That is a good point. If you consider movement of a stock as some kind of random walk, both seem equally likely. I guess my counter-argument would be as follows:
Suppose a stock is equally likely to lose 50% as gain 50%. Let T(1) be time zero. Write down the stock price at time T(1). Let T(2) be the first time when the price of the stock has halved or doubled. Write down the new price. When the stock next doubles or halves from this new price, call it time T(3) etc. By time T
, the stock will have halved about n/2 times and doubled about n/2 times. Its price will be approximately:
((0.50)^(n/2)) x ((1.50)^(n/2))
And here's the crunch:
0.50 * 1.50 = 0.75.
So its average price at time T
is (0.75 ^ (n/2)) times its original price, so on our original assumption, the company would most certainly go broke.
This is why I was complaining in my original thread that the same argument applies to me and implies that I am almost certainly going to lose all my money. But that's not fair. Surely something's wrong in the above argument. And I think I have resolved my own paradox.
Yes, I am almost certainly going to lose all my money in the long time because of the above.
BUT if you go through the above argument more carefully and work out the _AVERAGE_ price at time T
we find that it is actually the same as the original price. (Use binomial co-efficients to work out probabilities; sorry for technicalities)
Basically, I am almost certainly going to go broke, but I have only a finite amount to lose and an infinite amount to gain, so I may be broke with probability 99%, but with prob. 1% I have 100 times as much money.
SO, how to avoid going broke!?
(I've been betting on small amounts and lost 90% of what I had two weeks ago having doubled my money in the two weeks before that.)
Sorry for the rant.