Probability Riddle

Honestly, you can believe WW all you want, but the simple fact of the matter is, he is wrong. This is a widely known Probability problem that occurred in the mid 70's amongst mathematicians around the world. The answer I stated is based on the logic and reasoning of the smartest people's combined efforts inthe world. Take it or leave, but its true.

If you were take an introductory Error, Uncertainty, and Probability course in college you would see why he is wrong. Also, if it so appeases you, check out a Stochastic Systems course, and you will surely come across this problem.

I still think he's right, so take that.

im junior taking ap stats, we are going to start talking about this in a week -_-

i understand what titus means though about 2/3rd chance..

Honestly, you can believe WW all you want, but the simple fact of the matter is, he is wrong. This is a widely known Probability problem that occurred in the mid 70's amongst mathematicians around the world. The answer I stated is based on the logic and reasoning of the smartest people's combined efforts inthe world. Take it or leave, but its true. If you were take an introductory Error, Uncertainty, and Probability course in college you would see why he is wrong. Also, if it so appeases you, check out a Stochastic Systems course, and you will surely come across this problem.

I still think he's right, so take that.

man thats my favorite chess move, how could you?

My initial thoughts were as follows:

Probability Principles:
Bayes' Rule

The probability of A and B equals the probability of A, given B, multiplied by the probability of B.

P(A,B) = P(A|B)*P(B)

With the given information, you automatically know that all the doors have the same probability, 1/3. Next, you are given additional informaiton, which will have an outcome on the solution, hence, where Bayes' Rules comes into play, stated above.

Now, the probability of A, given that door C is wrong, is 1/2. Also, at the same token, the probability of B is just straight up, 1/2. This means that P(A|B) = 1/2, and P(B) = 1/2, therefore, P(A|B)*P(B) = 1/4.

This shows that by switching, you now have a new probability of 1/4 opposed to the initial probability of 1/3 you had in the beginning.

This, however is wrong, but my professor is still yet to give me a concrete answer on why... Asshole.

Who knows, maybe I proved hundreds of Mathematical geniouses wrong!

I hate when I post at the ass end of a thread and no one see's my post...

I hate when I post at the ass end of a thread and no one see's my post...

Who said that?

Honestly, you can believe WW all you want, but the simple fact of the matter is, he is wrong. This is a widely known Probability problem that occurred in the mid 70's amongst mathematicians around the world. The answer I stated is based on the logic and reasoning of the smartest people's combined efforts inthe world. Take it or leave, but its true. If you were take an introductory Error, Uncertainty, and Probability course in college you would see why he is wrong. Also, if it so appeases you, check out a Stochastic Systems course, and you will surely come across this problem. I still think he's right, so take that.

YATTA!

I wonder what Adrena was talking about?

Where?

And everybody say:

YATTA!

Sue itchy guys uh huh.

Whoo dong hide?

A perfect visual representation for those who couldn't verbally understand the theory. Good find Doofus.

Math Class repeats itself, you guys, this is simple shit, PROBABILITY NO NEED IN trying to teach it, if you don't understand it you never will. And good job Doofus, haha its funny, how can somebody not understand this basic jr. high stuff???



Example: There are twohomosexual behind 2 doors,and one GOAT behind the other door. Slaynish's probability of picking something he would fuck is 3/3. His probability of picking the goat is 1/3. Fags 2/3. If door one has the goat, it would make MORE sense to pick one of the others because your probability of picking the one he wants higher. But he would have stayed with the goat because you know.... you know...... hahaha



j/k

I love it how no one cares what you say D3V. Flame inside the flame forum please. It is starting to annoy me.

I like how he said these a couple of days ago:

Just to let you know I'm not going to be a dick head. I will TRY to speak when spoken to. I will respect other members, and their beliefs(besides zaggon).

And...WW said that when a door is opened, the problem is restated anew. If it does start anew, then he is correct.

With that theory, even if there are 100 doors, the chance of finding a prize behide one of them is always equal whenever a door is opened. 1/100, 1/99, 1/98...1/2.

But of course, if the question is based on some developed theories or rules, then "yes", switching is good.

I dont see how opening one door opens 98 others...

I think what people are trying to say is that, if you have 100 doors and you open 98 and there is nothing, what are you supposed to do, follow the rule of 3 and switch since there is increased probability that way... or keep since what are the chances that you will get the prize by switching since the other 98 doors had nothing..

I mean its really hard to figure it out, since its alot more randomized chance then theory. What are your chances of getting the prize from those 2 doors?

The way I think it, it should be 50:50 % chance since the car can only be in one of the two so, which to choose? A or B

The way a statistician looks at it, is that he/she would count all 3 doors, in the 3 door problem, since they were there originally and even though one is taken away, it still effects your chances. This is because you chose a door from 3 doors and not two.

That should clear up any doubts.