to bet twice as low. And if one had a ton of money but ended up with a sum close to zero (even if it's not technically zero), I'm sure that for the purposes of this discussion (that is, to show that a person is likely to lose even when there's no house edge) one could call it 'busted'.
As for what deisik pointed out, it's quite interesting because I have to say intuitively I would agree with those thinking that a person eventually gets even with the casino rather than loses when there's a pure 50/50 chance. bbc.reporter's remark about balancing one's losses with another one's wins was helpful for me to understand what's the confusion here.

[deisik]

In my time at the University, I studied probability theory.
My teacher had a joke example that I remembered: What is the probability of meeting a dinosaur in the city? 50-50 - either you meet him or you don't

FFS, this is not a joke, this is a cliché

The infinite toss theory assumes that you also have infinte bankroll for the very long timescale to occur and for the wins and losses to always become 50-50

That's not the case here

It is indeed about a very finite and, moreover, equal bankrolls. So the question is that whether one of the players is going to bust or it will be a "zero-sum game", i.e. no one winning or losing anything (give or take) provided the player's bankrolls are big enough to allow for a very large number of tosses

We have two polar opposite outcomes which exclude each other, and I'm utterly curious to find out which one is actually going to play out in real life (as it may have very important and practical implications in other domains). So far it looks more like one of the gamblers will invariably bust eventually

I reckon there might be another misunderstanding. It is a zero sum game. A zero sum game is a player's wins is balanced by another player's loses to the winning player

There's no misunderstanding here

The definition of a zero-sum game also includes the possibility of a no-win situation for both players. In fact, any such game would be a zero-sum game whether there is a house edge or not, whether bankrolls are equal or otherwise, or whatever. And throughout this and the other thread I have been enclosing this phrase in double quotation marks to show specifically that a) its meaning is somewhat different (narrower) from the established (wider) one, and b) it was not me who started to use it in this sense when referring to such an outcome. However, it wasn't that hard to get the meaning from the context (read, you could have done that too)

[spike420211]

Very weird calculated data.
Provided that the victory of one of the players is 50%, over a long distance their gain will be approximately 50 to 50, and the greater the distance, the more equal the number we will see.

In a short distance, even a thousand shots or more, there is a high probability of skewing to one side. However, with a distance of a million or more, the ratio of victories for both players will equalize.
This is commonplace math.

[deisik]

What's the people's claim, by the way, that this truth is in contrast to?

That no one is going to bust, and, consequently, no one will score a definitive win

But there's more to this than it appears at the first glance. It is extremely amusing to see some people claim that martingale is a losing strategy and those same people assert that "the variance will balance out to zero" with this setup. They don't seem to understand how tightly these two cases are interconnected. If the player with a finite bankroll is set to bust eventually when utilizing martingale, the same is equally applicable to the two players betting against each other (i.e. one of them will bust) as it is same variance at play here

As for what deisik pointed out, it's quite interesting because I have to say intuitively I would agree with those thinking that a person eventually gets even with the casino rather than loses when there's a pure 50/50 chance. bbc.reporter's remark about balancing one's losses with another one's wins was helpful for me to understand what's the confusion here

So is martingale a losing strategy for a finite bankroll on an infinitely long timescale or what?

[Darker45]

Very weird calculated data.
Provided that the victory of one of the players is 50%, over a long distance their gain will be approximately 50 to 50, and the greater the distance, the more equal the number we will see.

In a short distance, even a thousand shots or more, there is a high probability of skewing to one side. However, with a distance of a million or more, the ratio of victories for both players will equalize.
This is commonplace math.

Are you certain about this? That after a million or more shots, the victories of both possibilities will equalize? I don't think so.

A couple of fair toss coins, for example, will tell you of a 50:50 heads or tails probable outcome. But that is only the theoretical probability. However, theoretical probability does not equate to actual outcome. Which means to say that if you toss a coin in real life, it won't actual give you a result strictly based on your theoretical computation. A 50:50 toss coin may actually give you a 2:0 outcome for heads.

[deisik]

Very weird calculated data.
Provided that the victory of one of the players is 50%, over a long distance their gain will be approximately 50 to 50, and the greater the distance, the more equal the number we will see

It's called variance, get used to it

And on an infinite timescale it can be any, with the implication being that you are going to see losing streaks long enough to wipe out no matter how big a balance. That's technically the same reason why martingale is a losing strategy under similar conditions. But unlike martingale, you don't even need one losing streak to bust with a finite bankroll as just enough losing streaks interspersed with some wins will do exactly the same to your or your opponent's balance (read, martingale is better in this regard). It's so for the simple reason that the balance is limited (finite), while you have unlimited time to see a sequence of losses sufficient to drain any such balance dry (read, one of the players will inevitably bust)

This is commonplace math

This looks more like a commonplace misconception

[o_e_l_e_o]

This is essentially an arguments about infinites.

If bankrolls are infinite, then obviously no one is going to bust.
If time is infinite and bankrolls are finite, then someone will bust. Regardless of how large my bankroll is, and provided I can't subdivide it infinitely, as long as it is less than infinite then I will eventually bust given infinite time.

All of the above situations are meaningless in real life though.

Since both time and bankrolls are finite, then the chance of someone going bust will increase with time, and the person with the lower starting bankroll is more likely to bust.

A definition of a zero-sum game also includes the possibility of a no-win situation for both players.

Yes, but that is not all it includes. A zero sum game is any game where there is no net gain or net loss between all participants. Sure, if no one wins or loses anything, that is a zero sum game, but so is someone going bust if the other person wins all their money. All gambling is a zero sum game. You can't use a clearly defined term to refer specifically to only a subset of that term and not expect confusion. If you mean a situation where there is no individual gain or loss, as opposed to no net gain or loss, then you should say that.

So is martingale a losing strategy for a finite bankroll on an infinitely long timescale or what?

Categorically. Any strategy with a finite bankroll on an infinite timescale will bust.

[Lakai01]

Are you certain about this? That after a million or more shots, the victories of both possibilities will equalize? I don't think so.

No, you can absolutely not assume that. wolframalpha.com even has quite good online calculators for this. I tried it for 1 million throws and wanted to calculate the probability that exactly 500k heads are thrown. This probability is actually surprisingly low:

Here ist the link to the calculator if you want to give it a try: https://byjus.com/coin-toss-probability-calculator/

[serjent05]

Very weird calculated data.
Provided that the victory of one of the players is 50%, over a long distance their gain will be approximately 50 to 50, and the greater the distance, the more equal the number we will see.

In a short distance, even a thousand shots or more, there is a high probability of skewing to one side. However, with a distance of a million or more, the ratio of victories for both players will equalize.
This is commonplace math.

Are you certain about this? That after a million or more shots, the victories of both possibilities will equalize? I don't think so.

A couple of fair toss coins, for example, will tell you of a 50:50 heads or tails probable outcome. But that is only the theoretical probability. However, theoretical probability does not equate to actual outcome. Which means to say that if you toss a coin in real life, it won't actual give you a result strictly based on your theoretical computation. A 50:50 toss coin may actually give you a 2:0 outcome for heads.

That is because each toss probability is independent of each other.   Each time we made a toss, the probability of getting head or tail is reset.  So the more toss we made the ratio of 50:50 or equalizing the number of the head results and the number of the tail results is getting thin.


In a finite bankroll with infinite time, I agree with OP that some point in time, one will get busted no matter how huge is their bankroll.

[deisik]

If time is infinite and bankrolls are finite, then someone will bust. Regardless of how large my bankroll is, and provided I can't subdivide it infinitely, as long as it is less than infinite then I will eventually bust given infinite time

Okay, now we have two people who seem to have changed their mind

All gambling is a zero sum game. You can't use a clearly defined term to refer specifically to only a subset of that term and not expect confusion. If you mean a situation where there is no individual gain or loss, as opposed to no net gain or loss, then you should say that

And that's the exact reason why I was putting the term in double quotes

Was it not enough to give a hint that there's something wrong with its usage? Regardless, in case you missed (some part of) my previous reply or missed the other no-house-edge thread in its entirety, it was not me who had first used the term in this meaning ("no individual gain or loss"). So why did you decide to come up with this post now, and throw it at me? But seriously, where have you been before and what were you waiting for?

So is martingale a losing strategy for a finite bankroll on an infinitely long timescale or what?

Categorically. Any strategy with a finite bankroll on an infinite timescale will bust

Now everyone should stop and think where they really stand

[o_e_l_e_o]

Okay, now we have two people who seem to have changed their minds

Where? This is exactly what I said in the other thread:

This is what statistics says when you have a finite resource, which in your example is each person's bankroll. When you have infinite bankrolls, then statistics says that over infinite time the variance will be 0.

Infinite bankrolls and infinite time means the variance will trend towards zero.
Finite bankrolls and finite time means the person with the lower bankroll is more likely to bust, and the chances of someone busting increases as time goes on, but with finite time the game could end (or a player could choose to end it) before either party busts.

And that's the exact reason why I was putting the term in double quotes

And my point is that putting a word or phrase in double quotes doesn't magically mean you can redefine it to mean something that it doesn't mean. If you are going to use words and terms incorrectly, it is no wonder that you are getting conflicting answers.

[deisik]

Okay, now we have two people who seem to have changed their minds

Where?

You had it coming:

Statistically speaking, a zero house edge casino would not be profitable. Over infinite time, the variance trends towards zero, meaning the casino would pay out exactly what they took in

[o_e_l_e_o]

Which is true, when we are talking about infinite bankrolls over infinite time.

My post above is talking about finite bankrolls.