Super fun maths quiz

It hink its only the fact that the gameshow host knows what is beyond each door's he is opening that changes your chances.

It would eb a different scenario if you were trying to find someone behind three doors and you chose door 3, and just as you were about to open it someone else comes out of door 1. In this case switching to door 2 wouldn't improve your chances. It would be a 50-50 chance between door 2 and 3.

 

Eh ? what ballacks ! If one of the options is removed there are only two remaining to choose from. We'll have to get a supercomputer to tell us that the answer is 42... or summert !

 

In fact I understand more about the Infinite improbability drive

The principle of generating small amounts of finite improbability by simply hooking the logic circuits of a Bambleweeny 57 Sub-Meson Brain to an atomic vector plotter suspended in a strong Brownian Motion producer (say a nice hot cup of tea) were of course well understood — and such generators were often used to break the ice at parties by making all the molecules in the hostess's undergarments leap simultaneously one foot to the left, in accordance to the theory of indeterminacy.

Many respectable physicists said that they weren't going to stand for this, partly because it was a debasement of science, but mostly because they didn't get invited to those sorts of parties.

Another thing they couldn't stand was the perpetual failure they encountered while trying to construct a machine which could generate the infinite improbability field needed to flip a spaceship across the mind-paralyzing distances between the farthest stars, and at the end of the day they grumpily announced that such a machine was virtually impossible.

Then, one day, a student who had been left to sweep up after a particularly unsuccessful party found himself reasoning in this way: If, he thought to himself, such a machine is a virtual impossibility, it must have finite improbability. So all I have to do in order to make one is to work out how exactly improbable it is, feed that figure into the finite improbability generator, give it a fresh cup of really hot tea... and turn it on!

He did this and was rather startled when he managed to create the long sought after golden Infinite Improbability generator. He was even more startled when just after he was awarded the Galactic Institute's Prize for Extreme Cleverness he was lynched by a rampaging mob of respectable physicists who had realized that one thing they couldn't stand was a smart-arse.

 

You're meant to explain things like this to me, not the other way around. I understood it when I read the Ian McEwan novel. Being an arty type and that.

 

The probability of a coin landing heads is .5 (1/2), 2 successive heads is .25 (1/4), 3 is .125 (1/8), etc

So the probability of exactly all 10 landing heads is 0.00098 or 1/1024.

 

Yep, that's exactly right.

Fun fact: I've got an exam next week for a module of my maths degree called 'Applications of Probability'. I came on here to procrastinate.

 

My favorite application of probability was a scientific report linking global warming and piracy. The best way to stop climate change is to become a pirate... http://www.forbes.com/sites/erikaan...he-lack-of-pirates-is-causing-global-warming/

 

I make this about 0.0139, or 1.4%. That seems a bit high to me, though, so I could have made some mistakes in my calculations. I'd imagine there's a nifty way of combining the probability distributions, but I'd need to have a think about that, so I just tried to brute force it!

 

From memory the answer is about 7%. That was through Bayes Theorem and a binomial calculator to take the nastiness out of most of the working.

 

I was always better at Geography

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Sounds like you might enjoy this website!

 

Reight, been thinking about this fcuker all day. The marbles and the box and you fishing about in the box are all a red herring, they're superfluous to the question. The question is: What is the probability that all 10 marbles in the box are white? Which is exactly the same as asking what is the probability that someone flipped a coin and it came up heads 10 times in a row. A white marble is equal to a head, so we can completely ignore the marbles and the box, we just need the probability of flipping 10 heads in a row.

That's pretty simple probability that you learn in the first year at high school or maybe before these days. A coin toss has two outcomes, heads or tails. So the probability of it landing on heads is 1 in 2 or 0.5 or a half or 50% or however you want to express it. Flip it again and the first toss has no bearing on the second, the second is independent, so again you have a 1 in 2 chance of throwing heads. From the two coin tosses you have a 1 in 4 chance of throwing two heads. You can show this by representing the possible outcomes:

HH HT TH TT

To get this result it's a half multiplied by a half. You can represent it again after three coin tosses.

HHH HHT HTH HTT THH THT TTH TTT

This time there are 8 possible outcomes, only one of which is all heads. So the probability of heads coming up every time after 3 coin tosses is 1 in 8. A half multiplied by a half multiplied by a half. And now we've got our formula. Half or 0.5 to the power of n, where n is the number of coin tosses.

In this question there are 10 coin tosses, so it's 0.5 to the power of 10, which gives:

0.0009765625 or 1/1024 or 0.097%

That answer was actually given hours ago. If that's wrong, I'd like you to explain to me why because I can't see anything untoward.

 

It works on the same principle as the other 2 - Bayes Theorem, which is to do with conditional probability. It accounts for situations where you are given something with a known probability but you are then given further information which affects it. The theorem is for when you want to work out the probability of event A being true given the fact you know event B (effectively the further information). In logic notation (to stop me having to write it out every time) that is written as p(A|B) with the | sign simply standing for "given". The theorem states that p(A|B) = p(B|A) x p(A)/p(B)

Applying it to the switch decision in the 3 doors problem (assuming you chose door A) gives p(car being behind the Door C | the host revealed a dog turd behind Door B) = p(the host revealed a dog turd behind Door B | the car was behind Door C) * p(car being behind door C) / p(host revealing a dog turd behind Door B)

Taking each in turn - p(the host revealed a dog turd behind Door B given the car was behind Door C) equals 1, as if the car is behind door C then the host will always reveal a dog turd behind B.

p(car being behind door C) = 1/3, as there was a 1/3 chance that the car was behind Door C at the start of the game.

p(host revealing a dog turd behind Door B) = 1/2 as if you have chosen Door A, it is 50/50 as to whether the host will open B or C to show the dog turd.

This gives p(A|B) = 1*(1/3)/(1/2) which equals 2/3, showing that you should switch.

The numbers from the disease question can also be plugged in in the same manner. Now for the marble one:

p(all marbles in the box being white|you pull out 10 white marbles) = p(you pulling out 10 white marbles | all the marbles were white) * p(all the marbles in the box being white) / p(you pulling out 10 white marbles)

p(you pulling out 10 white marbles | all the marbles were white) is 1, as if all the marbles were white you would be guaranteed to pull out 10 white ones.

p(all the marbles in the box being white) = 1/2*1/2*1/2.... which is 0.0009765625

p(you pulling out 10 marbles) is the tricky one, this is the TOTAL probability of you pulling out 10 marbles and has to account for every situation where you might pull out 10 white marbles. So it includes the probability of there being 1 white marble in the box and you pulling it out 10 times, 2 white marbles being in there etc. So you need to do p(only one head coming up) * p(that one white marble being pulled 10 times) + p(2 heads coming up) * (some combination of those 2 marbles being pulled out 10 times) + .......p(all flips being heads). It's a bugger to work out manually but using a binomial calculator it comes out as 0.01391303

so p(all marbles being white | you pulled 10 white marbles) = 1* 0.0009765625 / 0.01391303 = 7.01%