Conjunction fallacy
This is the most often-cited example of a fallacy in reasoning about probabilities originating with Amos Tversky and Daniel Kahneman called the conjunction fallacy.
In studies, 90% of people choose ‘Linda is a bank teller and is active in the feminist movement’ rather than ‘Linda is a bank teller’. But if you turn on your System 2 analytic thinking, it’s clear that the probability of two events occurring together (in “conjunction”) is always less than (or equal to) the probability of either one occurring alone.
For example, even choosing a very low probability of Linda being a bank teller, say Prob(Linda is a bank teller) = 0.05 and a high probability that she would be a feminist, say Prob(Linda is a feminist) = 0.95, then Prob(Linda is a bank teller and Linda is a feminist) = 0.05 × 0.95 or 0.0475, lower than Prob(Linda is a bank teller).
The two child problem
One ‘intuitive’ way of reasoning is as follows: A man has a child, and it’s a girl. The probability of that happening was 1 in 2 (50%). He then has another child. The probability that the second child is a girl is also 1 in 2 or 50%. It’s like saying: if I toss a coin and it comes down heads, what are the odds that it will come down heads next time?
But this problem is quite different – and this here is why: If a man has two children, he could have had:
first a boy, then a girl – BG
first a boy, then another boy – BB
first a girl, then a boy – GB
first a girl, then another girl – GG
These are the only possible ways in which a man can have had 2 children, and they are all equally probable. The probability of each is therefore 1 in 4, or 25%. In the puzzle, we know that at least one of the children is a girl. That cuts out just one of the 4 possibilities – BB – leaving 3 possibilities: GB, BG, GG (each equally likely). Of those 3, only one – GG – satisfies the condition that the other child is a girl as well. So the answer is 1 in 3, or 33.3%
Note that if a man has two children, he can have two boys, two girls or one of each, i.e., three possibilities, each equally likely. But they’re not equally likely, because ‘one of each’ is not one possibility, but two. If you think of all the people you know with 2 children, you’ll most likely find that about a quarter of them have two boys, a quarter have two girls, and half of them have one of each.
Mindware strategy tip
One way of helping solve this problem is to represent it in a way that highlights the ‘space’ of possible outcomes:
A man has two children. At least one of them is a girl. What is the probability that both children are girls?
Formulating it like this means the solution is more easily accessible, a bit like representing the Prisoner’s Dilemma problem in a ‘tree’ that helps find the solution.
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