Hello there guys
Wondering if someone can help me out with this doubt. Where I play we have these rules (not all of them, just the ones relevant to the question):
- No Hole card dealt
- DAS
- One card per A when you split, nRSA
- If dealer gets a BJ you lose only your original bet. No splits or doubles lost....
....that sounds like american BJ where they check first and you can lose 1 bet only, right? But wait, there's a catch!. This applies to the bets standing on the table, since busted hands were already lost. Which busted hands, you say? You can't bust on doubling 11 or 10 against A or 10, or splitting A against A
But what about the pair of 8's or even pair of 9's if you get the right index according to your count system?
If you split pair of 8's you can lose all the hands due to busting or only one bet or two....
Whatever is left on the table when the dealer gets a BJ, they check if your ORIGINAL BET is still on the table. If it is, you lose only that one. If it's not on the table since you busted, you lose nothing.
The closest BJ handling type I've found to met these rules is BB+1, but it's not exactly the same. Why? Because if I split pair of 8's, got a 7 on first hand, hit and busted, then got a 10 on second hand and stay with 18 and the dealer gets a BJ, I don't lose anything more, since my original bet is already gone from the table. And BB+1 would mean that they would have to take one aditional bet besides the ones I already lost.
Is this OBBO?
To be more clear, let's take this example. Pair of 8's split to 3 hands
1st hand: 8, 7, 7 busted
2nd hand: 8, 3, double, 6 = 17 doubled down
3rd hand: 8, 10 = 18
In this casino I would lose only the busted hand (1 unit from 1st Hand). They won't even look at hand 2 or 3
OBBO from what I understand means I will also lose the original bet from hand 2.
I'd appreciate the help, thanks!
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