# The Rope of Dreams Recut: Polynomials of the Third Order – Cubic Equations

The Rope of Dreams in 3D

The Rope of Dreams Recut – Cubic Equations

Did you make the most of The Rope of Dreams in my previous post “The Rope of Dreams : Polynomials of the Second Order – Quadratic Equations“. I hope so. Well now you have a chance to take your 120 meter rope and enter another Dimension with Cubic Equations.

# What is the Scenario?

Once again you are given the Rope of Dreams, and a Golden Scissors which is the only thing that can cut the Rope of Dreams. You can cut the Rope of Dreams twice (cross section, no longitudinal cuts), which will give you 3 lengths. These lengths will be laid out one for each dimension X, Y, Z (i.e. left-right, backward-forward, up-down), and whatever volume you enclose anywhere on Earth is yours to keep, or do with whatever you wish. You can enclose only a single volume, a single time, and then must return the rope and the scissors. What would you do? You may not think of a Polynomial of the Third Order – a Cubic Equation, but you probably should. In this post we’re talking about cuboids, we’ll leave spherical shapes aside for now.

# Do the Math!

This time we’ll do the mathematics first and then apply our findings to determine how best to maximise the volume we enclose. In the previous post we derived and proved the following quadratic equation:

Previously: $${ x^2 = (x-n)(x+n) + n^2 }$$

Multiply by $$x$$: $${ x^3 = x ((x-n)(x+n) + n^2)}$$

Multiplying this out: $$\boxed{ x^3 = (x-n) x (x+n) + xn^2}$$

# The Rope of Dreams : Polynomials of the Second Order – Quadratic Equations

Where would you go? What would you enclose?

The Rope of Dreams – Polynomial

Imagine you are given a length of rope that is 120 meters long, and told that you can go to any place on Earth and whatever you enclose with the rope – is yours to keep, or do with whatever you wish. You can enclose only a single area, a single time, and then must return the rope. What would you do? You may not think of a Polynomial of the Second Order – a Quadratic Equation, but you probably should.

# What would you do?

Say a big “Thanks”, take the rope, and start pondering the options. A likely plan is to think of where on Earth you want to go (a tropical island, a bustling city, a countryside retreat, maybe even Fort Knox – it’s your choice), and while en route to your destination figure out how to maximize the area the 120 meter rope can enclose. I’ll leave the destination to your own imagination (you can post in the Comments section below) and turn our attention for now to maximizing the area the rope can enclose once you get there. Did someone say Polynomial!

# How long is a piece of string, or 120 meters of rope?

A likely first question you might have is to get an idea of just how long 120 meters is, so some reference examples might help, note that ‘m’ is short for ‘meter’. A soccer pitch is between 90m and 120m in length; A rugby pitch is 100m – the same as the 100m sprint in Athletics (Usain Bolt, Carl Lewis etc.); An American football pitch is 110m long; A CLG/GAA (Cumann Lúthchleas Gael / Gaelic Athletic Association) pitch is between 130m – 145m in length. For petrol heads, 120m is about 24 Nascars end-to-end, or 21 Formula1 cars end-to-end – that’s almost the entire grid – are you heading to Monaco with your rope?

# Monty Hall Proof – The Formula

Get the free App – Monty Hall Game with Monty Hall Proof.

Monty Hall Proof – The Formula is here. My two previous posts described the Monty Hall Problem – Can You Solve This Maths Puzzle? and Monty Hall Solution – Advanced! Well, this is the next installment of the trilogy, a simple mathematical proof.

If you don’t like Maths (Mathematics, Math) then, well, you have serious problems – get some help :^) This isn’t difficult at all, it’s just a bit of simple probability and algebra, yep ALGEBRA ♥

The Probability that you will Win is the quotient of the Number of Cars, and (divided by) the Number of Doors. To represent that symbolically using algebra is simple:

$$P(W) = \frac{NC}{NDtot}$$ … Equation (1)

The Probability that you will Lose is a little more interesting, it is the quotient of the Number of Doors less the Number of Cars, and (divided by) the Number of Doors, in symbolic notation this is:

$$P(L) = \frac {NDtot – NC}{NDtot}$$ … Equation (2)

There’s one last equation we want, and it says the Probability that we either Win or Lose is 1 – since these are the only two possible events. In other words, we have to either win or lose – there are no other possible events (see my earlier post re the philosophical and physics debates on that general point). Anyway, to represent this symbolically:

$$P(W) + P(L) = 1$$ … Equation (3)

(Equation (3) is based on Kolmogorov’s second axiom i.e. $$P(\Omega) = 1$$)

# Monty Hall Solution – Advanced!

Get the free App – Monty Hall Game

This post “Monty Hall Solution” continues on from my previous post Monty Hall Problem – Can You Solve This Maths Puzzle? If you haven’t read that post, then read it now before reading this. Because I will now show you even more Monty Hall Solution coolness! We saw that you could increase (double) your chances of winning a car by understanding some maths, so let’s delve further into it and who knows, you might win something big (then again you might not, but hey!).

So what happens, if there are 4 doors instead of 3 doors? And what happens if there are 5 doors, 6 doors, hmmm, more code required – cool!

Checkout the code on GitHub.
Get my free Monty Hall Game on Google Play.

# Does The Monty Hall Solution Hold True For 4 or More Doors?

Door 2 now has a 2/3 chance

For 4 doors we would expect that the odds would be 1/4 (25%) for not changing Vs 1.5/4 (37.5%) for changing. “How did you get those figures?” I hear you ask. Well, with 4 doors, each door has a 25% chance of being correct. Our 1st chosen door has a 25% chance – the other 3 doors have a combined 75% chance. When Monty removes one of those 3 doors by opening it -the remaining 2 doors still have a combined 75% chance – which is now divided by the 2 remaining doors i.e. 37.50% chance each. Once again the figures from the 70 million simulations are very precise.

Doors in Game: 4

Unchanged Wins: 2498993, Changed Wins: 3750566
Unchanged Wins: 24.99%, Changed Wins: 37.51%

Unchanged Wins: 2498720, Changed Wins: 3750419
Unchanged Wins: 24.99%, Changed Wins: 37.50%

Unchanged Wins: 2500812, Changed Wins: 3748980
Unchanged Wins: 25.01%, Changed Wins: 37.49%

Unchanged Wins: 2499171, Changed Wins: 3747829
Unchanged Wins: 24.99%, Changed Wins: 37.48%

Unchanged Wins: 2499602, Changed Wins: 3749207
Unchanged Wins: 25.00%, Changed Wins: 37.49%

Unchanged Wins: 2497548, Changed Wins: 3748332
Unchanged Wins: 24.98%, Changed Wins: 37.48%

Unchanged Wins: 2499206, Changed Wins: 3749860
Unchanged Wins: 24.99%, Changed Wins: 37.50%



# Monty Hall Problem – Can You Solve This Maths Puzzle?

Get the free App – Monty Hall Game

The Monty Hall Problem is an interesting Maths Puzzle – with a hotly disputed answer. Monty Hall is a well known American TV Show presenter, and this particular maths puzzle gained some notoriety on his TV Show – hence the name “Monty Hall Problem”. The puzzle is quite simple to understand, but as is so often the case it is a little more tricky (and fun) to figure out the answer.

World famous mathematicians have gone to their grave disputing the answer. But I’ll explain it in easy to understand language, and demonstrate a proof using a computer simulation I wrote which played no less than 70 million games. The source code for the simulation is available on GitHub so you can review it yourself 🙂

# Show Me The Monty Hall Problem Puzzle!

Monty shows you 3 doors which are closed. You are told that behind one of the doors is a car, behind the other 2 doors lies a Goat. If you choose the correct door you win the car. The doors are labelled 1, 2, and 3. Let’s say you choose Door 1. In an unexpected twist Monty opens say Door 3 – behind which stands a goat! Monty then asks you “Do you want to stick with Door 1, or do you want to choose Door 2 instead?”.

The Monty Hall Problem

So do you stick or twist? You can stick with Door 1 or try your luck with Door 2. Monty is waiting. Your heart is racing. The crowd are shouting in equal measure “Door 1” and “Door 2”, some jokers are even shouting “Door 3” ]:> 😀

# Show Me The Monty Hall Problem Answer.

Not so fast. The fun is figuring out the correct answer – if there is a correct answer. What I’m going to do is put forward an answer and explanation, then I’ll write a software program to simulate this scenario 70,000,000 (70 million) times and see if it agrees with my proposed answer. Is it better to stick or to twist – or does it make any difference? I’ll share the results (and the code) of this simulation with you. In the meantime, you can try to solve the puzzle by yourself. But before that, I’ll propose an answer – spoiler alert – don’t read the next paragraph if you don’t want to see the proposed answer.