Assuming same drawing order, the probability that the two draws are identical is 0.047%, or 1/2100.
Assuming unknown (different) drawing order, the probability that the two draws are identical is 0.018%, or 1/5463.

PLEASE REPLY IF YOU FIND ANY MISTAKES, AND I WILL CORRECT THEM ASAP

Hey everyone,

if you were not aware, apparently it has surfaced that the actual Champions League draw for the knockout phase that happened this afternoon came out identical to the practice draw that happened last night (thanks /u/OnAGoat!). Pictures of the practice draw had circulated since last evening in online forums, causing much confusion as to how the draw could have been known since yesterday if it only happened today.

Some said that because of the restrictions of the draw there was a legitimate chance that both draws would come out the same, so I was intrigued, and worked out the math. Here is a step-by-step explanation of my calculations.

Disclaimer: I'm not claiming absolute authority, and only present my calculations under correction. If you find that I made mistakes anywhere, let me know and I'll correct them. Also I'm not claiming any foul play; I don't have all the info, so this is just in the spirit of an exercise. Besides, how stupid would it be to announce to the whole world you are rigging the draw in this fashion?

a. Teams were split into two separate pots after the group stage, winners and runners-up, to be drawn against each other. Teams from the same group or the same association cannot be drawn against each other.

Winner pot (qualifying group): Paris Saint-Germain (A), Schalke (B), Málaga (C), Borussia Dortmund (D), Juventus (E), Bayern (F), Barcelona (G), Manchester United (H).

Runner-up pot (qualifying group): Porto (A), Arsenal (B), Milan (C), Real Madrid (D), Shakhtar Donetsk (E), Valencia (F), Celtic (G), Galatasaray (H).

b. For the following calculation, I assume that the order of drawing was preserved between the practice draw and the actual draw. That is, Galatasaray was drawn first in both draws, followed by Celtic in both draws, and so on. If that assumption does not hold, the probability that both draws are identical is probably even smaller. More on that later.

On to the actual calculation steps, in the order the draw happened.

Galatasaray can draw any of the winner group teams, with the exception of Manchester United (both were in Group F). Therefore 1/7 chance of drawing any team in the winner pot. Chance of drawing Schalke: 1/7

Celtic can draw any of the winner pot teams, except Schalke (already drawn), and Barcelona (both in Group G). 6 teams remain in the pot, therefore chance of drawing Juventus: 1/6

Arsenal cannot draw {Schalke, Juventus} (already drawn), or Manchester United (same association). 5 teams remain, therefore chance of drawing Bayern Munich=1/5

Shakhtar cannot draw {Schalke, Juventus, or Bayern Munich} (already drawn). They shared Group E with Juventus, that has already been drawn. Therefore they can draw any of the remaining 5 teams. Chance of drawing Dortmund = 1/5

Milan cannot draw {Schalke, Juventus, Bayern Munich, Dortmund} (already drawn), or Malaga (Group C). Therefore its chance of drawing Barcelona would nominally have been 1/3. However, any other pairing than Barcelona at this stage would necessarily render one of the subsequent pairings invalid due to teams from Spain remaining in the pool, and so Milan was assigned Barcelona, with probability 1. (much obliged to /u/Herpinho and /u/marcus898 who pointed this out)

Real Madrid cannot draw {Schalke, Juventus, Bayern Munich, Dortmund, Barcelona} (already drawn), or Malaga (same association). 2 teams remain (Man U and PSG). Therefore chance of drawing Manchester United=1/2.

Valencia cannot draw {Schalke, Juventus, Bayern Munich, Dortmund, Barcelona, Manchester United} (already drawn) or Malaga (same association). Therefore only Paris Saint-Germain remains eligible, and so chance of drawing PSG=1.

The last pairing has already been determined, since only Porto and Malaga remain. Therefore chance of drawing Malaga=1

This means that the probability of this specific draw happening again in the identical way it happened last night is the product of the above probabilities, or 1/2100, which comes out to 0.047%, or roughly 5 in 10,000.

Case where drawing order not preserved: Like I said above in b, this calculation assumes the drawing order was preserved between the practice draw and the actual draw. That means they decided the order of the draw, and did both draws after. The probability of having identical draws if the drawing order is not preserved is probably vastly smaller, since we have 8!=40320 possible drawing orders (to see how, consider how many possibilites we have for selecting the first team? 8. How many for selecting the second? 7, because we've already excluded one, and so on). However, the same draw can happen in many different drawing orders, so the calculation is a little complicated.

The calculation of the exact probability is tedious, because for each of those drawing orders we'd have to do the steps outlined above. However, a rough estimate is the above probability for obtaining a specific draw times 1/40320, multiplied by the number of times the same draw can have resulted from many drawing orders. This comes out to 3.94e-9, roughly 0.0000000004, or 4 chances in a billion, or 0.000004%. I'm striking this out until I work out the actual number. If anyone has worked with rook polynomials, drop a line.

Edit for this case: I worked it out using rook polynomials myself, and the number of possible arrangements appears to be given by the generating polynomial

1 + 14x + 75x2 + 200x3 + 286x4 + 220x5 + 87x6 + 16x7 + x8 , which works out to be 5463 individual arrangements. To those of you who found out the number individually using code, you rock. I just wanted a formal verification. So if the drawing order is not maintained, and we assume a uniform picking of a draw among all possible draws, the probability is 1/5643, or 0.018%.

Whew.