A local casino is running a promo right now where if a player gets three blackjacks in a row at their spot, they're paid a $500 bonus. If they get a fourth, it's a $10,000 bonus (I think they get the $500 as well). If the dealer gets a blackjack at the same time as the player, the player still gets credit for that blackjack for their "in a row" bonus, although they push their wager.
I was able to use the blackjack calculator on the Wizard of Odds site to determine the game's current edge (without the bonus), but I don't know how to factor in these bonuses. I guess I'm curious as to how much this promo decreases the house edge.
For what it's worth, the minimum bet is $5, and it's a 6-deck shoe game, dealer hits soft 17, player can double after split, can double on any two cards, re-split up to 4 hands, can resplit aces, no surrender option, and blackjack pays 3-to-2.
Thanks for any insight as to how to determine the effect on the house edge!
Card counting doesn't help with the promo since it is a fixed cash amount. It isn't high enough that sophisticates can take advantage of the system.
Is it at Canterbury Park ?
The odds of getting 3 blackjacks in a row is therefore .047451 ^ 3 = 0.01068% with an expected value of $0.05089 (you must take away the 4th blackjack odds).
The odds of getting 4 blackjacks in a row is .047451 ^ 4 = 0.000507% with an expected value of $0.05079.
So the total EV of the promotion is $0.1016.
On a $5 table this is a 2.03% player edge.
On a $10 table this is a 1.02% player edge.
On a $25 table this is a 0.41% player edge.
So if this is a normal 3-2 blackjack game there is a player edge for bets up to about $15.
The odds of getting blackjack is 4.8265% for a single deck, down to 4.7451% for 8 decks. Assuming 6-8 decks and six players at the table, you can negate the effects of card removal.
Maybe This is discussed elsewhere and I've missed it, but why is that there is a greater chance for a B/J on a single-deck game than there is in a multi-deck game? There is the same ratio of Aces/10s in 1 deck as there is 100 decks right?
You are correct, but the difference accounts for the fact that the dealer will more often get blackjack as well when you do, since there are more Aces/10's in a multiple deck than in a single deck. This leads to more pushes/losses depending on whether you have a blackjack or not.
It is true that you will get more pushes in a multi-deck, but it's NOT why it's more likely for a player to get a blackjack.
The reason has to do with card removal and conditional probability. As an example, let's say you are first dealt a 10. In a single deck game, you now have 4/51 chance (7.84%) of hitting your Ace for blackjack. In an 8 deck game, you have 32/415 chance (7.71%). This is slightly less than in the single deck game. The reverse applies as well when you first get an Ace. Together, the effect is more blackjacks when playing single deck than multideck, regardless of what the dealer gets.