# Puzzles

## Bending a straw

Two points along a drinking straw are picked at random. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?

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A triangle will be made if none of the segments of straw is longer than the other two added together. This is the same as requiring that each segment must be less than half the straw.

Let the length of the straw be 1 unit. Call the points \(x\) and \(y\). A triangle is made if either:

- \(x\lt y\), \(x\lt\tfrac12\), \(y-x\lt\tfrac12\), \(1-y\lt\tfrac12\); or
- \(y\lt x\), \(y\lt\tfrac12\), \(x-y\lt\tfrac12\), \(1-x\lt\tfrac12\).

For the second condition, the allowable region is shown below.

This region covers \(\tfrac18\) of the whole square. By switching \(x\) and \(y\) it can be seen that the first condition's region is the same size as the second's, plus they don't overlap. Therefore the probability of making a triangle is \(\tfrac18+\tfrac18=\tfrac14\).

#### Extension

One point along a drinking straw is picked, then a coin is flipped. If the coin shows heads, a second point above the first is chosen; If tails, a second point below the first is chosen. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?

## The sixth cent

You toss 6 fair coins, and I toss 5 fair coins. What is the probability that you get more heads than I do?

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As You have one more coin than me, if you don't throw more heads than me, then you must throw more tails than me.

This means that the probability of you throwing more heads then me plus the probability of you throwing more tails then me is equal to one.

By symmetry, the probabilities for heads and tails must be equal and so the probability that you throw more heads is \(\frac{1}{2}\)

#### Extension

You toss \(n\) fair coins, and I toss \(m\) fair coins. What is the probability that you get more heads than I do?

## Marbles

A bag contains \(m\) blue and \(n\) yellow marbles. One marble is selected at random from the bag and its colour is noted. It is then returned to the bag along with \(k\) other marbles of the same colour. A second marble is now selected at random from the bag. What is the probability that the second marble is blue?

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Let \(M_1\) be colour of the first marble and \(M_2\) be the colour of the second marble

$$\mathbb{P}(M_2=\mathrm{blue})=\mathbb{P}(M_2=\mathrm{blue}|M_1=\mathrm{blue})\mathbb{P}(M_1=\mathrm{blue})\\+\mathbb{P}(M_2=\mathrm{blue}|M_1=\mathrm{yellow})\mathbb{P}(M_1=\mathrm{yellow})\\
=\left(\frac{m+k}{m+n+k}\right)\left(\frac{m}{m+n}\right)+\left(\frac{m}{m+n+k}\right)\left(\frac{n}{m+n}\right)\\
=\frac{m(m+k)+mn}{(m+n)(m+n+k)}\\
=\frac{m(m+n+k)}{(m+n)(m+n+k)}\\
=\frac{m}{m+n}
$$

#### Extension

A bag contains \(m\) blue and \(n\) yellow marbles. One marble is selected at random from the bag and its colour is noted. It is then returned to the bag along with \(k\) other marbles of the same colour and \(l\) marbles of the other colour. A second marble is now selected at random from the bag. What is the probability that the second marble is blue?

## 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?

## The blue-eyed sisters

If you happen to meet two of the Jones sister (two sisters chosen at random from all the Jones sisters), it is exactly an even-money bet that both will be blue-eyed. What is your best guess of the total number of Jones sisters?

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If there are \(n\) sisters and \(k\) of these sisters have blue eyes, then the probability of the two sisters having blue eyes is:

$$\frac{\left(\begin{array}{c}k\\2\end{array}\right)}{\left(\begin{array}{c}n\\2\end{array}\right)}
\\
=\frac{k(k-1)}{n(n-1)}=\frac{1}{2}
$$

This means that:

$$2k(k-1)=n(n-1)$$

The smallest integer solution of this is when there are 4 sisters, 3 of whom have blue eyes.

The next smallest integer solution is when there are 21 sisters, 15 of whom have blue eyes. There are unlikely to be as many as 21 sisters, so 4 sisters are the most likely.

#### Extension

What is the next integer solution after 21 sisters?

## Equal opportunity

Can two (six-sided) dice be weighted so that the probability of each of the numbers 2, 3, ..., 12 is the same?

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Let \(p_1\), \(p_2\), ..., \(p_6\) be the probabilities of getting 1 to 6 on one die and \(q_1\), ..., \(q_6\) on the other. The probability of getting a total of 2 is \(p_1q_1\) and the probabilty of getting a total of 12 is \(p_6q_6\). Therefore \(p_1q_1=p_6q_6\).

If \(p_1\geq p_6\) then \(q_1\leq q_6\) (and vice-versa) as otherwise the above equality could not hole. Therefore:

$$(p_1-p_6)(q_1-q_6)\leq 0$$
$$p_1q_1-p_6q_1-p_1q_6+p_6q_6\leq 0$$
$$p_1q_1+q_6p_6\leq p_1q_6+p_6q_1$$

The probability of rolling a total of 7 is \(p_1q_6+p_2q_5+...+p_6q_1\). This is larger than \(p_1q_6+p_6q_1\), which is larger than (or equal to) \(p_1q_1+q_6p_6\), which is larger than \(p_1q_1\).

Therefore the probability of rolling a 7 is larger than the probability of rolling a two, so it is not possible.

#### Extension

Can two \(n\)-sided dice be weighted so that the probability of each of the numbers 2, 3, ..., 2\(n\) is the same?

Can a \(n\)-sided die and a \(m\)-sided die be weighted so that the probability of each of the numbers 2, 3, ..., \(n+m\) is the same?

## Downing Street

A knot of spectators in Downing Street was watching members of the Cabinet as they arrived for a critical meeting.

"Who's that?" I asked my neighbour, as a silk-hatted figure, carrying rolled umbrella, rang the bell at No. 10. "Is it the Minister of Maths?"

"Yes," he said.

"Quite right," said a second spectator. "The Minister of Maths it is. Looks grim, doesn't he?"

The first of the speakers tells the truth three times out of four. The second tells the truth four times out of five.

What is the probability that the gentleman in question was in fact the Minister of Maths?

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The probabilities can be summarised as follows:

| First person truthful | First person lying |

Second person truthful | \(\frac{3}{4}\times\frac{4}{5}=\frac{12}{20}\) | \(\frac{1}{4}\times\frac{4}{5}=\frac{4}{20}\) |

Second person lying | \(\frac{3}{4}\times\frac{1}{5}=\frac{3}{20}\) | \(\frac{1}{4}\times\frac{1}{5}=\frac{1}{20}\) |

As they both agree, only both lying and both truthful are possible. Hence the chance of them lying is ^{1}/_{13} and the chance of them telling the truth, and it indeed being the Minister of Maths, is ^{12}/_{13}

#### Extension

If the first person said it was the Minister of English and the second said it was the Minister of Maths, what is the probability that it was the Minister of Maths?