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Red7» Forums » Rules

Subject: Same color winning condition - 1 pair vs 2 pairs rss

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Kamil Śmigiel
Poland
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Hello.

Stright to my question.
Rule: same card (orange).

Player 1 cards (2 pairs): 1R 1B 4R 4B

Player 2 cards (1 pair): 6R 6B

Which player wins in this situation?
Thank you in advance.
 
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Scott Russell
United States
Clarkston
Michigan
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Player 2
The max number of same cards is 2.
The largest set of 2 contains the 6's.
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Kamil Śmigiel
Poland
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That's what I thought, thank you for the confirmation!
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James R. Gracen
United States
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Rules v1.2 wrote:
When comparing two cards, compare value first, then color.

In your example, there are three pairs: Player One with 1's and 4's, and Player Two with 6's. Scott is correct, the 6's are the highest pair, so they win.

If instead, Player Two had 4Y and 4G (instead of the 6's), and Player One had the same hand (1R, 1B, 4R, 4B), then you determine the winner like this:

First you compare the values (both player's highest pair is 4's), then you compare colors along the ROYGBIV spectrum (Player One has the Red 4, which is higher ranked than the other 4's). Player One wins.
 
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Gillum the Stoor
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The key point is that the principal criterion (for orange) is the size of a set of cards with the same number.

Thus, a player with 3 1s will win over a player with a pair of each of 2s and 3s.

If more than one player has the largest set, ties are initially broken by the value in the set.

Thus, a player with 2 4s (and no other pair) will win over a player with a pair of each of 2s and 3s.

The number of sets (e.g., pairs) is irrelevant. If you have more than one set of largest size, only the one with highest value is relevant.
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