I know this has been a long time coming but we finally have the answer for the April 2014 puzzle.

Though Buster's logic is incorrect, he is right that they should switch and rob the other train. The important piece of evidence that they’re ignoring is the fact that they have useful prior information from three previous robbery attempts. The average value they have seen in the past is 11.1 million in gold with a standard deviation of about 3. We can use this information to make the right decision.

If they’ve picked the right train on their first guess, the amount planned for each train was 5 million and so they are seeing 10 million on this train. If they have not picked the right train, it means that the original value was 10 million and so the other train now has 20 million. So the original value is either 5 or 10.

Which is more likely? Well, both 5 and 10 are less than the past average of 11.2 but 10 is certainly closer. The least amount they had seen before is 7 so 5 is about 2 standard deviations below the mean. So, it seems that the original amount being 5 would be the less likely scenario. It is more likely that it was 10 and that the other train now has 20.

What would true professionals do? They would calculate the cut-off value in advance so that they can make an immediate decision after counting the gold on the first train. What is the cut-off value? You might think that it is 11.1 but that actually isn't correct. In fact, it requires a few more assumptions about the form of the prior distribution and the use of Bayesian statistics. If we assume it is a normal distribution, you can show that the cut-off value is 4/3 times the mean of 11.1 or 14.8 and that switching will be the right choice about 80% of the time. The full solution is posted here. https://github.com/dave31415/sonny/blob/master/sol.pdf

Congrats to our best train robbers for coming to the right conclusion. We will send you a book by a ThoughtWorker of your choice.

Sign up to receive the latest edition of P2 Magazine.