# Puzzles

## 5 December

Carol rolled a large handful of six-sided dice. The total of all the numbers Carol got was 521. After some calculating, Carol worked out that the probability that of her total being 521
was the same as the probability that her total being 200. How many dice did Carol roll?

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The totals that are equally likely add up to 7 times the number of dice.

This can be seen by using the fact that the opposite sides of a dice add up to 7: for each way of making a given total with \(n\) dice, there is a way of making \(7n\) minus that total
by looking at the dice from below. Therefore \(T\) and \(7n-T\) are equally likely. (This also holds true (but is harder to explain) if you rearrange the faces of the dice so that the opposite
faces no longer add to 7.)

Therefore today's number is \((521+200)/7\), which is **103**.

## Cube multiples

Six different (strictly) positive integers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of 6.

What is the smallest possible sum of the six numbers?

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Consider the top, front, and right sides of the cube.

If the top number is \(a\) more than a multiple of six, then the front and right numbers must both be \(a\) less than a multiple of 6 (so that when added to the top number they make a multiple of 6). But when the front and right numbers are added, they make \(2a\) less than a multiple of 6; but this must also be a multiple of 6.

This is only possible if \(a=0\) or \(a=3\). So the numbers must be either all multiples of 6, or all 3 more than multiples of 6.

The smallest set of numbers that are all 3 more than multiples of 6 is 3,9,15,21,27,33. The sum of these is 108. The smallest set of numbers that are all multiples of 6 is the set with each number three more than these, so **108** is the smallest possible total.

#### Extension

Six numbers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of \(n\).

What is the smallest possible sum of the six numbers?

## Fair dice

Timothy and Urban are playing a game with two six-sided dice. The dice are unusual: Rather than bearing a number, each face is painted either red or blue.

The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they're different. Their chances of winning are equal.

The first die has 5 red faces and 1 blue face. What are the colours on the second die?

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Let \(A\) and \(B\) be the outcomes of the two dice. Let \(p\) be the probability that the second die lands on red. The probability of the dice being the same is:

$$\frac{1}{2}=\mathbb{P}(A=r)\mathbb{P}(B=r)+\mathbb{P}(A=b)\mathbb{P}(B=b)\\
=\frac{5}{6}p+\frac{1}{6}(1-p)\\
=\frac{1}{6}+\frac{4}{6}p
$$

This means that:

$$\frac{4}{6}p=\frac{1}{2}-\frac{1}{6}\\
=\frac{1}{3}\\
p=\frac{\frac{1}{3}}{\frac{4}{6}}=\frac{1}{2}$$

#### Extension

If the first die has \(n\) red faces and \(6-n\) blue faces, what colours are on the second die?

## Tetrahedral die

When a tetrahedral die is rolled, it will land with a point at the top: there is no upwards face on which the value of the roll can be printed. This is usually solved by printing three numbers on each face and the number which is at the bottom of the face is the value of the roll.

Is it possible to make a tetrahedral die with one number on each face such that the value of the roll can be calculated by adding up the three visible numbers? (the values of the four rolls must be 1, 2, 3 and 4)

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Let \(a\), \(b\), \(c\) and \(d\) be the numbers on the four faces. The following simultaneous equations must hold:

$$a+b+c=1$$
$$a+b+d=2$$
$$a+c+d=3$$
$$b+c+d=4$$

These can be solved to find that the numbers on the faces must be \(-\frac{2}{3}\), \(\frac{1}{3}\), \(\frac{4}{3}\) and \(\frac{7}{3}\).

#### Extension

Is it possible to make a six-sided die with one number on each face such that the value of the roll can be calculated by adding up the five visible numbers?

Is it possible to make an \(n\)-sided die with one number on each face such that the value of the roll can be calculated by adding up the \((n-1)\) visible numbers?

Is it possible to make a die with one **integer** on each face such that the value of the roll can be calculated by adding up the visible numbers?