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 !!!