In cell B0T2, a Banker holding two-card total 0 (10+K) faces a Player fifth card valued at 2 (2). Under BANKER_TABLEAU[0][2], the Banker DRAWS, resolving in 6 cards with a 27B / 27P / 6T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B0T2 (Banker 0 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(0, 2) === true on pair 10+K.
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 0 fares against the 6 Player 3-card totals for P3=2.
Sample sixth card A (1) verified through resolveHand().
Under [B0T2], Banker holds two-card total 0 (10+K) while Player draws fifth-card 2 (point value 2) onto an initial 0–5 total. Across those six legal Player starts, [B0T2] 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 [B0T2] sits inside the active draw window for Banker 0, the dealer deals a sixth card to 10+K: across all 60 (P0..P5 × sixth-card 0..9) combinations in [B0T2], Banker captures 27 wins, Player captures 27 wins, and 6 branches tie.
| B0T2 Player Start | B0T2 5th Card | B0T2 Banker 0 Action | B0T2 10-Card Split | B0T2 Sample Result |
|---|---|---|---|---|
| [B0T2:P0] 10+K=0 | [B0T2:5th] +2(2) -> P=2 | [B0T2:6th] B=0 draws (+A->1) | [B0T2:Split] 7B/2P/1T | [B0T2:Cards] 6c (P) |
| [B0T2:P1] 7+4=1 | [B0T2:5th] +2(2) -> P=3 | [B0T2:6th] B=0 draws (+A->1) | [B0T2:Split] 6B/3P/1T | [B0T2:Cards] 6c (P) |
| [B0T2:P2] 10+2=2 | [B0T2:5th] +2(2) -> P=4 | [B0T2:6th] B=0 draws (+A->1) | [B0T2:Split] 5B/4P/1T | [B0T2:Cards] 6c (P) |
| [B0T2:P3] 9+4=3 | [B0T2:5th] +2(2) -> P=5 | [B0T2:6th] B=0 draws (+A->1) | [B0T2:Split] 4B/5P/1T | [B0T2:Cards] 6c (P) |
| [B0T2:P4] 2+2=4 | [B0T2:5th] +2(2) -> P=6 | [B0T2:6th] B=0 draws (+A->1) | [B0T2:Split] 3B/6P/1T | [B0T2:Cards] 6c (P) |
| [B0T2:P5] A+4=5 | [B0T2:5th] +2(2) -> P=7 | [B0T2:6th] B=0 draws (+A->1) | [B0T2:Split] 2B/7P/1T | [B0T2:Cards] 6c (P) |
In 6-card cell [B0T2], Banker 0 (10+K) can reach a 3-card 7 (Dragon 7) by drawing rank 7 or a 3-card 6 (Super 6) by drawing rank 6 after Player's 2.
Once [B0T2] resolves (27B/27P/6T 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.
Train Third-Card Rule decisions, depth counting, lookback column references, and derived-road speed recognition.
Visual step-by-step breakdowns with closed captions, chapter markers, and full searchable transcripts.
Macau Cantonese, French, and mathematical table terms explained in plain English with road diagrams.
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