Probability in a knock-out tournament Maths newbie so please go gently. Imagine if you would:
4 teams in the semi-final of a soccer tournament A,B,C,D.
A,B and C all have 20% chance of winning the tournament. D however, is the favourite with a 40% chance of winning. 
The lunch time game is between A and B where A wins.
So...the question is how do I calculate the new probabilities of B,C or D now winning before the evening semi between C and D
Second question if I can? what the probabilities after D wins the evenings semi match?
Thanks in advance.    
 A: With a nod to @TonyK that this real world problem is impossible, I will treat it as a pure mathematical question.
Before the game was played:
$$P(A)=P(B)=P(C)=0.2$$
and
$$P(D)=0.4$$
It was a given that in order to win, Team D must win its semi and would play either A or B in the final. Assuming that its chances against either are the same, what has changed by now knowing that Team A won the semi?
Nothing
Team D still has $P(D)=0.4$.
By similar reasoning Team C still has $P(C)=0.2$.
Which means $P(A)=0.4$
Shame on those who downvoted @Raja!
A: Theoretically, this kind of question is impossible to answer, because there may be unknown factors. Suppose Team D has a weakness against Team A, so that it is more likely to lose against Team A than Team B, even though Team A will only beat Team B half of the time. This skews the probabilities, making an exact calculation impossible.
But let's make the approximation that Team A and Team B are equally difficult to beat (which is reasonable on the basis of their pre-tournament chances). Then obviously it doesn't make any difference to Team D's chances whether A or B advances to the final, so its probability of winning is still 40%.
A: Before the evening semi final between C and D, Probability of A winning is 40% (sum of probabilities of A and B), C winning is 20% (unchanged) and D winning is 40% (unchanged).
Probability of D winning the evening semifinal should be 66.66%. ( It has double the chances as compared with the opponent C)

Let us understand this situation with an example. let us assume 60 parallel worlds. (60 because that is convenient to me.) These worlds are similar but the outcomes of an experiment can be different.    Now, according to question, out of 60 worlds, A, B and C win in 12 worlds each, and D wins in the tournament in 24 worlds. Also, A wins the semi-final against B in 30 worlds and B wins a semi-final against A in rest of the 30. Similarly C wins a semi-final against D in 20 worlds, whereas D wins Semi-final in 40 worlds. Now, C plays final in 20 worlds, so it plays 10 against A and 10 against B.
  Now, C has to win in 12 worlds, so it wins 6 against A and 6 against B. This means A wins only 4 matches out of the 10 it plays with C. 'A' also has to complete its 12 victories. A plays with D in the Final in 20 worlds, out of which, A wins 8, D wins 12. D wins 12 again with B, hence completing its 24 matches.  Now, in the question, B loses the semi-final against A. So we can ignore all the 30 worlds in which B wins against A. That leaves ust with just 30 worlds, in which A defeats C in 4 worlds, C defeats A in 6, C wins in 6 worlds out of 30, A defeats D in 8 worlds, and hence gains a grand sum of 12, and finally D wins against A in 12 worlds. So, A wins 40%, C 20% and D 40%.
A: Updated
$\newcommand{\P}[1]{\operatorname{\mathsf P}\left[#1\right]}$
We want to know the probabilities that A, C, or D will win, given that we now know that B will not, and that A was matched with B in the semi-finals.
We shall use Bayes' Theorem to calculate the conditional probabilities.
Let $A_B$ mean A is victorious in a game with B, and $M_{ABCD}$ mean A is matched with B and C is matched with D in the semi-finals.
If we assume that the three ways to match the teams have equal weighting, then $\P{M_{ABCD}}=\P{M_{ACBD}}=\P{M_{ADBC}}=\frac 1 3$
Further since the probabilities of A, B, and C ultimately winning are equal, then by symmetry:
$\P{A_B}=\P{A_C}=\P{B_C}=\P{C_B}=\P{C_A}=\P{B_A}=\frac 12$
$p=\P{D_A}=\P{D_B}=\P{D_C}=1-\P{A_D}=1-\P{B_D}=1-\P{C_D}$
$\therefore \P{D} = p^2= 0.4$
We wish to find the probabilities that:
$$\begin{align}
\text{Team $D$ wins the final } & \text{ on the condition that $A$ faced and won against $B$ in the semi final round.}
\\
\P{D\mid M_{ABCD}, A_B} &= \frac{\P{M_{ABCD}, A_B, D_C, D_A}}{\P{M_{ABCD}, A_B}}
\\ & = \frac{\P{M_{ABCD}}\P{A_B}\P{D_C}\P{D_A}}{\P{M_{ABCD}}\P{A_B}}
\\ & = \P{D_C}\P{D_A}
\\ & = 0.4
\\[3ex]
\text{Team $C$ wins the final } & \text{ on the condition that $A$ faced and won against $B$ in the semi final round.}
\\
\P{C\mid M_{ABCD}, A_B} & = \P{C_D}\P{C_A}
\\ & = \frac{1-\sqrt{0.4}}2
\\ & = 0.183772...
\\[3ex]
\text{Team $A$ wins the final } & \text{ on the condition that $A$ faced and won against $B$ in the semi final round.}
\\
\P{A\mid M_{ABCD}, A_B} & = \P{A_C}\P{C_D}+\P{A_D}\P{D_C}
\\ & = \frac 12 (1-\sqrt{0.4})+(1-\sqrt{0.4})\sqrt{0.4}
\\ & = \frac {0.2+\sqrt{0.4}}2
\\ & \approx 0.416228...
\end{align}$$

Remarks
Two separate influences acted to change the chances of victory.  One condition was which teams were matched against D in the semi-final.  The other condition was how many further challenges awaited the team.
Team D's chance of victory remained at $0.4$ because they still faced two challenges of equal difficulty to those they had before the matching was decided.  The identity of their opposing teams did not affect their chances, so the outcome of Teams' A and B match did not affect their chance of victory in either of their challenges.
Team C's chance of victory was decreased because had become certain that they must face team D.  Their chances were decreased once the match up with D was decided; it was not influenced by the team A's victory.
Team A's chance of victory was increased because they had only one more challenge to overcome. (It had also become less certain that they would face team D, the chance of that depending on the outcome of teams C and D's match).

Addendum
I should note that this symmetry argument only works if we assume A, B, and C were equally likely to be matched against D in the semi-finals.  If we assume instead that the match between A and B, C and D was predetermined before the initial probabilities were assessed, then we will have six unknowns in four linear equations.
A: Team D is still twice as likely to win as A and C, even though B is now out of the running. A and C have equal odds still. Letting $P_{X}$ denoting the probably that team X wins, then we can set up the equation $P_{D}+P_{A}+P_{C}=1$. But again we already know $P_{D}=2P_{A}=2P_{C}$, so we can replace $P_{A}$ and $P_{C}$ in the first equation with $\frac{1}{2}P_{D}$. Now we have $P_{D}+\frac{1}{2}P_{D}+\frac{1}{2}P_{D}=1$ and solving for $P_{D}$ yields $P_{D}=\frac{1}{2}$, or 50%. Now that we know $P_{D}$ we can solve for $P_{C}$ and $P_{A}$, and we'll find that $P_{A}$=$25$% = $P_{C}$. You can repeat this process to figure out the remaining part of your question.
