In cell B5T1, a Banker holding two-card total 5 (2+3) faces a Player fifth card valued at 1 (A). Under BANKER_TABLEAU[5][1], 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 B5T1 (Banker 5 vs Player 3rd card 1).
This bead is at Row 4. Column 1 stopped at Row 3, one row above → blue (≠).
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 4 has 1 bead. Column 2 has 4 beads. Different depth → blue (≠).
Common table reading watchout: Comparing the wrong columns. Compare Column 4, the one that just finished, with Column 2.
This bead is at Row 3. Column 2 has a bead at Row 3 too → 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.
Verified via bankerShouldDraw(5, 1) === false on pair 2+3.
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 5 fares against the 6 Player 3-card totals for P3=1.
Sample sixth card 5 (5) verified through resolveHand().
Under [B5T1], Banker holds two-card total 5 (2+3) while Player draws fifth-card A (point value 1) onto an initial 0–5 total. Across those six legal Player starts, [B5T1] 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 [B5T1] falls outside the draw window for Banker 5, the dealer stands pat on 2+3 (5) at 5 cards: across the 60 (P0..P5 × unpulled 6th card 0..9) branches in [B5T1], standing yields 40 Banker wins, 10 Player wins, and 10 ties without wraparound risk.
| B5T1 Player Start | B5T1 5th Card | B5T1 Banker 5 Action | B5T1 10-Card Split | B5T1 Sample Result |
|---|---|---|---|---|
| [B5T1:P0] 6+4=0 | [B5T1:5th] +A(1) -> P=1 | [B5T1:Pat] B=5 stands at 5 | [B5T1:Split] 10B/0P/0T | [B5T1:Cards] 5c (B) |
| [B5T1:P1] 10+A=1 | [B5T1:5th] +A(1) -> P=2 | [B5T1:Pat] B=5 stands at 5 | [B5T1:Split] 10B/0P/0T | [B5T1:Cards] 5c (B) |
| [B5T1:P2] 8+4=2 | [B5T1:5th] +A(1) -> P=3 | [B5T1:Pat] B=5 stands at 5 | [B5T1:Split] 10B/0P/0T | [B5T1:Cards] 5c (B) |
| [B5T1:P3] 10+3=3 | [B5T1:5th] +A(1) -> P=4 | [B5T1:Pat] B=5 stands at 5 | [B5T1:Split] 10B/0P/0T | [B5T1:Cards] 5c (B) |
| [B5T1:P4] A+3=4 | [B5T1:5th] +A(1) -> P=5 | [B5T1:Pat] B=5 stands at 5 | [B5T1:Split] 0B/0P/10T | [B5T1:Cards] 5c (T) |
| [B5T1:P5] 2+3=5 | [B5T1:5th] +A(1) -> P=6 | [B5T1:Pat] B=5 stands at 5 | [B5T1:Split] 0B/10P/0T | [B5T1:Cards] 5c (P) |
In 5-card cell [B5T1], Banker 5 (2+3) 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 [B5T1] resolves (40B/10P/10T across 60 branches), the winning side logs to the Big Road on shoe PPPBBBBPPB.
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