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

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

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"

Also, I reckon @deisik misunderstood zero sum also

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

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?

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.

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

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

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

Are you certain about this? That after a million or more shots, the victories of both possibilities will equalize?

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

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

Where?

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

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

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.

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

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 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

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.

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

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

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

This is commonplace math

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.

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

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