Conquer Club

I can't find this rule anywhere.
Scenario: you eliminate an opponent and end up with four spoils that includes a set.
Question: During that same turn, can you play that set? or must you have at least five cards?

You must have at least 5 cards unless you somebody out and ended up with more than 5 cards. If this occurs, you may cash twice or even three times if possible.

Gabriel13 wrote:You must have at least 5 cards unless you somebody out and ended up with more than 5 cards. If this occurs, you may cash twice or even three times if possible.

Or four times, right? If you are playing a freestyle game where someone say Player 1 had 4 cards at the of their turn and took out Player 2 with 5 cards, and then you (Player 3), saying you had 4 cards at the start of your turn, took out Player 1. That would be like 13 spoils, which would be the max maybe? So that could be like 4 cashes? Or maybe I did the math wrong. I don't play much freestyle.


--Andy

AndyDufresne wrote:

Gabriel13 wrote:You must have at least 5 cards unless you somebody out and ended up with more than 5 cards. If this occurs, you may cash twice or even three times if possible.

Or four times, right? If you are playing a freestyle game where someone say Player 1 had 4 cards at the of their turn and took out Player 2 with 5 cards, and then you (Player 3), saying you had 4 cards at the start of your turn, took out Player 1. That would be like 13 spoils, which would be the max maybe? So that could be like 4 cashes? Or maybe I did the math wrong. I don't play much freestyle.


--Andy

Technically you could nuke kill three opponents with 5 cards each. If you cashed and had two this gives you 17 cards. You would be required to play five times.

Highly unlikely but mathematically possible.

osujacket wrote:

AndyDufresne wrote:

Gabriel13 wrote:You must have at least 5 cards unless you somebody out and ended up with more than 5 cards. If this occurs, you may cash twice or even three times if possible.

Or four times, right? If you are playing a freestyle game where someone say Player 1 had 4 cards at the of their turn and took out Player 2 with 5 cards, and then you (Player 3), saying you had 4 cards at the start of your turn, took out Player 1. That would be like 13 spoils, which would be the max maybe? So that could be like 4 cashes? Or maybe I did the math wrong. I don't play much freestyle.


--Andy

Technically you could nuke kill three opponents with 5 cards each. If you cashed and had two this gives you 17 cards. You would be required to play five times.

Highly unlikely but mathematically possible.

So then combining your situation with Andy's, a max of 34, resulting in 10 mandatory cashes and an additional 11th.

Of course counting out the possibility of the player being eliminated before the cards would be cashed by a previously uninvolved one.

Now I'm actually curious about how far this can go and will most likely spend all day tomorrow thinking about it.

AndyDufresne wrote:

Gabriel13 wrote:You must have at least 5 cards unless you somebody out and ended up with more than 5 cards. If this occurs, you may cash twice or even three times if possible.

Or four times, right? If you are playing a freestyle game where someone say Player 1 had 4 cards at the of their turn and took out Player 2 with 5 cards, and then you (Player 3), saying you had 4 cards at the start of your turn, took out Player 1. That would be like 13 spoils, which would be the max maybe? So that could be like 4 cashes? Or maybe I did the math wrong. I don't play much freestyle.


--Andy

If this happened in a Battle Royale, you could theoretically have a pretty large number of cashes