Dan Mayoh serves up a dicey problem for Volume 23 of the Critical Line. Plus, Oliver shares the solution to Volume 22 and its winner – who stood out by submitting an animated solution to the problem!

The problem this month is not of my own creation. It is based on a question that as I understand it originally appeared in the Online Math Olympiad. I quite like it though because there are a number of different ways of arriving at the solution, most of which should be familiar to those trained in the actuarial sciences, and it doesn’t require a computer.

At exactly 12:00 noon, you begin a game with 6 regular 6-sided dice. Nothing much happens for the first 60 seconds. At exactly 12:01, you instantly roll all 6 dice. Any dice with a value of 4, 5 or 6 are considered ‘locked’ and set aside, not to be rolled again. Any remaining dice will be re-rolled after waiting 1 minute. At exactly 12:02, you instantly roll the remaining dice. This process is repeated, where dice with a value of 4, 5 or 6 are locked and set aside, and remaining dice re-rolled after 1 minute, until eventually all 6 dice are locked.

What is the expected value of the time at which the final dice becomes locked? Exact answers only to the relevant fraction of a second please!

For bonus marks (but no additional book voucher), what is the expected value if instead of 4, 5 or 6, the dice require a value of only 5 or 6 to become locked?

#### Solution to Critical Line 22 – by Oliver Chambers

We had 13 submissions this month and 7 correct entries. Congratulations to Rob Deutsch, Tatiana Potemina, Mike Fowlds, Kelvin Duong, Ean Chan, Will Gardner and Andrew Parker.

This month’s winner is Ean Chan – who not only gave a correct solution but also provided a program to test different strategies.

The strategy is to put the bug into a race with the spiders, and then it is guaranteed that the spiders will catch the bug because they are faster. One way to do this is as follows – as illustrated below have one spider follow the black path $$A \rightarrow B \rightarrow D \rightarrow C \rightarrow A$$. We can keep the bug on this circuit by having the other two spiders guard the red edges ($$\overline{AD}$$ and $$\overline{BC}$$). They move between the vertices at half speed and meet the first spider as it passes either vertex.

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