How the dealer have advantage in black jack?

234 views

“The dealer must continue to take cards until the total is 17 or more, at which point the dealer must stand”.
It seem too risk for the dealer, am I wrong?

In: 5

15 Answers

Anonymous 0 Comments

Under that set playing strategy, even if the player plays perfect strategy (without making any adjustments based on the composition of the remaining shoe), the dealer pays out less than the bets being placed.

(It’s about a 30% chance of a dealer busting)

Anonymous 0 Comments

The dealer/house has the advantage here because the dealer goes last and you don’t know what both of the dealers cards are. You have to try to beat what you think the dealer might have. The dealer just plays according to pre-defined rules for how they must play and those rules are designed to give the house the edge.

Anonymous 0 Comments

The dealer plays _last_ in Blackjack – that is the advantage.

If you go bust when you play, it doesn’t matter if the dealer busts later as well – you lose your money right then and there.

Anonymous 0 Comments

A dealer at a blackjack table has a statistical advantage over the player because of the rules of the game. One key rule is that the player must act before the dealer, which means that the player may bust (go over 21) before the dealer has to make a decision. Additionally, the dealer must follow a specific set of rules for how to play their hand, which limits their options and increases the chances that they will make a favorable play. Finally, if the player busts, the dealer wins regardless of the value of the dealer’s hand, giving the dealer an additional advantage.

Anonymous 0 Comments

The rules give dealers an advantage. When a player busts (goes over 21), the dealer wins, even if the dealer also busts. They may also win in certain other tie scenarios, depending on how much edge the casino wants.

Anonymous 0 Comments

That about the optimal strategy given no knowledge of the deck. If you were to “hit” on 17, you have the options of getting ace-4 and not busting. 5/13 cards is less than a 40% chance of not busting. It is too risky to hit, it’s less risky to stay.

Because the players play first, the dealer wins every time the players bust, while the players always have to be lucky to win (excluding card counting).

Plus, giving the dealer express instructions for every value they have keeps the dealer from being an unknown variable.

Anonymous 0 Comments

When both the player and the dealer busts the house take the money. So even though the score is the same the house end up with the cash. That gives them the advantage.

Anonymous 0 Comments

Honestly just look at the odds, if the dealer and the player play perfect hands ie take a card or not the dealer is very slightly ahead.
The dealer will play all night following these rules, the player may try get lucky and try for a good card against the odds and loose.

Anonymous 0 Comments

It’s because the player plays first.

Lets imagine a player who plays with the exact same rules that a dealer does (when to hit, when to pass, ect). Lets create a situation where the dealer and player get basically the same cards.

Dealer deals K and 7 to the player. Dealer deals Q and 7 to dealer. (both have 17)

Player plays first, knows they have a 17 so they hit. Get a J, so bust.

The game is now over, the dealer does not draw a card, player has lost and dealer has won.

Anonymous 0 Comments

If you both bust the dealer still wins.

It’s like rock paper scissor but you lose on the tie.