Interesting fact about the interesting facts at the bottom: each point is reached an infinite number of times only when a random walk is performed in one or two-dimensional space. For dimensions >=3, an infinite walk visits each point a finite number of times, which means that, while the random walker may return to the starting point one or more times after starting, it will eventually wander off forever. This has been described as, "A drunk man will find his way home, but a drunk bird may get lost forever."
I had forgotten that one. You inspired me to visit Wikipedia where I learned that the probability of a 3D random walk returning to the origin (given an unbounded number of steps) is about .34 (in 2D, as you note, this probability is 1). The probability of eventual return goes down as dimensionality increases. As you say, get lost forever.
"A moral of this is that any lucky series will always eventually be reversed
and an expected outcome of a random walk is always 0.
Or in other words - don’t play roulette ;)"
The apriori expected outcome of a random walk is zero. The expected outcome of a random walk from a given position is the position itself. So no, 'any lucky series' will not 'always eventually be reversed'. I'm afraid that's not how probability works.
The way I understand this, is that for a random walk of infinite length, your expected outcome is always 0. It doesn't matter that you started very lucky by winning 10 games in a row, because if you play an infinite number of times, you will balance this by loosing 10 games in a row (you'll do it an infinite number of times).
In real life people usually don't play an infinite number of times, but the more you play, the better the theory approximates your final outcome. After very large number of games, you will very likely encounter a bad luck series.
This is a bit counter intuitive, because according to the theory, there is a difference between a person that enters a fair game with 10$ and a person that enters with 1$ and wins 9$ in the first 9 games. The first person is expected to finish with 10$, the second with 1$, even though both have the same amount of money at some point. But laws that involve infinity are often counter intuitive.
>The way I understand this, is that for a random walk of infinite length, your expected outcome is always 0. It doesn't matter that you started very lucky by winning 10 games in a row, because if you play an infinite number of times, you will balance this by loosing 10 games in a row (you'll do it an infinite number of times).
You understand it incorrectly.
>This is a bit counter intuitive, because according to the theory, there is a difference between a person that enters a fair game with 10$ and a person that enters with 1$ and wins 9$ in the first 9 games. The first person is expected to finish with 10$, the second with 1$, even though both have the same amount of money at some point. But laws that involve infinity are often counter intuitive.
That is exactly the example I would give of why your understanding is flawed. This is not 'counter-intuitive', it is a clear contradiction of basic logic. You are making a fairly common error in confusing a priori probability with conditional probability: http://en.wikipedia.org/wiki/Gamblers_fallacy. A standard phrase used by gamblers to refer to this truth is that "the dice have no memory."
If you start a random walk from Chicago, you are expected to end up in Chicago. If you start a random walk from New York, you are expected then to end up in New York. If you, later on that same walk, notice that you are currently in Chicago, you can start expecting yourself to end up in Chicago, despite that your random walk started in New York: you have observed the actual events (collapsed the waveform), and now you have just taken a walk.
This is the core of the misconception: "The first person is expected to finish with 10$, the second with 1$, even though both have the same amount of money at some point." The first person was expected to finish with 10$, and the second was expected to finish with 1$, but after the second player has won 9 games, he is then expected to finish with 10$, and anyone standing around watching who still has faith in their earlier expectation is uninformed.
