In cell B2T0, a Banker holding two-card total 2 (10+2) faces a Player fifth card valued at 0 (10). Under BANKER_TABLEAU[2][0], the Banker DRAWS, resolving in 6 cards with a 39B / 15P / 6T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B2T0 (Banker 2 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(2, 0) === true on pair 10+2.
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 2 fares against the 6 Player 3-card totals for P3=0.
Sample sixth card 7 (7) verified through resolveHand().
Under [B2T0], Banker holds two-card total 2 (10+2) while Player draws fifth-card 10 (point value 0) onto an initial 0–5 total. Across those six legal Player starts, [B2T0] 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 [B2T0] sits inside the active draw window for Banker 2, the dealer deals a sixth card to 10+2: across all 60 (P0..P5 × sixth-card 0..9) combinations in [B2T0], Banker captures 39 wins, Player captures 15 wins, and 6 branches tie.
| B2T0 Player Start | B2T0 5th Card | B2T0 Banker 2 Action | B2T0 10-Card Split | B2T0 Sample Result |
|---|---|---|---|---|
| [B2T0:P0] 10+K=0 | [B2T0:5th] +10(0) -> P=0 | [B2T0:6th] B=2 draws (+7->9) | [B2T0:Split] 9B/0P/1T | [B2T0:Cards] 6c (B) |
| [B2T0:P1] 7+4=1 | [B2T0:5th] +10(0) -> P=1 | [B2T0:6th] B=2 draws (+7->9) | [B2T0:Split] 8B/1P/1T | [B2T0:Cards] 6c (B) |
| [B2T0:P2] 10+2=2 | [B2T0:5th] +10(0) -> P=2 | [B2T0:6th] B=2 draws (+7->9) | [B2T0:Split] 7B/2P/1T | [B2T0:Cards] 6c (B) |
| [B2T0:P3] 9+4=3 | [B2T0:5th] +10(0) -> P=3 | [B2T0:6th] B=2 draws (+7->9) | [B2T0:Split] 6B/3P/1T | [B2T0:Cards] 6c (B) |
| [B2T0:P4] 2+2=4 | [B2T0:5th] +10(0) -> P=4 | [B2T0:6th] B=2 draws (+7->9) | [B2T0:Split] 5B/4P/1T | [B2T0:Cards] 6c (B) |
| [B2T0:P5] A+4=5 | [B2T0:5th] +10(0) -> P=5 | [B2T0:6th] B=2 draws (+7->9) | [B2T0:Split] 4B/5P/1T | [B2T0:Cards] 6c (B) |
In 6-card cell [B2T0], Banker 2 (10+2) can reach a 3-card 7 (Dragon 7) by drawing rank 5 or a 3-card 6 (Super 6) by drawing rank 4 after Player's 10.
Once [B2T0] resolves (39B/15P/6T across 60 branches), the winning side logs to the Big Road on shoe BBBBPBPBPP.
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