Question 976608
Here's a chart showing all of the possible combinations and probabilities for a 5 game series. 
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Since each team is equally matched, assume that the chance of A winning a game is 0.5. 
So the probability of each row is just the product of the probabilities. 
A three game series would have a probability of,
{{{P=(1/2)*(1/2)*(1/2)=(1/2)^3=1/8}}}
Four game series,
{{{P=(1/2)^4=1/16}}}
Five game series
{{{P=(1/2)^5=1/32}}}
Since A already won one game, start in the second column and add up the probabilities of A winning. 
The sum of those probabilities is 0.34375.
Now this is out of 0.5 instead of 1.0 since half the games have been knocked out since A won the first. 
So the probability of A winning given they won the first is,
{{{P=0.34375/0.5=11/16}}}