Maths paradox in LEGO stop-motion

In the animation they already determined who was going to die before the prisoner asked, so its still 33%

The video is wrong. Where it is wrong is the idea that probability cares about who knows what and how they know it. Probability is not so judgmental. You could frame the same problem and leave interaction with the prisoners out. Have fourth person, an independent conscientious observer pose the question.

Learn to conditional probability.

P(A|B) = probability of A occurring given B is known to be true
P(A and B) = probability that A and B are true
P(A|B) = P(A and B) / P(B); find text book or google it

Now:
A = Orange dies = 1/3
B = Green Dies = 1/3
C = Pink dies = 1/3
~C = Pink does not dies = 2/3; either 1 - P(C) or P(A) + P(B)
P(A and ~C) = P(A) = 1/3; because A is a subset of C; in other words they are not independent.

P(A | ~C) = P(A and ~C) / P(~C) = (1/3)/(2/3) = 1/2
P(B | ~C) = 1/2; same math but could be arrived by 1 - P(A | ~C)
P(C | ~C) = 0 because P(C and ~C) = 0

It amazes me how dumb some people are on this board.

If you don't understand how wrong what you just wrote is you need to go back to 10th grade math. There is no difference between saying the next roll will be 6 or I rolled the dice already under the table, what is the probability that when you look it will be 6.

To further illustrate how ridiculous your statement is spin a roulette wheel. There are 42 numbers on a roulette wheel. I spin it and you are blindfolded. What is the probability I got 0. If you think it is 0.5 because I already spun it, place your bet and I will take your money all night long.

the act of spinning a roulette wheel obviously affects the probability of the various outcomes...
jeez bro, where did you go to school?

Hello newfriend!

The video is an alternate version of the Monty Hall problem, you arrogant retard.

You're stupid or not. In this case it's 100% the first option.

Read about Monty Hall before saying something.

It's funny that you're calling someone dumb when all of that extra information is entirely irrelevant. It has no influence on anything. It's just there. If you spin the wheel and you want to know the probability that it's on zero, it's 50% because there two outcomes: it's either on zero or it isn't. The value is zero or nonzero. Those are literally the only two outcomes. How do you not get that?

Not him but I don't.

How do I not dumb, Veeky Forums.

How do I not dumb.

But you are so wrong, sir. You have no idea about probability.

Just because there are two outcomes it doesn't mean the chances are 50-50. Take a rigged coin. On a normal coin you have 50-50 but a rigged one gives, say heads (pun intended) more often than tails. So it's not 50-50 man!

Can Veeky Forums stop being fucking autistic by taking the massive b8 spill in this thread?

True! Still, the film is correct,

That's Monty hall restated.
en.m.wikipedia.org/wiki/Monty_Hall_problem

You can literally, physically, design a physical monty hall problem, execute it, gather statistics and results, and prove it's 66%, not 50%.

How are people this fucking stupid?

Seriously make a fucking diorama and educate yourselves you fucking dumbfucks.

Exactly!

Nooice! Keep up the good work!

I don't get it!

That's a quality troll, friendo

If there are only 2 possible choices, why is possibility 2/3? It's either x or y at the moment.

Yes but if you have a rigged coin and toss it and let's assume it shows heads 1/3 of all cases. Then you have 2 outcomes where one occurs with probability 1/3 and the other 2 /3

Or take a dice. You have two options. Either it shows 1 or it doesn't. However it shows 1 with probability 1/6 and not one with probability 5/6. Do an experiment if you want to and count the number of occurrences of 1 and not 1 and divide by the number of all tosses. See what you get.

That's quite different from Monty Hall's problem. If there are 3 choices with x occuring at 1/3 possibility and y at 2/3 possibility and one y choice is removed so only x and y choices are left then x and y may occur with same 1/2 possibility at the exact moment.
Nevermind, I'm dumb.

No. It's exactly the same as monty hall. Except no one is asked to change the gate and you only assess the chances of each prisoner (gate) being selected. Historically three prisoners problem was stated first and after seal years restated as the Monty hall problem.

I strongly encourage you to read the wiki entry on this
en.m.wikipedia.org/wiki/Three_Prisoners_problem

Wiki quote on the intuitive understanding of this:
Prisoner A only has a 1/3 chance of pardon. Knowing whether "B" or "C" will be executed does not change his chance. After he hears B will be executed, Prisoner A realizes that if he will not get the pardon himself it must only be going to C. That means there is a 2/3 chance for C to get a pardon. This is comparable to the Monty Hall Problem.

Since we're on the topic of logical paradoxes, can anyone explain what makes the Boy or Girl Paradox, a paradox? It seems so simple and trivial to me. Am I wrong to say that it's 50% for both cases? I can't see how it's anything different.

Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

The paradox is in the different assumptions people make about what it means when someone tells you at least one child is a boy. Since we do not know the process which led to this person knowing that, the question is ambiguous. For example, if the person looked at the two children and saw one or two boys, then the answer would be 1/3. If the person only looked at one of the children and saw a boy then the answer would be 1/2.

Ah I see. I wasn't thinking about it from that perspective, where either child has its own individual probability of being a boy or girl.

Regarding the term paradox...well all this is counter intuitive. And that's that. It's not a paradox per se but rather paradox-like. It contradicts the intuition.

So what's an actual paradox in math? Can you name one?

>when this faggot still uses the gender binary system
This is Veeky Forums and gender is unscientific and a spectrum

Lego animation is a great idea but I don't understand it. Maybe watching it twice will help.

Maybe for discrete finite shit

How is this Monty Hall? The one who receives the information can't do anything with it. It's essential to the Monty Hall that the contestant has the action to change doors.

>It's essential to the Monty Hall that the contestant has the action to change doors.
Not really. The question is merely which door/prisoner is more likely to have the car/execution.

Well, it's essential to the analysis. So, let's say they're wearing different coloured burkas and they can change them, should he? If he was picked for execution, then switching burkas saves him. If pink was picked, then swapping.kills you. Orange wasn't picked as we've been told.
I think the difference is that there is a difference between the "doors". In Monty Hall, "one of the other doors" reveals no car, but in this case, it's a specific burka.

Consider an intestine bug which lives in the guts of prisoner Green. Given the piece of info Green is given is it better for the bug to change its host to Pink or stay in Green? That's exactly monty hall mate.

It doesn't matter if the bug changes or not. There are only 2 situations to consider, Pink gets killed and Green gets killed and they're equally likely. Let me sketch Monty Hall's three situations
Your door is right: switching loses.
First other door is right: switching wins.
Second other door is right: switching wins.
in this case we have two situations
Pink dies, switching kills you.
Green dies, switching saves you.

We can make the intestine bug story more coherent. Assume there's another intestine bug which prays on dead bodies and will die within a day if it doesn't eat dead person. He sits in Green and hears the conversation with the trooper. Now should the bug change its host to Pink? Yes it should

Maybe I need to start earlier in the process. So the chance green was chosen is 1/3, then the chance that the guard tells him it's orange is 1/2 so it's 1/6 total. The chance that pink was chosen is 1/3 but then the chance that the guard tells him it's orange is 1, because he can't tell him his own fate, with a total chance of 1/3.

The problem with appeal to intuition is that intuition can lead you wildly astray, your various reformulations did nothing to help me understand it.

My aim wasn't appeal to your intuition but rather translate the problem to the Monty hall setting so that you see its the same. Lemme try once again. You are a bug which chooses a host to die and you will pray on it. The choice is random and you choose each host with probability 1/3. Let's assume you chose Green. Your host approaches the trooper and the trooper says that he cannot say anything about Green but says Orange lives (it's like saying door 3 is without the car if you chose door 1). Now you as the bug are given a choice to either stay with your host or change the host to Pink. What do you do? It's better to change just like in monty hall.

Why nobody at my school taught maths via LEGO?!!

You should have attended lessons at our school ;).

Is there anything continous in this world ;)?