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Subject: Any interest or suggestions regarding a solitaire variant? rss

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Olav Fakkeldij
Netherlands
Beverwijk
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I have been thinking about this game as a solo game. I hope to give it a try soon and will share my thoughts.
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Friedemann Friese
Germany
Bremen
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My ideas:

You play against a robot.
You sort the cards on the street and put the new drawn cards right from the old ones.

A turn of the robot:
Step 1: If he has not having 10 "0"s yet, he will take the left most "0". If there are "0" he will take the left most card for drawing new ones. To get a "0". If there is no possiblity to get a new "0" go to step 2.
Step 2: If he has not having 7 "1"s yet, he will take the left most "1". If there is no "1", he will take the left most card for drawing new ones. If this is impossible, he will take the left most "0". If this is impossible take the left most "2". If not the "3"...
Step 3: If he has not having 5 "2"s yet, do this similar to step 2 with the "2"s
Step 4: If not having 3 "3"s yet, ....
Step 5: Take the highest points. Tiebreaker "leftmost" card. Only turn over new cards if there is not the highest card left in the game visible.


I have not tried this, but I will do. I was never before thinking of a solo-version, thanks.

friedemann
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Friedemann Friese
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After 4 plays with different values. Getting harder every time. Here is the actual rule:

You play against a robot.
You sort the cards on the street and put the new drawn cards on the right of the old ones.

A turn of the robot:
Step 1: If he has not having 7 (count the 4 starting "0" towards this) "0"s yet, he will take the left most "0". If there are no "0"s he will take the left most card for drawing new ones. To get a "0". (As often as rules allow)
Rare: If there is no possiblity to get a new "0" go to step 2.
Step 2: If he has not having 5 "1"s yet, he will take the left most "1". If there is no "1", he will take the left most card for drawing new ones. If this is impossible, he will take a red "0". If not there take the left most "0".
Rare: If this is impossible take the left most "2". If not the "3"...
Step 3: If he has not having 4 "2"s yet, do this similar to step 2 with the "2"s. If not getting a "2" just take the most valuable lower card. (Tie breaker left)
Step 4: If not having 2 "3"s yet, ....
Step 5: Take the highest points. Tiebreaker "leftmost" card. Only turn over new cards if there is not the highest card left in the game visible.


The closest game was 89 for me 82 for the robot. This should be hard enough, because I think I know the game very well and play it good.

I will try more games later. No time by now.
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Friedemann Friese
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Hi Phil,

step 5 means, he takes the higest point value, which is normally a 4 and searching for the 4 in red. After all "4"s are gone or not visible, he will of course take a "3". Prefering a red "3"

Bitteschön!!
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Justus
United States
Las Vegas
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Just found a ruleset on Rio Grande's website

http://www.riograndegames.com/games.html?id=397

cheers!
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Jeff Eberlin
United States
Portland
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So this sounds like a stupid question...

but does this mean initially the robot will continue to discard the leftmost card (let's say it's a 3) and draw 3 cards to the right side. If none of them are a 0 (and there are still 0's left) then he'll do it again until he finds a 0, or will he do it once and then go to the next rule?

Thanks!
Jeff
 
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