Lowest beatable limit?

[AAmaz0n]

well, let's take a rough cut at the CA rake.

Let's deal with the $3 house drop, the $1 jackpot rake is essentially a side bet.

In 100 hands, the house takes in $300. assuming you are playing 9 handed, that makes your contribution $33.Ok, if you are a tight player, your end is a bit lower, but let's look at this number.

At 2/4 $33 is a bit more than 8 BB, so you would need to make more than 8 BB/100 to break even, not counting the $1 jackpot sidebet that is taken out.

If the rake is the same $3 at 3/6 and 4/8. Then the numbers are 5.5 BB/100 and 4 BB/100 to break even.

Ok, this is just an initial hack, I'm ignoring the fact that you play tight enough that you pay less than the table average of rake, so the actual number should be a bit lower.

It also doesn't count the "no flop no drop" situations that come up every so often, but those shouldn't affect the bottom line too much.

Also, I'm treating the jackpot drop as a side bet that should even out over time, although it will take quite a while to get into "the long run" where folks at your table (hopefully you!) hit the thing. They are taking $100 over 100 hands for this which 9 handed translates into 3 BB/100 at 2/4, 2 BB/100 at 3/6 and 1.5 BB/100 at 4/8 until that glorious day when the jackpot is hit at your table.

I play very tight, and my contribution to the rake online seems to be about 2/3 of table average. If we apply that number to the above analysis, it would be:

5.3 BB/100 at 2/4
3.7 BB/100 at 3/6
2.7 BB/100 at 4/8

just to break even.

I think that they take $4 at 6/12, which by the same set of calculations would be 3.7 BB/100 for table average rake and 2.5 BB/100 for someone fairly tight.

by contrast, the 10/20 game with the same $3 drop would be 1.6 BB/100 by the table average, and about 1 BB/100 for a tight player to break even.

I think that this is a pretty fair appoximation. Hope that others agree.

Shauna