Theorem on the presence of memory in random sequences - page 43

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Dmitry Fedoseev:
There's no reason to look for meaning where there can't be any.

expert, man... where it may or may not be...

Is this a revelation to you from above? ;))))))

 
Олег avtomat:

expert, man... where it may or may not be...

Is this a revelation to you from above? ;))))))

Read it and what? I agree. Didn't find any contradictions with my ideas. You many here would do well to pay attention to some of the thoughts from those pages.
 
Dmitry Fedoseev:
Read it and what? I agree. I didn't find any contradictions with my ideas. You would do well to pay attention to some of the thoughts on those pages.
Well, that's good...
 

I wonder if anyone has noticed the error? After 43 pages of discussion...

I haven't read the whole discussion, of course. But since the error hasn't been corrected, I assume no one has noticed it.

What are the rules of the strategy? They are as follows:

  • If x1 > x2, then bet $1 on all numbers smaller than x2.
  • If x1 < x2, then bet $1 on all numbers greater than x2

First line: x1=2, x2=3, x3=5.

Since x1 < x2, put $1 on all numbers greater than x2=3, i.e. 4, 5, 6. Since x3=5, i.e. a 5 fell out, we get 6-3=3. Not -2$ at all...

Further, why is this expected payoff calculated by summing up the probabilities of all outcomes? Shouldn't each outcome be multiplied by its probability?

But these are all, in fact, not fundamental errors. I'm interested in another thing: what does the theorem state? That the conditional expectation may not be equal to the full expectation? So it is clear to the hedgehog.

Strategy is the above two conditions. To find the total expected payoff of the strategy, one should consider the strategy's profit in all possible outcomes. The outcomes are the following:

x1 x2 x3

1 1 1

1 1 2

1 1 3

1 1 4

1 1 5

1 1 6

1 2 1

1 2 2

...

6 6 6

Add up the profits of all outcomes and make sure that the sum is equal to zero.

1...3637383940414243
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