this doesn't make sense to me, smart people? (infinity)

[seven-deuce]

Why would 0.999... be not really a number? Why can't 1 be not really a number, as it's just a short-hard way to write 0.999...? They are just 2 equivalent ways of describing one quantity.

I think what you're trying to ask is if 0.000...01 is equal to zero, and the answer is yes.

0.000...01 = 0

0.999.. = 1

1.000...01 = 1

0.57999... = 0.58

I wouldn't say i understand this, how can they be equivalent quantites if there's 0.0000...01 of a difference? Surely they are different quantities no matter how minute the difference?

[daviddem]

0.000...01 = 0

0.999.. = 1

1.000...01 = 1

0.57999... = 0.58

I wouldn't say i understand this, how can they be equivalent quantites if there's 0.0000...01 of a difference? Surely they are different quantities no matter how minute the difference?

since 0.0000...1 equals zero, then there is zero difference

The difference is not "minute". The difference is infinitely small, which is another way of saying that there is no difference at all. It's not the same thing as a difference tending to zero.

[seven-deuce]

since 0.0000...1 equals zero, then there is zero difference

The difference is not "minute". The difference is infinitely small, which is another way of saying that there is no difference at all. It's not the same thing as a difference tending to zero.

The penny just dropped, the difference is infinitely small so there is no difference, well explained.

Also using a bit of logic; 0.00....01 = 0 so that can be rewritten as 0 = 0 which shows a lot more clearly that there is no difference.

[Lukie]

since 0.0000...1 equals zero, then there is zero difference

The difference is not "minute". The difference is infinitely small, which is another way of saying that there is no difference at all. It's not the same thing as a difference tending to zero.

I think most people kind of, sort of understand the concept of infinity. It seems like less understand the concept of infinitely small, or 1/infinity. For example those questions about having a positive bankroll and an edge at a game, you will not go broke a certain percentage of the time (depending on the specifics) even given infinite chances.

[daviddem]

I think most people kind of, sort of understand the concept of infinity. It seems like less understand the concept of infinitely small, or 1/infinity. For example those questions about having a positive bankroll and an edge at a game, you will not go broke a certain percentage of the time (depending on the specifics) even given infinite chances.

There is never an absolute 0 probability that you will loose all your bankroll, no matter how high your average winrate is, unless you play some game where your chance of winning is exactly 100% every time. You can have a bankroll of a million buy ins, and be dealt AA 1 million times in a row and get it in a million times preflop with your opponent holding 27o and loose every single time.

While very, very highly improbable, it is not impossible.

[Lukie]

There is never an absolute 0 probability that you will loose all your bankroll, no matter how high your average winrate is, unless you play some game where your chance of winning is exactly 100% every time. You can have a bankroll of a million buy ins, and be dealt AA 1 million times in a row and get it in a million times preflop with your opponent holding 27o and loose every single time.

While very, very highly improbable, it is not impossible.

Right, which is basically the same thing I am saying, only I wasn't all that clear.

My point was that some people think that if you have a positive edge, you still have to go broke 100% of the time if you play any non-100% edge infinite times, which is not true.

[daviddem]

Right, which is basically the same thing I am saying, only I wasn't all that clear.

My point was that some people think that if you have a positive edge, you still have to go broke 100% of the time if you play any non-100% edge infinite times, which is not true.

You are right. The problem you are mentioning is called "risk of ruin" and can be simply described as follows:
- assume a gambler starts with an initial number of buy-ins BI
- assume he endlessly (unless he goes broke) plays a game where his probability of winning is P and his probability of loosing is 1-P, with P > 0.5
- his chance of going broke is ((1-P)/P)^BI. In the other cases he will become infinitely rich.
- as an aside, if P<=0.5, the gambler always eventually goes broke

So for example if a gambler has $5 and he repeatedly plays a $1 game where he has 60% chance to win (P=0.6), then his chance of going broke is:
(0.4/0.6)^5 = (2/3)^5 = 32/243 = ~13.2%

The mathematical proof can be found here:
http://www.columbia.edu/~ks20/FE-Not...7-Notes-GR.pdf

You can do further calculations to apply this to a poker player who has a given winrate, standard deviation and some number of buy-ins. See here: http://archives2.twoplustwo.com/show...st682045683150
and here for calculators based on these formulas:
http://www.castrovalva.com/~la/bank.htm
http://www.reviewpokerrooms.com/poke...uirements.html
(note that the winrate and standard deviation are in bb/100 and the bankroll is in bb, not in BI. A typical poker player's standard deviation is 80bb/100, check yours out in PT or HEM).