Question for Don: RoR and SCORE for different games

**MJ1** · 04-29-2022, 07:11 AM

Hello Don, I was contemplating the following questions.

Suppose a counter plays two games of BJ with different rules and decks. Using CVCX, he devises a betting schedule such that RoR is 5% for each game. Let's also assume he plays both games for an equal number of rounds. Does that mean his overall RoR will not exceed 5%?

Second question is similar to the first, except this time it pertains to N0. If two different games each have an N0 of 25000 rounds, assuming the counter plays both games for an equal number of rounds, does that mean overall the N0 for both games is STILL 25000 rounds?

Thanks for any answers!
MJ

**DSchles** · 04-29-2022, 09:08 AM

Originally Posted by MJ1

Suppose a counter plays two games of BJ with different rules and decks. Using CVCX, he devises a betting schedule such that RoR is 5% for each game. Let's also assume he plays both games for an equal number of rounds. Does that mean his overall RoR will not exceed 5%?

Theoretically, ROR changes with every hand you play. 5% ROR means that, if you never alter your bet size and are prepared to play that way forever, you have a 5% chance of going bust, and a 95% chance of eventually winning all the money in the world (or die first, trying

). The fact that you're playing two different games doesn't change anything, so long as they both have the same risk parameters.

Originally Posted by MJ1

Second question is similar to the first, except this time it pertains to N0. If two different games each have an N0 of 25000 rounds, assuming the counter plays both games for an equal number of rounds, does that mean overall the N0 for both games is STILL 25000 rounds?

Yes.

Don

**MJ1** · 04-29-2022, 10:05 AM

Originally Posted by DSchles

Theoretically, ROR changes with every hand you play. 5% ROR means that, if you never alter your bet size and are prepared to play that way forever, you have a 5% chance of going bust, and a 95% chance of eventually winning all the money in the world (or die first, trying

). The fact that you're playing two different games doesn't change anything, so long as they both have the same risk parameters.

Thanks, I would like to pose a follow-up question. Suppose in the first game the counter has a RoR of 5% and in the second game RoR is 10%. 20% of rounds played are in game 1 and 80% of rounds played are in game 2. I would think that the overall RoR from playing both these games off the same bank is somewhere between 5% and 10%, leaning closer to 10%. Is that a fair assumption?

Going back to the second question where N0 is 25k for game 1 and game 2 respectively, does that mean playing 12.5k rounds of each game will result in EV = SD?

MJ

**DSchles** · 04-29-2022, 10:19 AM

Originally Posted by MJ1

Thanks, I would like to pose a follow-up question. Suppose in the first game the counter has a RoR of 5% and in the second game RoR is 10%. 20% of rounds played are in game 1 and 80% of rounds played are in game 2. I would think that the overall RoR from playing both these games off the same bank is somewhere between 5% and 10%, leaning closer to 10%. Is that a fair assumption?

Yes, but now the plot thickens and the math gets complicated, because the formulas are not linear, and the individual RORs cannot simply be weight-averaged. Variances are linear and can be added and averaged, but s.d.s and RORs cannot.

Originally Posted by MJ1

Going back to the second question where N0 is 25k for game 1 and game 2 respectively, does that mean playing 12.5k rounds of each game will result in EV = SD?

Yes. In essence, it's no different than if you picked just one of the games and played all 25,000 rounds for just that one. Of course, when you say EV will equal SD, that's the theory. You understand that the actual result could be anything.

Don

**MJ1** · 05-01-2022, 04:16 PM

Originally Posted by DSchles

Yes, but now the plot thickens and the math gets complicated, because the formulas are not linear, and the individual RORs cannot simply be weight-averaged. Variances are linear and can be added and averaged, but s.d.s and RORs cannot.

Yes. In essence, it's no different than if you picked just one of the games and played all 25,000 rounds for just that one. Of course, when you say EV will equal SD, that's the theory. You understand that the actual result could be anything.

Thanks again, Don. Can N0 for different games be calculated using a weighted average? For example, suppose game 1 and game 2 each have a respective N0 of 20k and 30k. If 20% of rounds played are game 1 and 80% are game 2, then is overall N0 now 28k?

MJ

**DSchles** · 05-01-2022, 04:26 PM

Originally Posted by MJ1

Thanks again, Don. Can N0 for different games be calculated using a weighted average? For example, suppose game 1 and game 2 each have a respective N0 of 20k and 30k. If 20% of rounds played are game 1 and 80% are game 2, then is overall N0 now 28k?

Yes, I think so. Basically, I think that SCOREs can be added in this manner. It's an intelligent question that we don't see that often, and while I wouldn't bet my life on it, I think you can weight average them this way.

Don

**MJ1** · 05-01-2022, 06:05 PM

Originally Posted by DSchles

Yes, I think so. Basically, I think that SCOREs can be added in this manner. It's an intelligent question that we don't see that often, and while I wouldn't bet my life on it, I think you can weight average them this way. Don

Well if SCOREs can be weight averaged, then surely N0 can be. A long time ago, you said that N0 is the reciprocal of SCORE, with a *shitload* of zeros added on. So, they are essentially one in the same.

Can't the answer to my question be figured out if one knows the EV, variance, and number of rounds played for both games? Just take the weighted average of EV and variance for each game, add together, square root variance to convert to SD, then proceed to calculate overall SCORE. Then convert SCORE to N0. Check to see if the new N0 matches the weighted average N0.

MJ