World Cup probability question. What is the probability of none of the 32 teams of the World Cup bringing two consecutive draws in the first two games. Accept the fact that a win, draw or loss have the same probability to appear.

Assume each game has a probability $p$ of ending in a draw, and that the results of each game are independent of each other. Lets analyze one group with teams W,X,Y,Z. 
WLOG, the schedule for the first two games for each team is: 


*

*W vs X

*Y vs Z

*W vs Y

*X vs Z


with the last two group games being W vs Z and X vs Y. 
The ways for all four teams to avoid starting with two draws are: 
All four games do not end in a draw - probability $(1-p)^4$
Only one game ends in a draw - probability $4p(1-p)^3$
Game 1&2 end in a draw while 3&4 do not, or vice-versa - probability $2p^2(1-p^2)$. 
Thus, the probability that no one in this group starts with two draws is: $(1-p)^4 + 4p(1-p)^3 + 2p^2(1-p)^2 = (1-p)^2(1+2p-p^2)$
Therefore, the probability that no one across all $8$ groups starts with two draws is simply: $[(1-p)^2(1+2p-p^2)]^8$
If we assume $p = \dfrac{1}{3}$, then the desired probability is $\left(\dfrac{56}{81}\right)^8 \approx 0.05219$. 
