One of the classic problems of Mathemagistics, or Mathematical Magic, is the Bus Problem. It goes like this:

Question:

Every day, Fred gets the solitary 8 am bus to work. There is no other bus that will get him to his destination.

10 per cent of the time the bus is early and leaves before he arrives at 8 am.

10 per cent of the time the bus is late and leaves after 8.10 am.

The rest of the time the bus departs between 8 am and 8.10 am.

One morning Fred arrives at the bus stop at 8 am, sees no bus, and waits for 10 minutes without the bus arriving.

Now, what is the probability that Fred’s bus will still arrive?

Fred’s bus could yet arrive or he might have missed it. So there are two possibilities. So is it correct to assume that in the absence of further evidence the chance of each must be equal, so the probability at 8.10am that his bus will still arrive is 50 per cent?

But if that is the answer at 8.10am, was it also the correct answer at 8 am?

Or was 50 per cent the correct answer at 8am but not at 8.10am?

Or is it the wrong answer at both times, but was correct at 8.05am?

The solution is posted below.

Solution

When  Fred arrives at 8am, there is a 10 per cent chance that his bus will have already left. After Fred has waited for 10 minutes, he can eliminate the 80 per cent chance of the bus arriving in the period between 8 am and 8.10 am. So only two possibilities remain.

Either the bus has arrived ahead of schedule or it will arrive more than ten minutes late.

Both outcomes are unusual, but since the two outcomes are mutually exclusive and equally likely (10 per cent chance of each), and there are no other possibilities, we should update the probability that the bus will still arrive from 10 per cent (the likelihood, or prior probability, when Fred woke up) to 50 per cent, as there is (once the 80 per cent probability is eliminated) an equal probability (out of the remaining 20%) that the bus will still turn up and that he has missed it. So there is a 1 in 2 chance that he will still catch his bus if he has the patience to wait further, and a 1 in 2 chance that he will wait in vain. The follow-up question is how long he should wait. That’s for another day.

Puzzle Extra:

Every day, Fred gets the solitary 8 am bus to work. There is no other bus that will get him to his destination.

10 per cent of the time the bus is early and leaves before he arrives at 8 am.

30 per cent of the time the bus is late and leaves after 8.10 am.

The rest of the time the bus departs between 8 am and 8.10 am.

One morning Fred arrives at the bus stop at 8 am, sees no bus, and waits for 10 minutes without the bus arriving.

Now, what is the probability that Fred’s bus will still arrive?

Solution

When  Fred arrives at 8am, there is a 10 per cent chance that his bus will have already left. After Fred has waited for 10 minutes, he can eliminate the 60 per cent chance of the bus arriving in the period between 8 am and 8.10 am. So only two possibilities remain, that the bus has already left early or it will still arrive – more than ten minutes late.

Both outcomes are less likely than that the bus would arrive between 8 and 8.10am, but since the two outcomes are mutually exclusive (10 per cent chance and 30 per cent chance respectively), and there are no other possibilities, we should update the probability that the bus will still arrive from 30 per cent to something else, and the probability that it arrived early from 10 per cent to something else. Once the 60 per cent probability is eliminated, this probability should be distributed (using Bayesian principles) in the ratio of the prior probabilities of the remaining options (3 to 1 in favour of it arriving late), so 45 of the 60 per cent should be added to the prior probability of 30 per cent that it is still to arrive, and 15 of the 60 per cent should be added to the prior probability of 10 per cent that it arrived before 8am. So, at 8.10am there remains a 75 per cent that the bus will still arrive and a 25 per cent that it has already arrived and left.

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