assuming my move order is 12345 etc., and my opponent's is abcde etc.

I take 1234(OD)5
and he takes (OD)123
will his third order come before or after my 5th?

not sure if that's clear, what i'm trying to ask is if the delays that you put before using the card count the same way as the ones after using it, or if they're ignored in the count.

A1, A2,... will be your moves, his moves will be B1, B2...

The move order will be A1A2A3A4A5B1B2B3 or A1A2A3A4B1A1A1B2B3.

if i understand right, obviously after.
for this is important play OD don't before th move you want delay, but before all move that do not require one of first move.
AFTER OD (assuming both player use this) who have more delays will take last move.

PS if one use 2 delay card and other one only, the order will:
player 1 : ABCDEF
player 2 : 12345
1AB OD(player 1) OD (player 2) OD (player 1) 2345 CDEF

sorry, my question wasn't formulated, clearly, i'll give a better example

(OD)A1A2A3A4 are my orders
B1B2B3B4B5B6(OD)B7 are his orders
which comes first? B7 or A4?
my question is whether the moment you play OD matters, for example if you're doing some "dummy" moves does it matter if you play OD before or after htem

I answered your question already warlightvet, read my post. B7 always comes before A4.

B7 happen before A4.
Who have more delay or "dummy" move take last move.
How a top player on ladder don't know this?

Sze, was my post mystaken?

Order delay seperates order into an extra phase. The new phase restarts the same way a new turn would start. So make your dummy moves after you play the order delay, dummy.

BTW, the wiki is pretty clear about this. ALL moves after the card happen after ALL moves before the card. So yes, your 2nd, etc OD moves will all happen after his 1st OD move every time.

I've actually been playing as if that were the case, i was just curious really. it doesn't really influence how i take my turns, since i just would put the od as early as possible because that seemed logical. It was purely a matter of curiosity, please don't be mean gnuff and x, it makes me feel bad.
although i do enjoy being called a top player on ladder =3

another question that i should know the answer to: what are the chances of getting 2 +3 bonuses in 2 turns?
for example if you pick a territory in ant and another in CA, and the third pick is far away (I mean the west CA and north scan territories for example)

WLVet you phrased that example horribly. Assuming neither is a first turn bonus and you cluster one of them it is very likely. The calculations are boring and complicated but i did them once, and its almost as likely as getting a 5 territory bonus in 2 turns.

i was actually looking for a percentage. :P thanks for the help though, and you're right, i do have trouble phrasing these because it's hard to put practice into words.

another noob question: what is the chance of failing a 2 turn +4 bonus (for example east africa)?
the percentage is probably different if you start in somalia or ethiopia, what is the % for each?
sorry for question spam, just curious about these rare events :) i've had these fail a few times, and although it doesn't influence play it's a matter of curiosity. It also influences the game hugely if you fail to get a 2 turn bonus, effectively delaying you by almost a full turn very early in the game
obviously for both possibilities i mean that you are committing fully to taking the 2 turn bonus and do not have a double pick, on strat 1V1 settings

4v2= leftorvers 2 60% ; leftorvers 1 40%

1° case:
you can use leftovers for take other 2 place.
for fail bonus on turn 2, you have to have both leftovers of 1 only, so 16%, 1 times every 6.
after you attack with 3 and 4.
so you have 80% for take it and 20% of fail.
so 16%*20% = 3.2 % very low possibility.

2°case
you can use only leftovers of one spot.
so you attack with 3 and 5.
3v2 80%, so 20% of fail. if you fail, you have 60% of no need to deploi again, because 2 leftovers, and 40% you need +1 next turn.
5v2 = 3 leftovers 60% , 2 leftovers 40%.
if 3 leftovers +5 income of 2nd turn = 100%.

so:

A) 5v2= 3 leftovers and 3v2 fail with 1 only leftovers. 20% chance to fail
B) 5v2= 3 leftovers and 3v2 fail with 2 leftovers. 0% chance to fail
C) 5v2= 3 leftovers and 3v2 work. 0% chance to fail
D) 5v2= 2 leftovers and 3v2 work. 20% chance to fail
E) 5v2= 2 leftovers and 3v2 fail with 2 leftovers. 20% chance to fail
F) 5v2= 2 leftovers and 3v2 fail with 1 only leftovers 36% chance to fail

up: 1st case
now: 2nd case

thanks a lot! that's exactly what i meant :)