Marilyn vos Savant Conditional Probability Follow Up

In the Marilyn vos Savant problem I posted yesterday, I focused on the subtle shift from simple to conditional probability the writer of the question appeared to miss.  Two of my students took a different approach.

The majority of my students, typical of AP Statistics students’ tendencies very early in the course, tried to use a “wall of words” to explain away the discrepancy rather than providing quantitative evidence.  But two fully embraced the probabilities and developed the following probability tree to incorporate all of the given probabilities.  Each branch shows the probability of a short or long straw given the present state of the system.  Notice that it includes both of the apparently confounding 1/3 and 1/2 probabilities.

Marilyn_conditional2

The uncontested probability of the first person is 1/4.

The probability of the second person is then (3/4)(1/3) = 1/4, exactly as expected.  The probabilities of the 3rd and 4th people can be similarly computed to arrive at the same 1/4 final result.

My students argued essentially that the writer was correct in saying the probability of the second person having the short straw was 1/3 in the instant after it was revealed that the first person didn’t have the straw, but that they had forgotten to incorporate the probability of arriving in that state.  When you use all of the information, the probability of each person receiving the short straw remains at 1/4, just as expected.

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