In cell B4T9, a Banker holding two-card total 4 (A+3) faces a Player fifth card valued at 9 (9). Under BANKER_TABLEAU[4][9], 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 B4T9 (Banker 4 vs Player 3rd card 9).
Column 3 has 1 bead. Column 2 has 2 beads. Different depth → blue (≠).
Common table reading watchout: Comparing the wrong columns. Compare Column 3, the one that just finished, with Column 2.
This bead is at Row 3. Column 2 stopped at Row 2, one row above → blue (≠).
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.
Column 4 has 3 beads. Column 1 has 4 beads. Different depth → blue (≠).
Common table reading watchout: Comparing the wrong columns. Compare Column 4, the one that just finished, with Column 1.
Verified via bankerShouldDraw(4, 9) === false on pair A+3.
Exact resolution across 6 legal Player starts (0–5) × 10 sixth-card values (0–9).
Three-card Player totals produced when fifth card 9 (9) lands on P0..P5.
How standing Banker 4 fares against the 6 Player 3-card totals for P3=9.
Sample sixth card 4 (4) verified through resolveHand().
Under [B4T9], Banker holds two-card total 4 (A+3) while Player draws fifth-card 9 (point value 9) onto an initial 0–5 total. Across those six legal Player starts, [B4T9] produces three-card Player totals of P0+9->9, P1+9->0, P2+9->1, P3+9->2, P4+9->3, P5+9->4.
Because [B4T9] falls outside the draw window for Banker 4, the dealer stands pat on A+3 (4) at 5 cards: across the 60 (P0..P5 × unpulled 6th card 0..9) branches in [B4T9], standing yields 40 Banker wins, 10 Player wins, and 10 ties without wraparound risk.
| B4T9 Player Start | B4T9 5th Card | B4T9 Banker 4 Action | B4T9 10-Card Split | B4T9 Sample Result |
|---|---|---|---|---|
| [B4T9:P0] 10+K=0 | [B4T9:5th] +9(9) -> P=9 | [B4T9:Pat] B=4 stands at 4 | [B4T9:Split] 0B/10P/0T | [B4T9:Cards] 5c (P) |
| [B4T9:P1] 7+4=1 | [B4T9:5th] +9(9) -> P=0 | [B4T9:Pat] B=4 stands at 4 | [B4T9:Split] 10B/0P/0T | [B4T9:Cards] 5c (B) |
| [B4T9:P2] 10+2=2 | [B4T9:5th] +9(9) -> P=1 | [B4T9:Pat] B=4 stands at 4 | [B4T9:Split] 10B/0P/0T | [B4T9:Cards] 5c (B) |
| [B4T9:P3] 9+4=3 | [B4T9:5th] +9(9) -> P=2 | [B4T9:Pat] B=4 stands at 4 | [B4T9:Split] 10B/0P/0T | [B4T9:Cards] 5c (B) |
| [B4T9:P4] 2+2=4 | [B4T9:5th] +9(9) -> P=3 | [B4T9:Pat] B=4 stands at 4 | [B4T9:Split] 10B/0P/0T | [B4T9:Cards] 5c (B) |
| [B4T9:P5] A+4=5 | [B4T9:5th] +9(9) -> P=4 | [B4T9:Pat] B=4 stands at 4 | [B4T9:Split] 0B/0P/10T | [B4T9:Cards] 5c (T) |
In 5-card cell [B4T9], Banker 4 (A+3) never takes a sixth card, so 3-card Banker Dragon 7 is impossible while Player reaches 3-card 8 (Panda 8) whenever P=9 draws 9 (9).
Once [B4T9] resolves (40B/10P/10T across 60 branches), the winning side logs to the Big Road on shoe BBBBPPBPPP.
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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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