7 card trade myth

It was long ago i already thought theres something juicy about it till yesterday when it happened again in front of my eyes i decided to sit down and write all the options.

So, is it 7 cards 1 set trade in really just a balloni. YES IT IS.

You can ALWAYS trade in 2 sets with 7 cards, always, just look better and dont trade 1st set WRONG.

For example if u have 2 pairs 2 green and 2 blue cards and u get from a kill 3 diferent once.Now , if u trade 3 diferent once u end up trading only 1set BUT now if u trade 3 green ad 3 blue then u trade 2 sets and problem solved.

PS: if u think this is not true write down the option.
PSS: i went over all the options , always possible trading twice
PSSS: i should keep this to myself lol

5 color A and 2 color B, trade once and get a 4 cards no set!

now i feel dummie well at least i figured theres a lot of time mistradig since u can actualy make a mistake with trading ending up with only 1set when u actually can have 2

I've had a 7 card 1 set TP. I had 5 red and 2 green. And a bad mood

ya PsymonStark already enlightened me

dummie dummie dummmie i got love in my tummie

emjay I remember that game

7 cards is approximately 90% chance of 2 sets, which is considered safe

still sucks when you get in that 10% range

I also did this miscalculation long time ago and I also felt dummy to write it in the forum!

But once you know that 5+2 it's the only case you cannot turn in twice then you'll also know that the wild is not the only way to have a garanted double turn-in. If you simply hold one card of each color than you'll turn in twice with 7 cards, it doesn't matter which the other cards will be.

Damn I think I've had 9 in a row where I've been able to turn in twice. Guess im due for only one set next time

Bishop, it's impossible if and only if you are past the first round of turn ins and your opponents are smart enough not to hold a set past 4 cards in the lategame:

1+1 - if opponent holds 1+4 you would be unable to turn in (this could be considered impossible as the opponent would turn in)
2+0 - if opponent holds 0+5 you would be unable to turn in (this can be considered impossible as the opponent would turn in)
if opponent holds 3+2 you would be unable to turn in

So, if you hold 1+1 your chances of 7 cards 1 set can be blamed on opposing stupidity (or your own assumption of your opponent's non-stupidity, or the application of this rule on the first round of turn ins)

If you hold 2+0, it is more risky... I'll come up with time to do the math on that sometime.

Well nice observations, but... You only go for 5 cards Ultras?
Never gone for 4 cards? Or even 3, especially when you are stuck with 2+2. I mean there are other cases, for example if you hold 2+1 and you kill one holding 3+1 or you hold 3 of a kind and him 2+2. Plus the same cases but reversed with you holding 4 cards and your victim 3.

But of course one with 5 cards normally cannot have 4 of the same color otherwise he would have turned in at the previous round. Unless he's a suicider... O_o Since 2 cards of different colors plus 5 unknown cards normally* gives a garanted double turn-in. (* if you are sure there is no reason for your opponent to not turn in if he could) Again, nice observation.

I'm used to resume concepts in this way:
-if you hold 3 colors: you are safe, you'll turn in twice
-if you hold 2 colors: the risk is minimal but it can happen
-if you hold only 1 color: the risk is concrete, be careful!

I also would like to do the math giving precis odds. But it depends on the wild cards percentage that is not the same in every map, and also it depends on the map size, because the smaller is the card deck the smaller is the possibility of a 5+2. Since it's complicate!