interesting experiment on probability.

[NickTheKid 0]

I went on a camping trip with a good friend of mine, it's basically an excuse for 20 fathers, sons, daughters, and friends from different family's to get together for one weekend every year, us kids usually get to the beer and soft drugs for the weekend and leave it at that :wink:. Well anyway, of the people that go there, many are very educated professors, lawyers, judges, etc. and I heard this VERY interesting theological problem from a probabilty guy there.You are a contestant on some gameshow in the 60's. You get to the end of the game, and the host gives you a choice of ONE of three doors. Behind one of the doors has jewels, gold, rubies, treasure. Behind the other two there is, nothing.HOWEVER, the loop hole is that once you choose a door, the host will open one of the OTHER two unopened doors to reveal one of the empty ones. So once you choose a door, the host opens one of the loser doors (he will never open the one you chose), and he gives you the option to CHANGE your choice to one of the other doors.If you decide to change doors, what is the % you will win? (THIS is a MATHEMATICALY proven number, we even tested it with cups and quarters and it worked out) I'll see if one of you can get it. I will explain the answer after.

[Vade 0]

It's 50%Either you have the right door or you don't./wonders what the catch is

[NickTheKid 0]

Nope. It's not 50/50.

[econ_tim 0]

I don't want to spoil this, but this is often referred to in statistics as the "Monte Hall Problem" because Monte Hall hosted the gameshow "Let's Make a Deal." At the end of the show, contestants had a similar choice. The could picks from doors 1 through 3. Behind two of the doors were joke (or jopke) prizes, and behind one of the doors was a real prize. After they made their choice, Monte Hall would show one of the joke prizes.Anyway, the answer is counterintuitive, and many people won't believe it. But it is a good probability exercise.

[NickTheKid 0]

I don't want to spoil this, but this is often referred to in statistics as the "Monte Hall Problem" because Monte Hall hosted the gameshow "Let's Make a Deal." At the end of the show, contestants had a similar choice. The could picks from doors 1 through 3. Behind two of the doors were joke (or jopke) prizes, and behind one of the doors was a real prize. After they made their choice, Monte Hall would show one of the joke prizes.Anyway, the answer is counterintuitive, and many people won't believe it. But it is a good probability exercise.

econ, when I first heard this I was like, Whoaaaa!!!

[JaysonWeber 0]

If I am reading this right its 100%You open one door, Its wrong. So he opens another door (the other empty one) and you choose to change doors, going with the Last door.Am I mis-reading it?

[NickTheKid 0]

If I am reading this right its 100%You open one door, Its wrong. So he opens another door (the other empty one) and you choose to change doors, going with the Last door.Am I mis-reading it?

It cant be 100%, you dont know if your door is right or wrong.

[Breathing 0]

I believe the answer would be 1/4 where it seems like it would be 1/2, but since the first door could be right, its odds are now reduced from 1/3 to 1/2 because the third door is out. the odds of either door are 1/2 but since we know that at least 1 of the 2 doors not originally selected had nothing behind it, then the odds would now double on a door change. I may be wrong but this is what ran through my head at 2 am and it made sense at the time

[JaysonWeber 0]

Hmmm after Econ's response I was tempted to Google it.IF YOU GOOGLE THIS DONT REPLY!I haven't yet. Still waiting on a response... I dont know if I'm answering the question fully or reading it right :DI know the start of the problem is Door 1. 333Door 2 .333Door 3 .333

[econ_tim 0]

econ, when I first heard this I was like, Whoaaaa!!!

As my nickname indicates, I was an economics major as an undergrad, and now I'm a grad student. So I've had to take some statistics courses. And even "math people" often have problems with this question.

[NickTheKid 0]

It's not 1/4

[JaysonWeber 0]

Well, What are the rules? If he Opens the right door at first and the prize is behind there does he get the prize or what?

[Catamo 0]

Its not 1/2? Damn... Lemme think here...

[NickTheKid 0]

Well, What are the rules? If he Opens the right door at first and the prize is behind there does he get the prize or what?

No, if he has the right door, he can still change his discision and lose.

[uahphysics 0]

I'm very much looking foward to the answer to this problem. I too am stumped about the answer as all I can come up with is 50/50 odds. If one door is always shown to be a door without a prize, then you're left with only 2 doors, one of which is hiding the prizes. I wish I could figure out this problem, I'm a math major in college(albeit I just finished my freshman year, so no stat classes yet) and am generally regarded in my poker circle as "the math guy".

[NickTheKid 0]

Shall I give up the answer or tortue you guys?

[allinbluff35 0]

5/6 and whatever % that turns into is my guess

[econ_tim 0]

I know the start of the problem is Door 1. 333Door 2 .333Door 3 .333

This is an important part of the solution.

[Vade 0]

Can I guess again? I'll guess 1/3 this time. Dunno why, but I'll say the odds are unchanged

[NickTheKid 0]

Nope, okay here it is.It is 2/3 of the time.The reason is simple. Let's say you chose a door and decided not to change your mind. Well you know what that means? It doesn't matter if you were shown the goat or not, you blindly going in, then still deciding to stay with your door is 1/3 and whats 3/3 - 1/3? 2/3!

[Catamo 0]

I thought about this.. and i say 66%(2/3)...You Originally have a 1/3 chance of being right, Since you have a 2/3 chance of being wrong then if you picked the wrong one, the right one would become the door you can change to.Im good.

[Breathing 0]

i concur with that 2/3

[akishore 0]

i haven't looked at ANY of the replies, so this is an unbiased answer.i was confused by the question, so i'm assuming:1. you choose one of three doors.2. the host then opens a LOSER door (so if you pick door 1, and door 2 is the winner, he will NEVER open door 2--he will always open door 3 for you to see that it's empty).3. you then change doors if you wish.so...on the first pick, you have 1/3 chance of picking the winner. let's call the winner a value of 1, and the losers a value of 0 each. so your EV is 1/3 here.1/3 of the time, you pick the winner, so the host opens either of the two loser doors. if you decide to change doors, you will win 0% of the time, so your EV is 0 if you pick a winner (1/3) of the time and change doors.2/3 of the time, you pick the loser, so the host opens the OTHER loser. if you decide to change doors, you will win 100% of the time. so your EV is 1 if you pick a loser (2/3) of the time.SO, if you decide to change doors automatically--regardless of which door you picked--you will win 2/3 or 67% of the time.thus, changing doors is the correct strategy regardless.is that right?aseem

[NickTheKid 0]

Easy since I gave the answer, eh? :

[JaysonWeber 0]

Well, What are the rules? If he Opens the right door at first and the prize is behind there does he get the prize or what?

No, if he has the right door, he can still change his discision and lose.

So... In this experiment, We're just giving the Possibility that he can change doors and does so 100% of the time regardless of whether he is right on the first Guess?If that's the case I think I know it