Inspired by question $7a$ on the Euclid Maths competition $2023$, I have been stuck thinking about the following simplified version.
A bag contains $2$ blue marbles and $1$ red marble. You remove one at a time until the bag is empty. Each marble is chosen randomly from the remaining marbles. Given that the second marble removed is blue, what is the probability that the final marble removed is blue?
I have been presented with what appears to be two logical explanations with different answers.
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Given the second marble is blue there are only two scenarios. The first marble is blue and the final marble is red, or the first marble is red and the final marble is blue. Thus the probability that the final marble is blue is $1/2$.
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Notice that this question is equivalent to asking, what is the probability that the first marble picked is red. When the first marble is picked there are $2$ blue marbles, and $1$ red marble in the bag. Thus the probability that the first marble picked is red is $1/3$.
I am not sure whether this is a simple conditional probability question as explanation $1$ suggests, the Monty Hall problem in disguise as explanation $2$ suggests, or an ambiguously worded question that relies on interpretation of the language.
If anyone could provide some insight into why either one of the explanations is correct or incorrect it would be much appreciated.
Best Answer
There is no Monty Hall type of situation here.
The flaw in your reasoning is that for number 2., the question is not equivalent to merely asking if the first marble is red. It is instead asking if the first marble is red, conditioned on the fact that the second marble picked was blue. This conditional probability is given by
$$P(\text{first red}|\text{second blue})= \frac{P(\text{first red and second blue})}{P(\text{second blue})} = \frac{1/3}{2/3}=\frac{1}{2}\text{.}$$
Thus the two approaches do yield the same probability, once interpreted correctly.