# Walking Tour Puzzle Solutions

Solutions to the puzzles in the Discover Edinburgh's Mathematical History walking tour.

This page contains the solutions to the puzzles appearing in the Edinburgh Maths Walking tour we have created.  For details about the tour and the various stops, and to try the fun and engaging puzzles, visit the walking tour webpage

Witches' Well

There are three strategies they could follow:

1. Each could guess the same colour as her card;
2. Each could guess randomly;
3. One could guess the same colour as her card, the other can guess the opposite.

The first strategy means they will lose if they have the same colour cards, so a 50% chance.

For the second strategy, each witch has a ½ chance of guessing correctly, so their chance of guessing correctly is ¼, or 25%.

As for the third strategy, it turns out to be a winning strategy - they are released every time! These are the possible scenarios:

• 1 gets red, 2 gets red: 1 guesses red, 2 guesses black. 2 is incorrect and they are released.

• 1 gets red, 2 gets black: 1 guesses red, 2 guesses red. 1 is incorrect and they are released.

• 1 gets black, 2 gets red: 1 guesses black, 2 guesses black. 1 is incorrect and they are released.

• 1 gets black, 2 gets black: 1 guesses black, 2 guesses red. 2 is incorrect and they are released.

Why does this work?

The first player is guessing the colour she gets. In other words, she is only right if they both have the same colour card.

The second player, on the other hand, is guessing the opposite of the colour she gets. So, she is only correct if they have opposite colour cards.

This means that one of them is bound to be incorrect regardless of what colour cards they receive, and, with this strategy, both can live!

### Camera Obscura

The two triangles are similar, and so we can use the fact that we know the dimensions of the larger one to determine the size of the smaller one.  You could use Pythagoras’ Theorem to work out the missing side length of the big triangle, find the corresponding side of the small triangle (using that the sides of the large triangle are in the same ratio as those in the small one), and then find the horizontal distance, again using Pythagoras’ Theorem.  However, a more elegant solution is to realise that the horizontal ‘widths’ of the two triangles have the same ratio as the vertical sides, so the distance to the wall is 80/60m = 4/3m.

### Peter Guthrie Tait

This may seem impossible! The trick is to cross your arms before you pick the two ends of the scarf. In this way you tie a knot with your arms, and then “transfer” it to the scarf.

If you pick the two ends of the scarf as you would naturally do, without crossing your arms, your arms and the scarf would basically make a circle, and there is no way to knot a circle.

### James Clerk Maxwell Statue

Image 1 is Red, image 2 Blue, and image 3 Green.  One way to work this out is to notice that the wings are very blue (possibly with some green), and hence should be bright in the Blue image and very dark in the Red image.  The Green image is then the remaining one.

### Scott Monument

The pigpen translates to “Oh, what a tangled web we weave when first we practise to deceive!”, from Walter Scott’s poem “Marmion: A Tale of Flodden Field”.

### City Observatory

The speed of light in vacuum is c = 3 x 108 meters/second

So, light leaving Alpha Centauri would take (4.1315 x 1016)/(3 x 108 ) = 137716666.667 seconds

One year has: 365 days x 24 hours per day x 60 minutes per hour x 60 seconds per hour = 31536000 seconds

Thus, it would take 137716666.667/31536000 = 4.367 years for light leaving Alpha Centauri to reach earth.

That means an image of Alpha Centauri viewed on earth today would be approximately 4.367 years old!

### Holyrood Abbey

#### Game 1

1. You can only win for the first time on the 20-th round, so you need to have lost 19 times before (2^19 = 524288, 2^20 = 1048576).
2. The probability of losing 19 times in a row is 2^(-19) = 0.00000190734, or 0.00019073486%
3. Given that it is quite likely that you will win on the first couple of rounds, I would not pay much to play the game. It is also unlikely that any casino would offer such a game due to the infinite expected value of winnings for the player.

Given that you may be able to win an infinitely large amount of money, you may believe it is fair to pay any fee that I request. This problem is usually referred to as the St. Petersburg paradox: although you are expected to win an infinite amount of money, you most likely won’t, so should therefore not pay a very large fee to enter the game.

#### Game 2

You always make a profit of X, exactly doubling your initial bet!

So, what’s the catch? Well, the odds were in your favour in this game (2/3 to 1/3) which makes it more likely for you to win. In theory, this is a winning strategy if the odds are at least 50/50 (they never are at a casino), you have unlimited funds (good for you!) and there are no limits in bet size.

Disclaimer: we are not responsible if you decide to use this strategy and lose all your money.

### Scottish Parliament

On possible solution is to the right.  Can you find any others?  There are at least two different solutions!

### The World's End

#### Puzzle 1

You should find that you get closer to pi the more needles you drop.

#### Puzzle 2

You will (almost always) find that the circumference is much longer than the height of the glass.  Most people find this to be very surprising!

### The Oyster Club

This is the famous Monty Hall problem; there is a nice explanation of the solution on Wikipedia.

### Greyfriars Kirkyard

You can see an animation of the solution using Geogebra.

### Harry Potter

Minerva McGonagall = 49755941 4376516133

Minerva = 4 + 9 + 7 + 5 + 5 + 9 + 4 + 1 = 44

McGonagall = 4 + 3 + 7 + 6 + 5 + 1 + 6 + 1 + 3 + 3 = 39

Character number: 2 ( 44 + 39 = 83, 8 + 3 = 11, 1 + 1 = 2 )

Heart number: 5 (9 + 5 + 1 + 6 + 1 + 1 = 23, 2 + 3 = 5)

Social number: 6 ( 83 – 23 = 60 )

### Royal Mile

One way to do this is to work out how far the Lord travels between each of the servant's visits.  However, there is a much slicker way:

Since the lord travels at 0.5 miles per hour, it will take him 2 hours until he reaches the Parliament.

The servant will be constantly moving during these two hours at a constant speed of 5 miles per hour. Therefore, he must have travelled a total of 10 miles.

### Luckenbooths

Let's call the area A.

Picture 1: Depending on how you decide if a square is mostly inside the circle or not, there are between 4 and 16 squares, and the squares have area 1.  So 4 $$\leq$$ A $$\leq$$ 16 (=4*1 and =16*1).

The other two pictures use a similar method:

Picture 2: 4  $$\leq$$ A $$\leq$$ 9 (=16*0.25 and =36*0.25)

Picture 3: 5.92  $$\leq$$ A $$\leq$$ 8.32 (=148*0.04 and =208*0.04)  Using the formula A=pi*r2 with a radius of 1.5 gives A = 7.06858347058...

So the estimated areas are getting closer to the true area.

### Surgeons' Hall

First must convert the height to metres:

1 ft \approx 0.3m, 1inch \approx 0.025m

5ft = 1.5m

7 inches = 0.175 m

Height = 1.675m

Now use the BMI formula to calculate the weight:  22.1*1.6752= 62kg

Now use the dosage formula to get the maximum allowable volume: 7*62/10*1/0.5=7*6.2*2= 86.8 mL

### Bayes Centre

P(A) = probability that patient is disease free = 0.99

P(B) = probability that the test is positive = 0.10304

P(B|A) = probability of testing positive given that you don’t have the disease = 0.096

Now we calculate P(A|B) : the probability that the patient is disease free given they have received a positive test result using Bayes Theorem.

P(A|B)=P(B|A)P(A)/P(B) = 0.096*0.99/0.10304 = 0.9223602.

So, in this case, even though the patient has tested positive, there is an over 90% chance that they're disease-free!

### St Albert's Chapel

They started at the North Pole.

There are a few ways to do this.  For example:

• Start at the top right;
• Walk clockwise around the outside until you get back there;
• Now walk south-west to the middle bottom point (passing through the central point);