In cell B6T2, a Banker holding two-card total 6 (3+3) faces a Player fifth card valued at 2 (2). Under BANKER_TABLEAU[6][2], the Banker STANDS, resolving in 5 cards with a 40B / 10P / 10T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B6T2 (Banker 6 vs Player 3rd card 2).
Column 2 has 3 beads. Column 1 has 3 beads. Same depth → red (=).
Common table reading watchout: Comparing the wrong columns. Compare Column 2, the one that just finished, with Column 1.
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.
This bead is at Row 2. Column 2 has a bead at Row 2 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(6, 2) === false on pair 3+3.
Exact resolution across 6 legal Player starts (0–5) × 10 sixth-card values (0–9).
Three-card Player totals produced when fifth card 2 (2) lands on P0..P5.
How standing Banker 6 fares against the 6 Player 3-card totals for P3=2.
Sample sixth card 5 (5) verified through resolveHand().
Under [B6T2], Banker holds two-card total 6 (3+3) while Player draws fifth-card 2 (point value 2) onto an initial 0–5 total. Across those six legal Player starts, [B6T2] produces three-card Player totals of P0+2->2, P1+2->3, P2+2->4, P3+2->5, P4+2->6, P5+2->7.
Because [B6T2] falls outside the draw window for Banker 6, the dealer stands pat on 3+3 (6) at 5 cards: across the 60 (P0..P5 × unpulled 6th card 0..9) branches in [B6T2], standing yields 40 Banker wins, 10 Player wins, and 10 ties without wraparound risk.
| B6T2 Player Start | B6T2 5th Card | B6T2 Banker 6 Action | B6T2 10-Card Split | B6T2 Sample Result |
|---|---|---|---|---|
| [B6T2:P0] 10+K=0 | [B6T2:5th] +2(2) -> P=2 | [B6T2:Pat] B=6 stands at 6 | [B6T2:Split] 10B/0P/0T | [B6T2:Cards] 5c (B) |
| [B6T2:P1] 7+4=1 | [B6T2:5th] +2(2) -> P=3 | [B6T2:Pat] B=6 stands at 6 | [B6T2:Split] 10B/0P/0T | [B6T2:Cards] 5c (B) |
| [B6T2:P2] 10+2=2 | [B6T2:5th] +2(2) -> P=4 | [B6T2:Pat] B=6 stands at 6 | [B6T2:Split] 10B/0P/0T | [B6T2:Cards] 5c (B) |
| [B6T2:P3] 9+4=3 | [B6T2:5th] +2(2) -> P=5 | [B6T2:Pat] B=6 stands at 6 | [B6T2:Split] 10B/0P/0T | [B6T2:Cards] 5c (B) |
| [B6T2:P4] 2+2=4 | [B6T2:5th] +2(2) -> P=6 | [B6T2:Pat] B=6 stands at 6 | [B6T2:Split] 0B/0P/10T | [B6T2:Cards] 5c (T) |
| [B6T2:P5] A+4=5 | [B6T2:5th] +2(2) -> P=7 | [B6T2:Pat] B=6 stands at 6 | [B6T2:Split] 0B/10P/0T | [B6T2:Cards] 5c (P) |
In 5-card cell [B6T2], Banker 6 (3+3) never takes a sixth card, so 3-card Banker Dragon 7 is impossible while Player reaches 3-card 8 (Panda 8) whenever P=6 draws 2 (2).
Once [B6T2] resolves (40B/10P/10T across 60 branches), the winning side logs to the Big Road on shoe BBBPPPBBBB.
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About 64 minutes of lessons covering outcomes, the Bead Plate, Big Road columns, and dragon tails.
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Macau Cantonese, French, and mathematical table terms explained in plain English with road diagrams.
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