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.
Topic: No House Edge -- How Will It End? - page 2. (Read 786 times)
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
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)
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.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.
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
Quote
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
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.
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?
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?
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
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. 
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.
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.
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
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. In the OP however you are insisting this
Quote
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
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
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
Why all the fuss with 200 players?
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.
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 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, and it is not like everyone agrees with the conclusion set forth in the OP.
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
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
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 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
-snip-
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.
-snip-
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. 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.
-snip-
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).
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
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.
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.
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.
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.
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.
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 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?
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
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
So why would someone play a PvP game with 50% winning chance?
To find out who is luckier?
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."
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.
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