Board & Card Games Stack Exchange is a question and answer site for people who like playing board games, designing board games or modifying the rules of existing board games. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've heard that we can play X-wing with 4 players. Does anyone know if this requires an expansion, or is it perhaps a house rule?

share|improve this question
up vote 8 down vote accepted

The game is structured such that there are shared turn phases. Everyone's moves happen in a single phase and the order is determined by model statistics. Then everyone's shooting happens in a single phase and the order is determined by model statistics. You can basically add as many people to that system as you like and they just take their actions in order based on their model's stats.

To play in two teams of two all you need is to own enough models so that each team member has enough to control that the game is interesting.

To play in a free-for-all with more than two players the only house rule required would be determining the tie-breaker for who goes first when two models have tied statistics. (In the two player game it is always Imperials have the initiative if all other checks are tied, so if there are "imperial" players who are enemies you need a system to break the tie.)

share|improve this answer
To illustrate this, TableTop just had a 4 player episode of X-Wing using 2 teams of 2. You can watch it at – Tapan Zee Aug 27 '13 at 20:32
My experience with trying 2v2 has been it generally plays out as 2 1v1s with an occasional potshot taken at the other enemy when you mess up a maneuver and happen to end up with them in your sights. But we are not particularly competitive players :) – Affe Aug 27 '13 at 21:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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