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As this (https://bitcointalk.org/index.php?topic=5253970) 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"

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.

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.

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 (https://bitcointalk.org/index.php?topic=5253970.msg54578887#msg54578887) 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)

Yes, this is a correct statement. And now ? ;D

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.

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?

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.

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.

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

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 (https://www.gamblingcommission.gov.uk/PDF/RTP-graph.pdf). 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."

The variance can be quite large, but if the number of games is fairly large and the bankroll is big enough, both players should just end up with their starting balance.

So why would someone play a PvP game with 50% winning chance?

Also, who is running the site? No expenses, or is just a simulation?

I think PvP only becomes interesting if people believe they have an edge over the opponent, but if its clear from the start everyone has a 50% chance, there would be no fun involved.

So we have two conflicting and mutually exclusive ideas

And that's actually a good thing because it allows us to work out a constructive approach that will lead to a correct conclusion, whatever it might be. If what you say is true and we know that with small bankrolls (relative to bet amount) someone is going to bust, there should necessarily be a tipping point, a cusp, where the function (its derivative) changes the sign. In more mundane terms, there should be an exact bankroll to bet amount ratio above which it statistically becomes a "zero-sum game", no matter how long you are going to play. That's highly suspicious if you ask me


To find out who is luckier?

I reckon that there might be a misunderstanding.

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.

Place a 1% house edge to the infinite bankroll, infinte toss game, what result do you have?

The infinite theory is correct but I don’t see any point as why should we know how it will end when in fact nowadays no one like to play PvP as we prefer more to play against a casino.I think that when two players start with the same bankroll in a no house edge game like PvP there will be equal chances for both of them to go bust.What I mean is that the chances for them to go bust first is 50 percent.

Why too much of concerns about house edge? Will you ask for no-fee trading on exchanges or tax free transactions from your government? Basically, I am not seeing house edge will be having any big impact on gambler's experiences but if you remove, it will get big impacts on house's profit levels.

Moreover house edge is a concern for profit seeking gamblers but as per I have seen most gamblers here are playing only to entertain themselves and for them paying house edge is not a big thing.

People had already tried this by having a dedicated token: Edgeless (https://bitcointalk.org/index.php?topic=1718384).

I guess no-house-edge concept may not end now but it may open doors for many other innovations because some houses may plan up based on this so that they want to stand tall to attract gamblers.

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

The risk scenario is based purely on PvP, which will hardly ever happen in the "real world". This scenario here which tyKiwanuka posted reflects the world of online casino very well: 1 casino vs. an undefined number of players.

Let's think the scenario described here further, so the casino at the end of the day again has 1000 USD. But in reality it is much less:

• During the time of the casino operation employees, servers and infrastructure must be paid.
• There were ev. advertising campaigns like here in the forum by signature. So you have to pay the campaign participants.
• It was developed further on new games to keep the players happy in the future.
• ...

So you have not lost any money by just playing, but if you include the additional costs of a casino, you will see that at the end of the day you'll be left with less money than when you started and will eventually go bankrupt in no time if you don't generate any income (e.g. by getting a house edge in place).

Why all the fuss with 200 players?

The whole task can be reduced to just one player with 200 dollars. Then it is a bankroll of 1000 USD versus a bankroll of 200 USD, the point of discussion in this (https://bitcointalk.org/index.php?topic=5253970) thread (of which you are certainly aware). But that's all side issues as the main issue here is still not settled decidedly, i.e. whether it is a bust and a win or a definite no-win for either player if the bankrolls are the same as well as large enough in a PVP game like heads or tails (with square odds in mind). And yes, it is important to know the answer to this dilemma (especially if it is a "zero-sum game", after all), and it is not like everyone agrees with the conclusion drawn in the OP

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


That was in reference to your previous thread, which you referenced in the first post here. In the end it boils down to 2 players going against each other for the win of all the money in the game - no matter how many players joined the game initially.


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.

From a practical point of view......well.....I guess you know what I would write here ;D

Everyone doesnt have to agree with the conclusion but it is one of the conclusion that would happen. In the OP however you are insisting this


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


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

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


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

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.

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.

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.

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

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?

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.

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


These are not conclusions, these are possible outcomes


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

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.

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


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 (https://bitcointalk.org/index.php?topic=5253970) 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)

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.

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


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

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.

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 looks more like a commonplace misconception

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.

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.

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

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:
https://i.postimg.cc/SswBxc82/screenshot-188.png
Here ist the link to the calculator if you want to give it a try: https://byjus.com/coin-toss-probability-calculator/

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.

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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.

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


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?


Now everyone should stop and think where they really stand

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

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 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.

You had it coming:

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

My post above is talking about finite bankrolls.

Emphasis added:

You failed to quote my very next words where I said "Obviously, this is not the case", since neither the players nor the casino have an infinite bankroll.

On any specific timeframe (or within any given number of bets) that should be the case

That's another interesting aspect, which is worth looking into. But it doesn't take anything from the outcome described in the OP, of course, with one of the players busting in the end. Put differently, we don't know when and within which range this is going to happen. The only thing that we can be certain about is the inevitability of such an outcome on an indefinitely long timeframe. The example of martingale seems to be fitting here once more. You may roll a million times collecting dust, and then bust on a losing streak of 20 rolls, end of story

So Are there any other possible outcomes aside from those three?


Not just few games, I wrote X number of games in the previous post so it could be any number.


Having bigger balance is not an edge against the smaller balance player. If the wager amount stays the same each round, bigger balance player will just have longer time to lose everything compared with smaller balance assuming that those player keep losing


There is no certainty in exactly 50% chance game, as I have written before there are 3 conclusions outcomes in finite bankroll and infinite timescale

What's the point?


The key difference, though, is whether X represents any finite number. If it does, then it doesn't matter. And while we are at it, "several" is actually synonymous with "a few" (not to be confused with just "few", which has a slightly different meaning or connotation)


You are welcome here (https://bitcointalk.org/index.php?topic=5253970)


If you wager (synonymous with stake) all, and your opponent does the same, one of you is going to bust with 100% certainty, regardless of the individual odds (fifty-fifty or otherwise)

So do you agree that martingale is a losing strategy on an infinite timeframe with a finite bankroll?

This is just a theory. Unfortunately, in practice, the results may be far from ideal. Maybe 2:0 and 10:0 everything depends primarily on luck. Of course, the more bets you place, the more balanced the result will be, but the money may run out much sooner than the result will be equal.

That was why I reckon that there was a misunderstanding on the theory. An infinite coin toss cannot occur with a finite bankroll.

Also, I reckon @deisik misunderstood zero sum also.

Yeah, I understand that the distinction may not be clear to an inexperienced observer

But if you look through my posts where I referred to the notion of a zero-sum game in the course of the discussion, the phrase was double-quoted, and for a reason. So should I proceed to assume that it is actually you who failed to understand the purpose that double quotation marks (https://en.wikipedia.org/wiki/Scare_quotes) can serve, and how they can be used in a sentence?


I leave it to you to find out where the term was first misused. I guess you might be interested in that kind of thing

The result will always be one side will loose, it's just that you have to toss that coin for that much number of times until one looses their entire bankroll. But in the real world cases people have their own betting strategies that give them a little edge and if there won't be any house edge(which is never going to happen) then the gamblers will surely outplay the house in the long run.

That's unlikely in practice

For this to come true, we need all the gamblers to play flawlessly and profitably. However, if they just play randomly, i.e. without trying to take advantage of variance (which is possible with a properly managed martingale setup), then the house would still beat them all even on even odds (smells like a pun to me) provided the house bankroll by far exceeds the combined balances of the players

On the other hand, if all gamblers are capable of sticking to a safe martingale setup, they will be able to beat any real casino with a small house edge on any finite timescale. To avoid possible confusion and misunderstanding, by safe here I mean such setup that allows gamblers to earn more than lose collectively (read, some may bust, but the net result will still be positive on the part of the gamblers). To repeat, on finite timescales only