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.
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!
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.