Risk of Ruin help

Can some authority figure please settle this?

Can some authority figure please settle this? Is moo right? Do all non-resizing betting schemes have 100% RoR because the possibility of any sized negative flux is always there? Can Ken or Fred settle this? Anyone?

But I can tell you that for an infinite number of hands, your risk of ruin is 100%.

But you could never actually say that, let's even assume a -EV game, could you? All you would know for sure is somebody lost all their money in a billion or trillion hands. And that every time so far the first billion people eventually lost all their money in the first trillion hands. But you would never know for sure the next person couldn't play "forever" until he actually went broke. You'd assume it's real, real close to 100% but you'd never know for sure. Likewise, in a lifetime ROR we only know it's real real close to that ROR %age.

No big deal. If you want to know your ROR before doubling roll before either going bust or doubling it, just because you think it's more important, and I'm not saying it's not, that's fine but even then what does it mean to you if I tell you your ROR is 40%. How do we know what will happen after a billion hands when one has neither doubled or lost his roll? If he plays potentially "forever", does he have 100% of doubling or 100% of losing it all?

Would it satisfy you if it was defined that exactly a billion people each played a billion hands at most and that as it actually happened 13% of those billion people either had gone bust during the billion hands while the other 87% still had money that 13% ROR is a pretty good number for a lifetime ROR?

If 40% of those billion people had either doubled their roll at some point and then quit playing but the other 60% lost their roll in the first billion hands before doing so?

Just BSing lol.

But from now on, since you're so specific, just ask specific stuff like what are my chances of losing my roll in the next billion hands. Or, if I play at most a billion hands, what are the chances I will have either quit after doubling my roll or lost it all.

Or what are the chances of me doubling my roll in the next 1000 hands or losing it all or neither?

Just please stop saying your lifetime ROR is always 100%, in a +EV game no less, no matter what.

It makes you seem like you're a member of the Flat-Earth Society or something lol.

You'd have to have someone with a lot more math knowledge than I have to settle it. But I can tell you that for an infinite number of hands, your risk of ruin is 100%. If you counted cards in hell, and never re-sized your bets, eventually you'd lose a million hands in a row.

No no.

Let's take an extreme example. You're playing a game with a small edge (1%), but the variance is zero. That is, for every unit you wager, you will receive only and exactly 1.01 units in return. Your risk of ruin is 0%. Even if an uber-infinite number of wagers are made.

Now, let's take a game with a small advantage, but with normal variance, and an uber-infinite number of hands are played. And let's say that out of a thousand people, most are betting fixed bets, but very conservative compared to their starting point. Risk of Ruin can never be 0%, because a few of the poor saps really are going to lose a million hands in a row and bust out. But a very large majority of these guys are not. Or more accurately, they may lose a million in a row, but they will have won two million before. The variance truly can't catch up with them. As Don Schlesigner put it, they've got to get you early, or they aren't going to get you at all.

The reason that some people can reach escape velocity and never bust the bankroll is the same reason that players in a negative expecation game are required to have a 100% ROR, eventually the advantage or disadvantage dominates the variance.

We're arguing about this semantically, and while it's useful, it's not a mathematical proof. I know that the proof involves things that I can't calculate, so unless anyone on here has a math degree, I think we're not going to have a proof.

If you played "forever" in a negative EV game you would have to go broke, or else it wouldn't be negative EV. You would have to go broke at some point.

I know that the proof involves things that I can't calculate, so unless anyone on here has a math degree, I think we're not going to have a proof.

You'd have to have someone with a lot more math knowledge than I have to settle it. But I can tell you that for an infinite number of hands, your risk of ruin is 100%. If you counted cards in hell, and never re-sized your bets, eventually you'd lose a million hands in a row.

This is only true with a negative EV game. Yes you might lose a million hands in a row. But, in a positive EV game, you are more likely to have won enough hands in a row prior to losing a million hands in a row to withstand the loss and remain in the game. OTOH, in a negative EV game, your risk is 100% -- even with a Martingale and infinite bankroll.

Actually all we would (likely) know is that after a trillion people went broke within a trillion hands is that 100% of those trillion people went broke in the first trillion hands. Does that mean the trillionth + 1 person will eventually go broke with 100% certainty?

And, if one believes, for some silly reason, that one will always go broke with 100% certainty at some point in a -EV game, why wouldn't one believe that one will always win with 100% certainty in a +EV game?