This is a classic underspecified problem.
You are invited to a family party. A Boy opens the door for you. There are two children there. What is the probability that a boy opens door for you next time?
We can make many sets of assumptions about facts not mentioned, and get different responses.
As an example, what if we assume we live on an island where there are no girls. Then the probability next time it is opened by a boy is clearly 100%.
What if we assume children always alternate turns opening doors, and each child has a 50% chance of being a boy or a girl. Then the probabilty is 50%.
If we assume a random child opens the door, that children are randomly boys or girls, then the probability is 75%.
If we assume that, if there are two boys, neither opens the door and instead an adult does, and if there is a boy and a girl they take turns, then the probability is 0%.
So unspecified criteria about how the door is chosen to be opened and the distribution of children and how they interact permits the probability to be anywhere from 0% to 100%.
This is the same issue that the classic Monte Hall problem has: there are reasonable assumptions that can set the answer anywhere. People pick some set they consider reasonable and draw a conclusion with them.
Now, on a test, you'd just answer 75% and state your assumptions directly, then draw conclusions from those assumptions. But that is a test taking strategy, not a mathematical truth.
Assume each child is randomly of either gender.
Assume a random child always opens the door.
BB BG GB GG
BB 100% 0% 0% 0%
BG 25% 25% 25% 25%
GB 25% 25% 25% 25%
GG 0% 0% 0% 100%
on the left is the gender of the two children. Each of the rows has equal probability. We could simply divide every probability by 4, but a later step makes that not required as we are going to normalize it anyhow.
On the top is who opens the door 1st and second time. In the middle is the chance that this happens.
We now eliminate every case where a boy doesn't open it the first time:
BB 100% 0%
BG 25% 25%
GB 25% 25%
Now add up percentages:
75% chance that a boy opens the door the 2nd time.
If you prefer more math, we can use Bayes's theorem. Here are the initial probabilities (same chart as above, except divided by 4):
BB BG GB GG
BB 1/4 0 0 0
BG 1/16 1/16 1/16 1/16
GB 1/16 1/16 1/16 1/16
GG 0 0 0 1/4
now we want the probability of boy boy given boy first, or:
P(BB|BX) = P(BX|BB) * P(BB) / P(BX)
P(BB|BX) = 100% * (3/8) / (1/2)
Alternatively, eliminate the cases where a boy doesn't open the door first:
BB 1/4 0
BG 1/16 1/16
GB 1/16 1/16
GG 0 0
3/8 1/8 = 1/2
Giving is P(XB|BX) = (3/8)/(1/2) = 3/4