1

We’re playing the wizard jubilee edition and the rules seem to be unclear about this: the juggler has the effect that everyone in the game must pass a card to the left neighbor. Now the question is whether this event is triggered only when the juggler is played at the very end of the trick, so being the last card of the trick, or if the juggler can be played at any moment in the trick and still triggers its event?

Would appreciate if anybody has an answer to this. Thanks!

1

It triggers when the trick is finished, no matter which card it was. It doesn't have to be the last card played to the trick.

These google-translate-based rules are phrased ambiguously:

When a trick is completed with the Juggler, each player simultaneously gives one of his hand cards face down to his left neighbor.

However, the original German is not ambiguous. (Emphasis added)

Den Jongleur kannst du immer spielen, auch wenn du bedienen könntest. Er hat den Wert 7 1⁄2, ist also höher als eine 7 und niedriger als eine 8. Spielst du den Jongleur in einen Stich, sagst du deutlich an, welche Farbe (auch Trumpffarbe) die Karte annehmen soll. Ist dieser Stich beendet, gibt jeder Lehrling gleichzeitig eine seiner Handkarten verdeckt an seinen linken Nachbarn. Danach nimmt jeder seine neue Karte auf die Hand. Das Weitergeben der Karten entfällt, wenn der Stich mit dem Jongleur der letzte Stich einer Stichrunde war.

Which translates to

You can always play the Juggler play, even if you could follow suit. He has the Value 7 1⁄2, and so it is higher than a 7 and lower than an 8. When you play the Juggler to a trick, you state what suit (even Trump suit) the card is. When the trick is finished, each apprentice simultaneously passes one of his hand cards facedown to his neighbor on his left. Cards are not passed if the trick with the Juggler is the last trick of the round.

This video explanation agrees (T=0:58)

1
  • 1
    An example of where failing to distinguish between the instrumental and genitive case can lead to ambiguity. Jul 6 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.