Fyp, where B is bankroll. Bankroll doesn't increase by 1 with every coinflip, so the probability as a function of N is different than as a function of B. And you forgot it's not a fair coin so it's not 1/2 but 2/3. If it were a fair coin we'd go broke 100% of the time.

I think we both have a different example in our head. Regardless in words it meant; if your chance of something changes (decreases fast) with every additional N then it is not a guarantee to happen in an infinite series.

So, with a fair coin, we have a constant chance of going broke, because our ev is 0 for every flip, and over an infinite number of trials we must go broke, but with a coin biased in our favor, we have an ever decreasing chance of going broke, because our ev is positive for every flip, so our risk of ruin is not 100%.

Circling back to poker, against a pool of villains that have, essentially, an infinite bankroll, if we have a positive edge in a game, our chance of going broke is constantly decreasing, given that we play at the same stake with the same edge (for simplicity's sake, let's assume that "same stake" requires there to be a maximum bet that remains constant), and so we do not have 100% chance of going broke even over infinite trials.

Do I understand it correctly?

Yup

Same stake can simply mean you routinely cash out and then return later and rebuy for the max. Max wager is limited to your stack size and you can have a policy of cashing out when effective stack sizes reach a certain point.

But even if you sometimes move up in stakes, your RoR still isn't 100%, provided you don't move up too quickly. Especially if your edge were to somehow remain the same (but more importantly, it just needs to remain positive).

Re-circling back to the 60/40 coin example, your RoR is less than 100% as long as your progressive wager is less than 40% of your bankroll (2x the Kelly bet). Flat-betting $1 minimizes your RoR, but Kelly betting (which involves progressive betting) would give you a low RoR too (with the advantage of growing your bankroll at the fastest possible rate).

Of course not! Infinite bankroll, in theory, will always win. If we stick to math, reason and logic, there is no way a finite bankroll could ever compete with an infinite one. Why, because infinite bankroll has infinite number of options and the finite one doesn't. It's as simple as that. Pretty linear.

No.

It isn't possible for the player with an infinite bankroll to "win" if we define it as increasing his bankroll. It *IS* however, entirely possible for his opponent with a finite bankroll, to be ahead after any finite period no matter how long.

In fact, if you have a skill edge over your opponent it is to your advantage for him to have an unlimited bankroll. You will win at the same rate as you would if his bankroll were finite.

End of subject.