Endgame Puzzle – Can Player A stop Player B from winning ?
Player A rack FOXRIOT score 382 to play
Player B rack INST score 397.
It is Player A's turn. Can you see how Player B can win ? Can you stop it ?
Thoughts and analysis below.
On the first inspection, a 15 point gap holding an X should be a straightforward win for A. However B can play out on their next move in 3 places :
- Row 3 NITS or TINS making SAG (12 points)
- Row 5 NITS or TINS making AGS (10 points)
- Column L making LINTS (6 points).
Player A needs to score 15 points remove the deficit, another 6-12 points to outscore B’s play AND score enough to outscore the +/- of the remaining letters.
For example playing EX/EX in column 15 scores 34 points but leaves FORIOT (10 points) and doesn’t block B’s best play.
Thus the moves would go :
A : EX/EX 382 + 34 = 416
B : TINS/SAG 397 + 12 = 409 and out.
A : 416 – FORIOT (9) = 407.
B : 409 + FORIOT (9) = 418.
B wins by 11.
A similar result occurs if player A plays FAX/HOAX row 9.
Player A can’t block the AG (B’s highest scoring places) without enhancing B’s scoring.
Example OXO/AGO scores 16, but B can play TINS/SAGO.
Similarly RIFT/TAG by A, leaves B with TINS/TAGS.
But there is one move by A which will guarantee a minimum of a draw, and an outside chance of a win!!!!!
That is to play FIXT/TAG scoring 32. (A’s Total score is now 414 with ROO to be subtracted if B plays out, i.e 411 final score)
B’s best play is to play TINS/TAGS scoring 11 giving a total of 408 which with ROO to be added gives 411. A draw.
But what if B got greedy ?
The F of FIXT is two places below a triple word score, and SIFT scores a very reasonable 24.
It doesn’t finish the game so B would have a total of 421 (with an N left, so a final total of 420 assuming A plays out).
The score is now 414 for A, and 421 for B. A’s ROO will score well in column H making ROOT/ZO for 23 winning the game easily.
So in summary, if both players played to the best of their ability, the final score is a 411 draw !!!