Ask The Wizard #298
What is the house edge on this awful set of blackjack rules, known as "Party Blackjack" at the Adelaide casino:
- Six decks
- Dealer hits a soft 17
- No dealer hole card. Player loses total amount bet if dealer gets a blackjack.
- Double on 9 to 11 only.
- Double after split allowed.
- Player may re-split to three hands, except no re-splitting aces.
- Blackjack pays 6 to 5.
- Push on dealer 22.
Oy! Those rules are so bad that the hair on the back of my neck stood up reading them.
Using my blackjack house edge calculator, I get a house edge of 2.22% before considering the no hole card and push 22 rules. I'm assuming a continuous shuffler is used and the player can re-split to four hands. The no hole card rule costs the player another 0.11%. However, the most offensive rule is the push 22, which costs the player 6.91%. The bottom line is a house edge of 9.24%! That is simply an abomination to the game.
I invite my readers to let me know if there is a worse set of blackjack rules out there somewhere.
Normally in 6-5 blackjack the player can't take even money. What would be the effect to the house edge if he could?
Assuming six decks, allowing even money in 6-5 blackjack, which the player should take if a blackjack pays only 6-5, then it lowers the house edge by 0.06%.
I know that I am not supposed to play 6-5 blackjack, but should I find myself in this situation is there any time that it would be advantageous to double down on a blackjack instead of taking the 6-5 payout?
Using the Hi-Lo count, I show that against a dealer 5 and blackjacks pay 6 to 5, and if the true count is +18 or greater, then you should double on a blackjack. Not that you asked, but if a blackjack paid 3 to 2, then you would need to get to a count of +27 to double. These numbers should be considered as estimates. Against any other dealer up card, the count would need to be even greater for doubling to be the right play.
This question was raised and discussed in my forum at Wizard of Vegas.
The Rivers City casino in Pittsburgh has a promotion where the shooter in craps wins a bonus of $100,000 for at least 100 rolls of the dice and $1,000,000 for 155. What is the value of that promotion?
The probability of at least 100 rolls is 1 in 2,197,679. Given a win of $100,000, the value is $100,000/2,197,679 = 4.55 ¢.
The probability of at least 155 rolls is 1 in 7,515,200,125. Given an additional win of $900,000 (beyond the $100,000 bonus for 100) the value is $900,000/7,515,200,125 = 0.01 ¢.
So, the total value to the shooter is 4.56 ¢.