In cell B6T3, a Banker holding two-card total 6 (2+4) faces a Player fifth card valued at 3 (3). Under BANKER_TABLEAU[6][3], the Banker STANDS, resolving in 5 cards with a 30B / 20P / 10T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B6T3 (Banker 6 vs Player 3rd card 3).
This bead is at Row 3. Column 2 stopped at Row 1, more than one row above → 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.
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 5 has 1 bead. Column 2 has 1 bead. Same depth → red (=).
Common table reading watchout: Comparing the wrong columns. Compare Column 5, the one that just finished, with Column 2.
Verified via bankerShouldDraw(6, 3) === false on pair 2+4.
Exact resolution across 6 legal Player starts (0–5) × 10 sixth-card values (0–9).
Three-card Player totals produced when fifth card 3 (3) lands on P0..P5.
How standing Banker 6 fares against the 6 Player 3-card totals for P3=3.
Sample sixth card 6 (6) verified through resolveHand().
Under [B6T3], Banker holds two-card total 6 (2+4) while Player draws fifth-card 3 (point value 3) onto an initial 0–5 total. Across those six legal Player starts, [B6T3] produces three-card Player totals of P0+3->3, P1+3->4, P2+3->5, P3+3->6, P4+3->7, P5+3->8.
Because [B6T3] falls outside the draw window for Banker 6, the dealer stands pat on 2+4 (6) at 5 cards: across the 60 (P0..P5 × unpulled 6th card 0..9) branches in [B6T3], standing yields 30 Banker wins, 20 Player wins, and 10 ties without wraparound risk.
| B6T3 Player Start | B6T3 5th Card | B6T3 Banker 6 Action | B6T3 10-Card Split | B6T3 Sample Result |
|---|---|---|---|---|
| [B6T3:P0] 10+K=0 | [B6T3:5th] +3(3) -> P=3 | [B6T3:Pat] B=6 stands at 6 | [B6T3:Split] 10B/0P/0T | [B6T3:Cards] 5c (B) |
| [B6T3:P1] 7+4=1 | [B6T3:5th] +3(3) -> P=4 | [B6T3:Pat] B=6 stands at 6 | [B6T3:Split] 10B/0P/0T | [B6T3:Cards] 5c (B) |
| [B6T3:P2] 10+2=2 | [B6T3:5th] +3(3) -> P=5 | [B6T3:Pat] B=6 stands at 6 | [B6T3:Split] 10B/0P/0T | [B6T3:Cards] 5c (B) |
| [B6T3:P3] 9+4=3 | [B6T3:5th] +3(3) -> P=6 | [B6T3:Pat] B=6 stands at 6 | [B6T3:Split] 0B/0P/10T | [B6T3:Cards] 5c (T) |
| [B6T3:P4] 2+2=4 | [B6T3:5th] +3(3) -> P=7 | [B6T3:Pat] B=6 stands at 6 | [B6T3:Split] 0B/10P/0T | [B6T3:Cards] 5c (P) |
| [B6T3:P5] A+4=5 | [B6T3:5th] +3(3) -> P=8 | [B6T3:Pat] B=6 stands at 6 | [B6T3:Split] 0B/10P/0T | [B6T3:Cards] 5c (P) |
In 5-card cell [B6T3], Banker 6 (2+4) never takes a sixth card, so 3-card Banker Dragon 7 is impossible while Player reaches 3-card 8 (Panda 8) whenever P=5 draws 3 (3).
Once [B6T3] resolves (30B/20P/10T across 60 branches), the winning side logs to the Big Road on shoe BBPBBBBPPB.
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