Schools need to teach more probability and formal logic.
Relevant analogy :
If there is a country where every husband and wife decides to have children until they have a boy, what is the distribution of boys vs. girls in the country?
This is a great analogy. Let's see if a written approach can help:
Separating the trial-end condition from the trials themselves, we know that on average, the distribution should be equal. However, if it's not inherently obvious that the trial-end condition doesn't affect the probability, consider this:
Half of the pairs will have a boy on the first try.
The other half will have a girl on the first try.
Now, set aside the first try and the male births. Start over, ignoring the first birth, with all of the couples who had girls.
Half of them will have a boy on this try.
The other half will have a girl on this try.
Wash, rinse, repeat.
50% of the country has one male child.
25% of the country has one male and one female child.
12.5% ... 2 females, one male
6.25% ... 3 females, one male
3.125% ... 4 females, one male
etc etc
If there were 100 couples, it should be obvious from the definition that there would be 100 boys, since each couple has a boy. But how many girls would there be? On average, 100.
Mathematically:
The population of females is the sum of the sequence (n-1)/(2^n) from n=1 to infinity
The population of males is the sum of the sequence (1)/(2^n) from n=1 to infinity
I can go into more detail, but both of these sequences converge, and both of them are equal to 1. (Thus each parent will have, on average, 2 children.)
What most likely throws people off is the different "speeds" at which these functions change. Plug them into Wolfram Alpha and compare the graphs.
