'Double up blackjack' - outcome of my 20 vs dealer 10
- Thread starter jacblacc911
- Start date
jacblacc911 said:
What percentage of the time/how often does my inital 2 card total of 20 win vs a dealer 10? And how often does it push? Assume neutral deck.
Don
Yes, it is 6 decks.
In this BJ variant, using the example of a 2 card total of 20, I have the option of placing the ‘double up wager’ (equal to my original bet - pays 1/1). If however the dealer makes 20, I lose the double up wager and push the original.
So for every situation of a 2 card 20 vs 10, I stand to win 2 units (original 1 unit, double up wager 1 unit) if dealer makes 19 or less or lose 1 unit if the dealer makes 20.
In this BJ variant, using the example of a 2 card total of 20, I have the option of placing the ‘double up wager’ (equal to my original bet - pays 1/1). If however the dealer makes 20, I lose the double up wager and push the original.
So for every situation of a 2 card 20 vs 10, I stand to win 2 units (original 1 unit, double up wager 1 unit) if dealer makes 19 or less or lose 1 unit if the dealer makes 20.
"Yes, it is 6 decks."
OK.
"In this BJ variant, using the example of a 2 card total of 20, I have the option of placing the ‘double up wager’ (equal to my original bet - pays 1/1). If however the dealer makes 20, I lose the double up wager and push the original.
Sounds too good to be true. He obviously won't make 20 more than about 40% of the time. So, you lose one unit 40% and win two units 60%. Why wouldn't you want to do that all day long?
"So for every situation of a 2 card 20 vs 10, I stand to win 2 units (original 1 unit, double up wager 1 unit) if dealer makes 19 or less or lose 1 unit if the dealer makes 20."
No, you also win the two units if he breaks his 10, no? That happens 23% of the time!
Don
OK.
"In this BJ variant, using the example of a 2 card total of 20, I have the option of placing the ‘double up wager’ (equal to my original bet - pays 1/1). If however the dealer makes 20, I lose the double up wager and push the original.
Sounds too good to be true. He obviously won't make 20 more than about 40% of the time. So, you lose one unit 40% and win two units 60%. Why wouldn't you want to do that all day long?
"So for every situation of a 2 card 20 vs 10, I stand to win 2 units (original 1 unit, double up wager 1 unit) if dealer makes 19 or less or lose 1 unit if the dealer makes 20."
No, you also win the two units if he breaks his 10, no? That happens 23% of the time!
Don
It’s ENHC - dealer takes all original bets if he gets BJ. So if you doubled 11 against Ace or split Aces vs Ace and dealer got BJ then you would lose all. BUT....If you split Aces vs Ace and get AT AT, then place the 'double up wager' on each and the dealer gets BJ, you just lose your original bets, not the doubled up wagers.
DSchles said:
This needs to be simulated. A bit too involved to do analytically.
Anyone else want to make an intuitive guestimate before wizard or someone else does a simulation assessment?
xengrifter said:
I am revising my estimate upwards, I have this pegged at 1.3% starting house edge, with a modified optimal BS.
Anyone else want to make an intuitive guestimate before wizard or someone else does a simulation assessment?
Anyone else want to make an intuitive guestimate before wizard or someone else does a simulation assessment?
However what isn’t clear is whether or not the ENHC rule was included or not. If not, then can I just simply add the value of the ENHC disadvantage (-0.11) to this HE of 0.32% to get 0.43% HE?
Also, can I do the same with other values pertaining to different rule variations?
xengrifter said:
Okay, I was only off by 1%!
Post the wizard double up blackjack strategy here, please.
Post the wizard double up blackjack strategy here, please.