<p>No. With four of the same color, you'd have a set of 3 and one extra.</p>
<p>From the <a href="https://boardgamegeek.com/thread/2459561/compilation-rules-clarifications" rel="nofollow noreferrer">compiled FAQ on BGG</a>:</p>
<blockquote>
<p>If you have more properties of one color than is needed to make a complete property set, you must form as many complete sets as possible, and group the remaining properties of that color together.</p>
</blockquote>

CC BY-SA 4.0
2020-12-02T12:48:55.323Z
2020-12-02T12:48:55.323Z
L. Scott Johnson
<p>For example: 2 groups of green color in order to avoid deal breaker, of this way the set is not completed</p>

<monopoly-deal>
Can I play 2 sets of 2 cards each one of the same color to avoid a Deal Breaker?
Geovanny Vega
# Can I play 2 sets of 2 cards each one of the same color to avoid a Deal Breaker?

(Asked by: username_0 on 2020-12-02)

For example: 2 groups of green color in order to avoid deal breaker, of this way the set is not completed

**Tags:** monopoly-deal


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## 1 answers

### Answer by username_1 on 2020-12-02. Score: 2

No. With four of the same color, you'd have a set of 3 and one extra.

From the [compiled FAQ on BGG](https://boardgamegeek.com/thread/2459561/compilation-rules-clarifications):

> If you have more properties of one color than is needed to make a complete property set, you must form as many complete sets as possible, and group the remaining properties of that color together.


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