In cell B7T1, a Banker holding two-card total 7 (3+4) faces a Player fifth card valued at 1 (A). Under BANKER_TABLEAU[7][1], the Banker STANDS, resolving in 5 cards with a 60B / 0P / 0T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B7T1 (Banker 7 vs Player 3rd card 1).
Column 2 has 3 beads. Column 1 has 2 beads. Different depth → blue (≠).
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 2, more than one row above → red (=).
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 6 has 1 bead. Column 3 has 4 beads. Different depth → blue (≠).
Common table reading watchout: Comparing the wrong columns. Compare Column 6, the one that just finished, with Column 3.
Verified via bankerShouldDraw(7, 1) === false on pair 3+4.
Exact resolution across 6 legal Player starts (0–5) × 10 sixth-card values (0–9).
Three-card Player totals produced when fifth card A (1) lands on P0..P5.
How standing Banker 7 fares against the 6 Player 3-card totals for P3=1.
Sample sixth card 3 (3) verified through resolveHand().
Under [B7T1], Banker holds two-card total 7 (3+4) while Player draws fifth-card A (point value 1) onto an initial 0–5 total. Across those six legal Player starts, [B7T1] produces three-card Player totals of P0+1->1, P1+1->2, P2+1->3, P3+1->4, P4+1->5, P5+1->6.
Because [B7T1] falls outside the draw window for Banker 7, the dealer stands pat on 3+4 (7) at 5 cards: across the 60 (P0..P5 × unpulled 6th card 0..9) branches in [B7T1], standing yields 60 Banker wins, 0 Player wins, and 0 ties without wraparound risk.
| B7T1 Player Start | B7T1 5th Card | B7T1 Banker 7 Action | B7T1 10-Card Split | B7T1 Sample Result |
|---|---|---|---|---|
| [B7T1:P0] 6+4=0 | [B7T1:5th] +A(1) -> P=1 | [B7T1:Pat] B=7 stands at 7 | [B7T1:Split] 10B/0P/0T | [B7T1:Cards] 5c (B) |
| [B7T1:P1] 10+A=1 | [B7T1:5th] +A(1) -> P=2 | [B7T1:Pat] B=7 stands at 7 | [B7T1:Split] 10B/0P/0T | [B7T1:Cards] 5c (B) |
| [B7T1:P2] 8+4=2 | [B7T1:5th] +A(1) -> P=3 | [B7T1:Pat] B=7 stands at 7 | [B7T1:Split] 10B/0P/0T | [B7T1:Cards] 5c (B) |
| [B7T1:P3] 10+3=3 | [B7T1:5th] +A(1) -> P=4 | [B7T1:Pat] B=7 stands at 7 | [B7T1:Split] 10B/0P/0T | [B7T1:Cards] 5c (B) |
| [B7T1:P4] A+3=4 | [B7T1:5th] +A(1) -> P=5 | [B7T1:Pat] B=7 stands at 7 | [B7T1:Split] 10B/0P/0T | [B7T1:Cards] 5c (B) |
| [B7T1:P5] 2+3=5 | [B7T1:5th] +A(1) -> P=6 | [B7T1:Pat] B=7 stands at 7 | [B7T1:Split] 10B/0P/0T | [B7T1:Cards] 5c (B) |
In 5-card cell [B7T1], Banker 7 (3+4) never takes a sixth card, so 3-card Banker Dragon 7 is impossible while Player reaches 3-card 8 (Panda 8) whenever P=7 draws A (1).
Once [B7T1] resolves (60B/0P/0T across 60 branches), the winning side logs to the Big Road on shoe BBPPPBBBBP.
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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.
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