In cell B3T3, a Banker holding two-card total 3 (10+3) faces a Player fifth card valued at 3 (3). Under BANKER_TABLEAU[3][3], the Banker DRAWS, resolving in 6 cards with a 21B / 33P / 6T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B3T3 (Banker 3 vs Player 3rd card 3).
This bead is at Row 4. Column 1 stopped at Row 3, one row above → blue (≠).
Common table reading watchout: Thinking "no bead at this row" means blue. It is blue only when Column 1 stopped exactly one row above this bead.
Column 4 has 1 bead. Column 2 has 4 beads. Different depth → blue (≠).
Common table reading watchout: Comparing the wrong columns. Compare Column 4, the one that just finished, with Column 2.
This bead is at Row 3. Column 2 has a bead at Row 3 too → red (=).
Common table reading watchout: Thinking "no bead at this row" means blue. It is blue only when Column 2 stopped exactly one row above this bead.
Verified via bankerShouldDraw(3, 3) === true on pair 10+3.
Exact resolution across 6 legal Player starts (0–5) × 10 sixth-card values (0–9).
Three-card Player totals produced when fifth card 3 (3) lands on P0..P5.
How standing Banker 3 fares against the 6 Player 3-card totals for P3=3.
Sample sixth card 9 (9) verified through resolveHand().
Under [B3T3], Banker holds two-card total 3 (10+3) while Player draws fifth-card 3 (point value 3) onto an initial 0–5 total. Across those six legal Player starts, [B3T3] produces three-card Player totals of P0+3->3, P1+3->4, P2+3->5, P3+3->6, P4+3->7, P5+3->8.
Because [B3T3] sits inside the active draw window for Banker 3, the dealer deals a sixth card to 10+3: across all 60 (P0..P5 × sixth-card 0..9) combinations in [B3T3], Banker captures 21 wins, Player captures 33 wins, and 6 branches tie.
| B3T3 Player Start | B3T3 5th Card | B3T3 Banker 3 Action | B3T3 10-Card Split | B3T3 Sample Result |
|---|---|---|---|---|
| [B3T3:P0] 6+4=0 | [B3T3:5th] +3(3) -> P=3 | [B3T3:6th] B=3 draws (+9->2) | [B3T3:Split] 6B/3P/1T | [B3T3:Cards] 6c (P) |
| [B3T3:P1] 10+A=1 | [B3T3:5th] +3(3) -> P=4 | [B3T3:6th] B=3 draws (+9->2) | [B3T3:Split] 5B/4P/1T | [B3T3:Cards] 6c (P) |
| [B3T3:P2] 8+4=2 | [B3T3:5th] +3(3) -> P=5 | [B3T3:6th] B=3 draws (+9->2) | [B3T3:Split] 4B/5P/1T | [B3T3:Cards] 6c (P) |
| [B3T3:P3] 10+3=3 | [B3T3:5th] +3(3) -> P=6 | [B3T3:6th] B=3 draws (+9->2) | [B3T3:Split] 3B/6P/1T | [B3T3:Cards] 6c (P) |
| [B3T3:P4] A+3=4 | [B3T3:5th] +3(3) -> P=7 | [B3T3:6th] B=3 draws (+9->2) | [B3T3:Split] 2B/7P/1T | [B3T3:Cards] 6c (P) |
| [B3T3:P5] 2+3=5 | [B3T3:5th] +3(3) -> P=8 | [B3T3:6th] B=3 draws (+9->2) | [B3T3:Split] 1B/8P/1T | [B3T3:Cards] 6c (P) |
In 6-card cell [B3T3], Banker 3 (10+3) can reach a 3-card 7 (Dragon 7) by drawing rank 4 or a 3-card 6 (Super 6) by drawing rank 3 after Player's 3.
Once [B3T3] resolves (21B/33P/6T across 60 branches), the winning side logs to the Big Road on shoe PPPBBBBPPB.
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Macau Cantonese, French, and mathematical table terms explained in plain English with road diagrams.
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