In cell B7T0, a Banker holding two-card total 7 (2+5) faces a Player fifth card valued at 0 (10). Under BANKER_TABLEAU[7][0], 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 B7T0 (Banker 7 vs Player 3rd card 0).
Column 4 has 1 bead. Column 3 has 1 bead. Same depth → red (=).
Common table reading watchout: Comparing the wrong columns. Compare Column 4, the one that just finished, with Column 3.
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, 0) === false on pair 2+5.
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 7 fares against the 6 Player 3-card totals for P3=0.
Sample sixth card 2 (2) verified through resolveHand().
Under [B7T0], Banker holds two-card total 7 (2+5) while Player draws fifth-card 10 (point value 0) onto an initial 0–5 total. Across those six legal Player starts, [B7T0] 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 [B7T0] falls outside the draw window for Banker 7, the dealer stands pat on 2+5 (7) at 5 cards: across the 60 (P0..P5 × unpulled 6th card 0..9) branches in [B7T0], standing yields 60 Banker wins, 0 Player wins, and 0 ties without wraparound risk.
| B7T0 Player Start | B7T0 5th Card | B7T0 Banker 7 Action | B7T0 10-Card Split | B7T0 Sample Result |
|---|---|---|---|---|
| [B7T0:P0] 6+4=0 | [B7T0:5th] +10(0) -> P=0 | [B7T0:Pat] B=7 stands at 7 | [B7T0:Split] 10B/0P/0T | [B7T0:Cards] 5c (B) |
| [B7T0:P1] 10+A=1 | [B7T0:5th] +10(0) -> P=1 | [B7T0:Pat] B=7 stands at 7 | [B7T0:Split] 10B/0P/0T | [B7T0:Cards] 5c (B) |
| [B7T0:P2] 8+4=2 | [B7T0:5th] +10(0) -> P=2 | [B7T0:Pat] B=7 stands at 7 | [B7T0:Split] 10B/0P/0T | [B7T0:Cards] 5c (B) |
| [B7T0:P3] 10+3=3 | [B7T0:5th] +10(0) -> P=3 | [B7T0:Pat] B=7 stands at 7 | [B7T0:Split] 10B/0P/0T | [B7T0:Cards] 5c (B) |
| [B7T0:P4] A+3=4 | [B7T0:5th] +10(0) -> P=4 | [B7T0:Pat] B=7 stands at 7 | [B7T0:Split] 10B/0P/0T | [B7T0:Cards] 5c (B) |
| [B7T0:P5] 2+3=5 | [B7T0:5th] +10(0) -> P=5 | [B7T0:Pat] B=7 stands at 7 | [B7T0:Split] 10B/0P/0T | [B7T0:Cards] 5c (B) |
In 5-card cell [B7T0], Banker 7 (2+5) 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 [B7T0] resolves (60B/0P/0T across 60 branches), the winning side logs to the Big Road on shoe BBBBPBPBPP.
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