# The Critical Line – Volume 20

In honour of the 2018 Snooker World Championships which are being played this month, April’s puzzle has a snooker theme.

Basically, your task is to count how many ways there are of making a break of exactly 100 points. First, some quick information on the relevant rules of scoring a break for those unfamiliar with the game (if you are familiar, you can skip over this).

At the start of a frame a snooker table has 15 Red balls, worth 1 point each, and 6 coloured balls worth between 2 and 7 points as follows:

Yellow = 2 points

Green = 3 points

Brown = 4 points

Blue = 5 points

Pink = 6 points

Black = 7 points

A ‘break’ is the total score earned by a player during a single visit to the table. The visit is over once the player misses a ball (or clears the table). Making a century break (100 points or more) is a big deal. Assuming there is at least 1 red ball still on the table, a player’s first shot at the start of a visit will always be a Red ball. If the red ball is potted, it stays in the pocket and the player then attempts to pot a coloured ball (their choice which one). If the colour is potted, it is placed back on the table and the player then attempts another red ball, then a colour, then a red and so on. This goes on until the final red is potted, at which point the player pots a coloured ball of their choice, which is then placed back on the table, leaving the six coloured balls on the table. The player must then pot those **in order** from Yellow to Black, and they will stay in the pocket once potted. If at any point the player misses a ball, the break is over. The value of the break is the value of all balls potted, and a break can be characterised by the sequence of balls potted in the specified order. The maximum possible break (ignoring the free-ball rule, which this question is ignoring) is 147, earned by potting Red then Black 15 times in a row, and then the six colours in order. This earns (1+7)*15 + (2+3+4+5+6+7) = 147 points.

Note that every break greater than zero will commence with a Red ball, except when only coloured balls remain, in which case the break will begin with the lowest value coloured ball. From this you can see that every break of more than 27 must begin with a Red ball.

The primary challenge this month is to count the number of ways of possible breaks (characterised by the sequence of balls potted) that earn **exactly** 100 points (assume our player misses the next shot after scoring exactly 100, unless there are no balls remaining).

For example, there are exactly 5 sequences of 5 point breaks:

- Blue
- Yellow, Green
- Red, Brown
- Red, Green, Red
- Red, Yellow, Yellow

Note that sequences (a), (b) and (e) are special cases. (A) can only occur when the break begins with just the Blue, Pink and Black balls remaining. (B) can only occur when the break begins with just the 6 coloured balls remaining, and (e) can only occur when the break begins with exactly 1 red and the 6 coloured balls remaining.

Also, a sequence of Red, Blue, Red, Black (which we can write as 1-5-1-7) is a different sequence to Red, Black, Red, Blue (which we can write as 1-7-1-5). Order matters!

So, how many 100-point break sequences are there?

If that is too easy for you, then here’s an extension question. What value of break (from a minimum of 1 to a maximum of 147) has the most possible sequences? If you really want to impress me, show me a graph or table of the number of sequences for all possible break values from 1 to 147.

For your chance to win a $50 book voucher, send your solutions to ActuariesMag@actuaries.asn.au. You only need to answer the primary 100-point question to be eligible for the prize.

#### Volume 19 – Solution

We had 6 correct entries this month. Congratulations to Raymond Chow, Lawrence Uy, Dan Mayoh, Tim Yip, Andrew Parker, and Arvind Preetham. This month’s winner of the coveted book voucher is Tim Yip!

**Weighing Stones**

You have a $$N$$ stones in a bag and you wish to calculate the weight of the heaviest stone. You do this by randomly pulling the stones from the bag one at a time (without replacement) and weighing it. Each time the weight of the stone is the heaviest so far you record its weight. Assuming that the weights of each stone is identically distributed, what is the expected number of times that you will record the weight of a stone?

**Solution**

Consider the sequence of stones being pulled from the bag $$s_1, s_2, \dots, s_N$$. The expected value of the total number of recorded weights is the sum of the expectation that each stone $$s_j$$ is recorded (by linearity of expectation). The probability that $$s_j$$ is recorded (i.e. is the heaviest stone so far) occurs with probability $$\frac{1}j$$. So the expected value of the total number of recorded weights is $$\sum_{j=1}^{N}\frac{1}{j}$$

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