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Subject: Solo variant rss

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Igor Larchenko
Russia
Ivanteevka
Moscow region
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There are 3 teams - blue, red and grey - and no any assassin.
Teams have no players, but every team has 8 agent cards.

Every your turn consists of 3 phases:
1. Take from the top of the deck 6 cards, and add them in any order to teams with least amount of agents (you can choose if tie).
2. Select and remove from the game 1 card of every team (as if they was selected by opponent).
3. Think 3 good clues for exactly 2 agents from every team. Clue must avoid any other (this team or enemy) agents. Remove 2 selected agents from the game.

You lost game, if you can't think any clue.

You won game, if at any moment of the game 1 team hasn't agents.

Good game (when you successfully find 3 clues) lasts 8 turns. At the final turn every team has 1 agent, and, after you added 2 and removed 1 agent, you select 3 good clue for those 2 remain agents in every team.

You can try it in Tabletop Simulator mode, that I made (change table for standard one):

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Mia Neu
Germany
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This is an interessting variant. I like to play it with picture cards. But I really tried to play it, but cant, because I have some questions, maybe because of language problems:

In the starting round: Do I have to add all 6 cards to one team? Or could I add 1 card to team A and 3 cards to team B and the remaining 2 cards to team C? Could the number of agents at the end of this turn be unequal?

Do I need 3 clues for two agents in one team? Or do you mean, I need in total 3 clues, one clue (for two agents) for each team?
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Igor Larchenko
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badaee wrote:
Do I have to add all 6 cards to one team? Or could I add 1 card to team A and 3 cards to team B and the remaining 2 cards to team C? Could the number of agents at the end of this turn be unequal?

Take from the top of the deck 6 cards, and add them in any order to teams with least amount of agents (you can choose if tie). You choose 1 card from 6, and add it to team with least amount of agents. Then take 2nd card from remained 5, and add it to team with least amount of agents etc. The number of agents at the end of every turn will be almost equal.

badaee wrote:
In the starting round: Do I have to add all 6 cards to one team? Or could I add 1 card to team A and 3 cards to team B and the remaining 2 cards to team C? Could the number of agents at the end of this turn be unequal?

Since all teams have the same number of agents, all of them will receive 2 cards.

badaee wrote:
Do I need 3 clues for two agents in one team? Or do you mean, I need in total 3 clues, one clue (for two agents) for each team?

The second.
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Mia Neu
Germany
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Great! I was looking for a solo variant with picture cards for some month. Your game works very well! And it's challenging.

Update: I played the game several times and it ist often easy to win. I think it is easier to play the game in german, than in english/russian or in many different languages. Because there are more compound words and the possiblitiy to form a new compound word. (Therefore the german rulebook of codenames is a little bit longer in the explanation of dealing with this words.) But I still like this variant
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Joel Dettweiler
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Thank you for posting this variant. After playing it a little I came up with some modifications that make it more fun for me:

Start with two teams of 10 words each.
Every round:
1. Turn over the timer. You are trying to come up with clues for a pair of words from each team before the time runs out. Codenames rules apply. It's okay to only come up with one clue, or none. Remove those words from the board.
2. Draw two new words to add to each team.
3. Use a die (I suggest a d12) to randomly remove one word from each team. Or if you don't have a die handy, keep the teams in a line and remove the first word in line (farthest from newly added words) each time.

You win if either team has no words after step 1. You lose if either team has 12 words after step 3.

If the game is too easy, add a third team. If the game is too hard, lower how many words you start with or get a longer timer .
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