No House Edge -- How Will It End?

[deisik]

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

Emphasis added:

However, for this to be true in practice, you would need all customers to be playing perfectly, have perfect money management, and have a combined bank roll equal to that of the casino

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[o_e_l_e_o]

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.

[deisik]

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?

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

[arallmuus]

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 outcomes 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

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

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

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

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.

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

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

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

[deisik]

These are not conclusions, these are possible outcomes

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

What's the point?

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

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

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)

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.

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

You are welcome here

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

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

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?

[FontSeli]

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.

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.

[bbc.reporter]

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?

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.

[deisik]

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 can serve, and how they can be used in a sentence?

Scare quotes may indicate that the author is using someone else's term, similar to preceding a phrase with the expression "so-called"; they may imply skepticism or disagreement, belief that the words are misused, or that the writer intends a meaning opposite to the words enclosed in quotes

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

[lumeire]

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?


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"

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.

[deisik]

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