You hold a hard 11 (no ace counted as 11) and the dealer shows a 4. Below is the correct basic-strategy decision computed by the Blackjack Theorem engine, and how it shifts with the table rules.
| Hard | 3 | 4 | 5 |
|---|---|---|---|
| 10 | Dh | Dh | Dh |
| 11 | Dh | Dh | Dh |
| 12 | H | S | S |
Dealer upcard across the top. The outlined cell is the hand on this page.
Across every true count that actually occurs at a six-deck table, from -3 to +7, the correct play stays the same: double.
Double wins by 28.38 points. No true count you will ever see at a table moves a gap that size, so this play is not a judgement call.
Computed by the engine for a full six-deck shoe. The true-count column comes from the same count-dependent table that powers the strategy charts.
| Table rules | Correct play |
|---|---|
| 6-Deck · S17 · DAS | Dh Double if allowed, else Hit |
| 6-Deck · H17 · DAS | Dh Double if allowed, else Hit |
| 6-Deck · S17 · DAS · Surrender | Dh Double if allowed, else Hit |
| 2-Deck · S17 · DAS | Dh Double if allowed, else Hit |
| 8-Deck · H17 · DAS | Dh Double if allowed, else Hit |
The mathematically correct play is: double if allowed, else hit. This is for 6-deck, dealer stands on soft 17, DAS. It can change with other rules (see the table).
It can. The table above shows the play under both S17 and H17 so you can match your exact table.
No. Across every true count that occurs at a six-deck table the correct play stays double. It wins by 28.38 points, a margin no reachable count can close.
