Quote:
Originally Posted by NewOldGuy
No we don't. That's only if we play to infinity, which is a thought experiment that doesn't exist. In any finite timeframe, it is more likely that they don't cross again than that they do *.
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No it isn't. Let's switch to a random walk which I'm sure you can see is just a simple version of the poker version.
Suppose we lose 1 unit if we flip heads and gain 1 unit if we flip tails. We use a fair coin. We are currently x units below expectation (the EV line being 0 all the way along, of course). Are you saying that the probability we get back to 0 in k flips is < 1/2, for all x?
That's obviously false. If x = 1 and k = 3 then the probability we get back to 0 within k flips is 5/8, since it happens if we flip {H, anything, anything}, and {T, H, H}. So the probability that they cross within three steps is 5/8. The probability that they cross within a large number of steps is virtually 1.
It doesn't matter how far we are away, there is nothing special about being one step away. Polya's theorem shows that 0 is recurrent, that is, if we start at 0, then with probability 1 we return to 0 in a finite time. Some of the random walks starting from 0 will land on -x. But we know that all random walks starting at 0 get back to 0 within a finite time by Polya's theorem. Therefore all random walks starting at -x get back to 0 within a finite time by Polya's theorem.
Obviously the difference between the two lines isn't a simple random walk, but it seems 'obvious' that the same result will hold.
*your comment about it being true at infinity but not being true for any finite timeframe is impossible. The result 'at infinity' is
defined as the limit of the results of finite timeframes as we increase the timeframe arbitrarily. So it cannot be true that the probability for 'all finite timeframes' is < 1/2 and the probability for 'infinite timeframes' is 1.
Last edited by Pyromantha; 08-12-2014 at 07:06 AM.