What is it that actually goes negative when the house risks over 2x Kelly? And it can't be the expected bankroll growth, since that's simply the expected profit per bet summed (or averaged) over all the bets, which therefore is also a constant 1%.
It's the bankroll that goes negative. The casino starts losing money if the allowed bet is too high, it's a statistical certainty long term.
Well, the bankroll can never go negative - the house only ever risks a percentage of it, so at worst it approaches zero. Maybe you mean the site profit goes negative, but that's not necessarily the case either. It's possible for the house to over-leverage and still make a profit by getting lucky.
Thanks to Ryan's linked article I think I understand it now. The point I wasn't understanding is that "expected value" isn't the same as "expected bankroll growth".
Specifically, the expected value of each bet is 1% of the amount wagered, whether the house is over-leveraged or not. It's always positive so long as the house edge is positive.
Whereas the expected bankroll growth is more to do with the factor by which the bankroll grows with each bet.
For instance, suppose the house is risking 50% of its bankroll with every bet, that the payout is 2x, and the chance of the player winning is 49.5%. That's a standard 1% house edge bet, but massively over-leveraged.
Each time the house wins a bet like that, it adds 50% to its bankroll, or in other words it multiplies the bankroll by 1.5
Each time the house loses a bet like that, it loses 50% of its bankroll, or in other words it multiplies the bankroll by 0.5
In the long run, the house wins 0.505 times per bet, and loses 0.495 times per bet.
We can calculate the expected bankroll growth factor per bet as (growth when winning)^(probability of winning) * (growth when losing)^(probability of losing).
In this case, that's:
1.5^0.505 * 0.5^0.495 = 0.8708
In other words, when the house is risking 50% of its bankroll every roll with a 1% house edge on a 49.5% bet, the house expects to multiply its bankroll by 0.8708 each roll, on average.
If we change the maximum bet to allow players to only win 2% of the house bankroll each bet, the multiplies change to 1.02 for a win and 0.98 for a loss, and so the expected bankroll growth factor changes to:
1.02^0.505 * 0.98^0.495 = 1
In other words risking 2% of the bankroll with a 1% edge leads to a static bankroll.
What really made me see it, is imagine the case whether the house risks 50% of its bankroll each time, and starts with 100 units.
If the player wins their first bet, the bankroll is down to 50 units. Then if the player loses a bet, the bankroll only goes up to 75.
On the other hand, if the player loses their first bet, the bankroll goes up to 150 untis, but then if the player wins, the bankroll again goes down to 75.
So each time the player makes two bets where one wins and one loses, the bankroll drops 25%. In the long run, the number of wins and losses is going to be about the same (assuming 50% betting, and no house edge, let's say). We can pair each winning bet with a losing bet, and see that the house is going to lose 25% about n/2 times, where n is the number of bets that were made. There's a tiny chance that the player does exceptionally badly and ends up massively down, but by far the most likely outcome is that the house loses almost all of its bankroll.
