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.
 A: You are clearly assuming that all games have the same chance to draw, $1/3$.  We will do it more generally.  Let the chance of a game being drawn be $p$.  We do have to assume the chance of a draw is the same for every game and independent of the results of any other game.  Let us consider one group.  The chance of no draws in the first two games is $(1-p)^2$, in which case no team in the group will start with two draws.  The chance of one draw is $2p(1-p)$ and to avoid a team starting with two draws we need two games not to draw, so the chance is $2p(1-p)^3$.  The chance of two draws is $p^2$ and again we need two games not to draw to avoid a team starting with two, for a total of $p^2(1-p)^2$.  The chance that a group does not yield a team starting with two draws is then $(1-p)^2+2p(1-p)^2+p^2(1-p)^2=(1-p)^2(1+2p+p^2)=(1-p^2)^2$  We have eight groups, and need all of them not to have a team start with two draws, so the final answer is $(1-p^2)^{16}$.  For $p=\frac 13$, this is about $0.1519$  
If we ignored the correlations between the teams, each team would avoid two draws with probability $1-p^2$ and $32$ teams would avoid two draws $(1-p^2)^{32}$ of the time.
A: 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 $\left(\dfrac{56}{81}\right)^8 \approx 0.05219$. 
