In cell B7T5, a Banker holding two-card total 7 (3+4) faces a Player fifth card valued at 5 (5). Under BANKER_TABLEAU[7][5], 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 B7T5 (Banker 7 vs Player 3rd card 5).
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 3 has 1 bead. Same depth → red (=).
Common table reading watchout: Comparing the wrong columns. Compare Column 5, the one that just finished, with Column 3.
This bead is at Row 2. Column 3 stopped at Row 1, one row above → blue (≠).
Common table reading watchout: Thinking "no bead at this row" means blue. It is blue only when Column 3 stopped exactly one row above this bead.
Verified via bankerShouldDraw(7, 5) === 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 5 (5) lands on P0..P5.
How standing Banker 7 fares against the 6 Player 3-card totals for P3=5.
Sample sixth card 7 (7) verified through resolveHand().
Under [B7T5], Banker holds two-card total 7 (3+4) while Player draws fifth-card 5 (point value 5) onto an initial 0–5 total. Across those six legal Player starts, [B7T5] produces three-card Player totals of P0+5->5, P1+5->6, P2+5->7, P3+5->8, P4+5->9, P5+5->0.
Because [B7T5] 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 [B7T5], standing yields 30 Banker wins, 20 Player wins, and 10 ties without wraparound risk.
| B7T5 Player Start | B7T5 5th Card | B7T5 Banker 7 Action | B7T5 10-Card Split | B7T5 Sample Result |
|---|---|---|---|---|
| [B7T5:P0] 6+4=0 | [B7T5:5th] +5(5) -> P=5 | [B7T5:Pat] B=7 stands at 7 | [B7T5:Split] 10B/0P/0T | [B7T5:Cards] 5c (B) |
| [B7T5:P1] 10+A=1 | [B7T5:5th] +5(5) -> P=6 | [B7T5:Pat] B=7 stands at 7 | [B7T5:Split] 10B/0P/0T | [B7T5:Cards] 5c (B) |
| [B7T5:P2] 8+4=2 | [B7T5:5th] +5(5) -> P=7 | [B7T5:Pat] B=7 stands at 7 | [B7T5:Split] 0B/0P/10T | [B7T5:Cards] 5c (T) |
| [B7T5:P3] 10+3=3 | [B7T5:5th] +5(5) -> P=8 | [B7T5:Pat] B=7 stands at 7 | [B7T5:Split] 0B/10P/0T | [B7T5:Cards] 5c (P) |
| [B7T5:P4] A+3=4 | [B7T5:5th] +5(5) -> P=9 | [B7T5:Pat] B=7 stands at 7 | [B7T5:Split] 0B/10P/0T | [B7T5:Cards] 5c (P) |
| [B7T5:P5] 2+3=5 | [B7T5:5th] +5(5) -> P=0 | [B7T5:Pat] B=7 stands at 7 | [B7T5:Split] 10B/0P/0T | [B7T5:Cards] 5c (B) |
In 5-card cell [B7T5], 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=3 draws 5 (5).
Once [B7T5] resolves (30B/20P/10T across 60 branches), the winning side logs to the Big Road on shoe PPBBBBPBPB.
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