Actually, the casinos win in Las Vegas because the odds of almost every bet are in their favor.
In most cases the odds favor the house---I never claimed otherwise---and that certainly speeds up the inevitable process of cash extraction.
Larger capital allows you to affect the distribution of winnings, but not whether or not the underlying bet is a good one.
If the difference in bankrolls exceeds a tolerance related to the `odds', the quality of the bet is immaterial.
The direct implication of the weak law of large numbers is: a) the longer you play, the more certain you will experience a `run of bad luck'; b) the party with less money goes broke waiting for their `run of bad luck' to end. When one part goes broke, the game is over, even if the distribution of winnings does not match the theoretical expectations (and in the case of going broke, it can't ... or you wouldn't have played).
Every casino, in effect, takes on the whole world. As all the bets are independent, it doesn't matter if they are played by one player or by a new player every time. The world has much more capital. Yet the casinos consistently win.
No. The whole world doesn't go broke as a unit. Individuals stop playing, leaving their money in an unexpected distribution, when they _personally_ go broke.
Pretend the casino is run out of a church. "Parishioners" arrive and enter a confessional to place their bets. The "priest" cannot see who is placing each bet. Each "parishioner" plays until he or she is broke. "Parishioners" arrive at a steady rate and will do so indefinitely. How can the "priest" tell who is broke and who isn't? Why should he care? The chance of the "church" to win or lose is the same on every bet, regardless of who places it.
In fact, most gambling decisions are related in some way to cash resources of the participants. For example, I propose a hypothetical game where you (the player) flip a fair coin. If it comes up heads on the first toss, I pay you $2; game over. If it comes up heads on the second, I pay you $4; game over. $8, $16... How much would you pay me (the house) to play this game? The theoretical value is infinite; you could win any amount of money at this game -- 1/2 the time $2 dollars, 1/4 of the time $4, 1/8 of the time $8... expectations = Sum_{n \goesto \infty}{n \over n}.
Let's say I'm an actual casino, and could reasonably pay out winnings up to but not beyond $4.3 billion. You should pay no more $33 for a chance at that money. Derivation as an exercise for the reader. Consider this from the perspective of the house. The house is using the Martingale system against you, doubling its bet every time it loses until it gets that $33. That means that to launder $33, one party could conceivably lose $4.3billion. Obviously no mathematicians work at my casino. They all left to persue jobs that ensure a paycheck.
What you have constructed is an outcome where the house is almost certain to make a miniscule amount, but has a slim chance of a massive loss. This is not what "making money" means. Think about what the return on investment is likely to be. I don't think you will find an "inevitable process of cash extraction". That only occurs if the odds favor of the house.
These are _not_ my personal conclusions. This is sound, if disturbing, probability theory---known for at least 250 years. This particular effect goes by many names including "Gambler's Ruin". Given the odds, and the respective bankrolls, you can calculate the probability that any given party will go broke in extended play. The problem of "Duration of Play" was solved by Bernoulli and published posthumously in 1713.
I think you may be misapplying your reading. Duration of play is interesting, but I hope the actual issue is making money. They are not the same thing. This can be very confusing. I've seen two professional mathematicians and a futures textbook make this mistake. Peter