In cell B4T0, a Banker holding two-card total 4 (2+2) faces a Player fifth card valued at 0 (10). Under BANKER_TABLEAU[4][0], 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 B4T0 (Banker 4 vs Player 3rd card 0).
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, 0) === false on pair 2+2.
Exact resolution across 6 legal Player starts (0–5) × 10 sixth-card values (0–9).
Three-card Player totals produced when fifth card 10 (0) lands on P0..P5.
How standing Banker 4 fares against the 6 Player 3-card totals for P3=0.
Sample sixth card 5 (5) verified through resolveHand().
Under [B4T0], Banker holds two-card total 4 (2+2) while Player draws fifth-card 10 (point value 0) onto an initial 0–5 total. Across those six legal Player starts, [B4T0] produces three-card Player totals of P0+0->0, P1+0->1, P2+0->2, P3+0->3, P4+0->4, P5+0->5.
Because [B4T0] falls outside the draw window for Banker 4, the dealer stands pat on 2+2 (4) at 5 cards: across the 60 (P0..P5 × unpulled 6th card 0..9) branches in [B4T0], standing yields 40 Banker wins, 10 Player wins, and 10 ties without wraparound risk.
| B4T0 Player Start | B4T0 5th Card | B4T0 Banker 4 Action | B4T0 10-Card Split | B4T0 Sample Result |
|---|---|---|---|---|
| [B4T0:P0] 10+K=0 | [B4T0:5th] +10(0) -> P=0 | [B4T0:Pat] B=4 stands at 4 | [B4T0:Split] 10B/0P/0T | [B4T0:Cards] 5c (B) |
| [B4T0:P1] 7+4=1 | [B4T0:5th] +10(0) -> P=1 | [B4T0:Pat] B=4 stands at 4 | [B4T0:Split] 10B/0P/0T | [B4T0:Cards] 5c (B) |
| [B4T0:P2] 10+2=2 | [B4T0:5th] +10(0) -> P=2 | [B4T0:Pat] B=4 stands at 4 | [B4T0:Split] 10B/0P/0T | [B4T0:Cards] 5c (B) |
| [B4T0:P3] 9+4=3 | [B4T0:5th] +10(0) -> P=3 | [B4T0:Pat] B=4 stands at 4 | [B4T0:Split] 10B/0P/0T | [B4T0:Cards] 5c (B) |
| [B4T0:P4] 2+2=4 | [B4T0:5th] +10(0) -> P=4 | [B4T0:Pat] B=4 stands at 4 | [B4T0:Split] 0B/0P/10T | [B4T0:Cards] 5c (T) |
| [B4T0:P5] A+4=5 | [B4T0:5th] +10(0) -> P=5 | [B4T0:Pat] B=4 stands at 4 | [B4T0:Split] 0B/10P/0T | [B4T0:Cards] 5c (P) |
In 5-card cell [B4T0], Banker 4 (2+2) never takes a sixth card, so 3-card Banker Dragon 7 is impossible while Player reaches 3-card 8 (Panda 8) whenever P=8 draws 10 (0).
Once [B4T0] 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.
Direct answers to real student questions on dragon tails, tie handling, lookback rules, and table variants.
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