In cell B0T5, a Banker holding two-card total 0 (6+4) faces a Player fifth card valued at 5 (5). Under BANKER_TABLEAU[0][5], the Banker DRAWS, resolving in 6 cards with a 19B / 35P / 6T split across the 60 (P0..P5 × 6th-card 0..9) branches.
Verified scoreboard paired with tableau cell B0T5 (Banker 0 vs Player 3rd card 5).
Column 3 has 1 bead. Column 2 has 2 beads. Different depth → blue (≠).
Common table reading watchout: Comparing the wrong columns. Compare Column 3, the one that just finished, with Column 2.
This bead is at Row 3. Column 2 stopped at Row 2, one row above → blue (≠).
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.
Column 4 has 3 beads. Column 1 has 4 beads. Different depth → blue (≠).
Common table reading watchout: Comparing the wrong columns. Compare Column 4, the one that just finished, with Column 1.
Verified via bankerShouldDraw(0, 5) === true on pair 6+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 0 fares against the 6 Player 3-card totals for P3=5.
Sample sixth card 4 (4) verified through resolveHand().
Under [B0T5], Banker holds two-card total 0 (6+4) while Player draws fifth-card 5 (point value 5) onto an initial 0–5 total. Across those six legal Player starts, [B0T5] 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 [B0T5] sits inside the active draw window for Banker 0, the dealer deals a sixth card to 6+4: across all 60 (P0..P5 × sixth-card 0..9) combinations in [B0T5], Banker captures 19 wins, Player captures 35 wins, and 6 branches tie.
| B0T5 Player Start | B0T5 5th Card | B0T5 Banker 0 Action | B0T5 10-Card Split | B0T5 Sample Result |
|---|---|---|---|---|
| [B0T5:P0] 10+K=0 | [B0T5:5th] +5(5) -> P=5 | [B0T5:6th] B=0 draws (+4->4) | [B0T5:Split] 4B/5P/1T | [B0T5:Cards] 6c (P) |
| [B0T5:P1] 7+4=1 | [B0T5:5th] +5(5) -> P=6 | [B0T5:6th] B=0 draws (+4->4) | [B0T5:Split] 3B/6P/1T | [B0T5:Cards] 6c (P) |
| [B0T5:P2] 10+2=2 | [B0T5:5th] +5(5) -> P=7 | [B0T5:6th] B=0 draws (+4->4) | [B0T5:Split] 2B/7P/1T | [B0T5:Cards] 6c (P) |
| [B0T5:P3] 9+4=3 | [B0T5:5th] +5(5) -> P=8 | [B0T5:6th] B=0 draws (+4->4) | [B0T5:Split] 1B/8P/1T | [B0T5:Cards] 6c (P) |
| [B0T5:P4] 2+2=4 | [B0T5:5th] +5(5) -> P=9 | [B0T5:6th] B=0 draws (+4->4) | [B0T5:Split] 0B/9P/1T | [B0T5:Cards] 6c (P) |
| [B0T5:P5] A+4=5 | [B0T5:5th] +5(5) -> P=0 | [B0T5:6th] B=0 draws (+4->4) | [B0T5:Split] 9B/0P/1T | [B0T5:Cards] 6c (B) |
In 6-card cell [B0T5], Banker 0 (6+4) can reach a 3-card 7 (Dragon 7) by drawing rank 7 or a 3-card 6 (Super 6) by drawing rank 6 after Player's 5.
Once [B0T5] resolves (19B/35P/6T across 60 branches), the winning side logs to the Big Road on shoe BBBBPPBPPP.
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About 64 minutes of lessons covering outcomes, the Bead Plate, Big Road columns, and dragon tails.
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