In cell B3T4, a Banker holding two-card total 3 (9+4) faces a Player fifth card valued at 4 (4). Under BANKER_TABLEAU[3][4], the Banker DRAWS, resolving in 6 cards with a 15B / 39P / 6T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B3T4 (Banker 3 vs Player 3rd card 4).
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(3, 4) === true on pair 9+4.
Exact resolution across 6 legal Player starts (0–5) × 10 sixth-card values (0–9).
Three-card Player totals produced when fifth card 4 (4) lands on P0..P5.
How standing Banker 3 fares against the 6 Player 3-card totals for P3=4.
Sample sixth card 10 (0) verified through resolveHand().
Under [B3T4], Banker holds two-card total 3 (9+4) while Player draws fifth-card 4 (point value 4) onto an initial 0–5 total. Across those six legal Player starts, [B3T4] produces three-card Player totals of P0+4->4, P1+4->5, P2+4->6, P3+4->7, P4+4->8, P5+4->9.
Because [B3T4] sits inside the active draw window for Banker 3, the dealer deals a sixth card to 9+4: across all 60 (P0..P5 × sixth-card 0..9) combinations in [B3T4], Banker captures 15 wins, Player captures 39 wins, and 6 branches tie.
| B3T4 Player Start | B3T4 5th Card | B3T4 Banker 3 Action | B3T4 10-Card Split | B3T4 Sample Result |
|---|---|---|---|---|
| [B3T4:P0] 6+4=0 | [B3T4:5th] +4(4) -> P=4 | [B3T4:6th] B=3 draws (+10->3) | [B3T4:Split] 5B/4P/1T | [B3T4:Cards] 6c (P) |
| [B3T4:P1] 10+A=1 | [B3T4:5th] +4(4) -> P=5 | [B3T4:6th] B=3 draws (+10->3) | [B3T4:Split] 4B/5P/1T | [B3T4:Cards] 6c (P) |
| [B3T4:P2] 8+4=2 | [B3T4:5th] +4(4) -> P=6 | [B3T4:6th] B=3 draws (+10->3) | [B3T4:Split] 3B/6P/1T | [B3T4:Cards] 6c (P) |
| [B3T4:P3] 10+3=3 | [B3T4:5th] +4(4) -> P=7 | [B3T4:6th] B=3 draws (+10->3) | [B3T4:Split] 2B/7P/1T | [B3T4:Cards] 6c (P) |
| [B3T4:P4] A+3=4 | [B3T4:5th] +4(4) -> P=8 | [B3T4:6th] B=3 draws (+10->3) | [B3T4:Split] 1B/8P/1T | [B3T4:Cards] 6c (P) |
| [B3T4:P5] 2+3=5 | [B3T4:5th] +4(4) -> P=9 | [B3T4:6th] B=3 draws (+10->3) | [B3T4:Split] 0B/9P/1T | [B3T4:Cards] 6c (P) |
In 6-card cell [B3T4], Banker 3 (9+4) can reach a 3-card 7 (Dragon 7) by drawing rank 4 or a 3-card 6 (Super 6) by drawing rank 3 after Player's 4.
Once [B3T4] resolves (15B/39P/6T across 60 branches), the winning side logs to the Big Road on shoe BBBBPBPBPP.
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