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Subject: Selling gemstones rss

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Carsten
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I do not understand the rules regarding the selling of gemstones when there are more of them in the safe.
Quote:
If he [the player] has another colour of Gemstones, he converts three cards into money, [...]

So if I have 8 cards of red and 3 of green, I think I would sell both rows of red for three money each. This is not the way the example in the rules works.

Do I count different colours in the safe or do I count rows of gemstones in the safe to reduce the money I get?

(Edit: corrected spelling errors)
 
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Yegor Sadoshenko
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You can't sell green, so can only sell both of red. First you sell one column of red for 2 money (because there are 2 more columns - red and green). Then you sell the second red for 3 money (since there's only green). So you get 5 money.

You have to count columns, not colors in the safe.
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Carsten
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Thank you. I find the quoted rule very misleading.
 
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George Leach
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No, if you had 8 of a colour/jewelry-type (why would you for a start) and one other colour/jewelry-type you sell each set for 3 coins.
 
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Aelfric Brewer
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Jugular wrote:
No, if you had 8 of a colour/jewelry-type (why would you for a start) and one other colour/jewelry-type you sell each set for 3 coins.


No, this is wrong, and Yegor had it right.

If you have more than a set of 4, the extras count as another type. To get 8 red in your safe at one time, you'd have to have 3 or fewer there already and buy 5 or more from a window - unlikely, but not impossible.

In the example as described, you'd have a set of 4 red, another set of 4 red, and a set of 3 green. You sell the sets sequentially. You sell the first set of 4 red for $2 because there are TWO other types present (the second set of 4 red, and the set of 3 green); then you sell the remaining set of 4 red for $3 because at that time there is ONE other type still present at that point (the 3 green).
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Carsten
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Jugular wrote:
No, if you had 8 of a colour/jewelry-type (why would you for a start) and one other colour/jewelry-type you sell each set for 3 coins.

After having played this game, I see now that my question is very theoretical. It is possible to get more rows of the same color, if someone puts that many cards in their storefront, but this is very unlikely.
 
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Aelfric Brewer
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jorl wrote:
Jugular wrote:
No, if you had 8 of a colour/jewelry-type (why would you for a start) and one other colour/jewelry-type you sell each set for 3 coins.

After having played this game, I see now that my question is very theoretical. It is possible to get more rows of the same color, if someone puts that many cards in their storefront, but this is very unlikely.


only theoretical in that you had two sets of the same color in your example. but we can rearrange it to a slightly more realistic case:

suppose you have 3 red, 3 green, and 3 orange in your safe. and you purchase from someone else's window 2 red, 2 green, and 1 orange. you now have a full set of 4 red, an extra set of 1 red, a full set of 4 green, an extra set of 1 green, and a full set of 4 orange.

you sell one of the full sets, for example, the 4 red. there are FOUR other sets (1 red, 4 green, 1 green, 4 orange), but the minimum you get for any sale is $1. so you get $1 and put the other 3 reds in discards.

then you another full set, for example, the 4 green. there are THREE other sets (1 red, 1 green, 4 orange), so your sale gets reduced to $1 again, and the other 3 green go into the discards.

finally, you sell the last full set of 4 orange. there are TWO other sets (1 red and 1 green), so you get to bank $2 and put the other two orange in the discards.

you have thus paid the other player $1, made $4, returned 8 cards for the next hand, and reduced the number of colors in your safe from 3 to 2.

if the other player is really trying to get the reds and greens back into the discards because she needs them, this might be a reasonable tactic for her (i'm presuming she got stuck with some odd red or green in her safe after putting those in her window). you could counter it slightly by doing the red or green last and thus only returning 2, instead of 3, to the discards (you would return 3 of the orange in this case).
 
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