The problem, which was given in our May [1907] number, is here repeated, together with the solution, which is as follows:–

Trick 1. A, 7 hearts; Y, king hearts; B, 3 spades; Z, 4 hearts.

Trick 2. B, jack spades; Z, 10 clubs; A, 3 clubs; Y, 2 clubs.

(If Z discards a diamond at trick 2, A does the same, and B then leads a diamond at trick 3.)

Trick 3. B, 4 clubs; Z, queen clubs; A, ace clubs; Y, 8 clubs.

Trick 4. A, ace hearts; Y, 9 diamonds; B, 2 diamonds; Z, 5 hearts.

Trick 5. A, 9 hearts; Y, jack clubs; B, 6 spades; Z, 10 hearts.

Trick 6. B, 10 spades.

(If Z now discards the queen of hearts, A discards a diamond; if Z discards a diamond, A discards the jack of hearts. In the latter case Y must either unguard diamonds or throw away the winning club, so that A makes all the remaining tricks.)