The Monty Hall problem.

[Rhizomorph13]

Imagine if monty hall played the game with live or dead cats beholden to a quantum triggered lethal device...

Rhizo

[Tiberius]

They aren't though, because your chances of being on the correct door are much lower than you being on the correct door. I wrote a computer program that played the game. If you tell it to always switch, it wins 2/3 of the time. If you tell it to always stick, it wins 1/3 of the time.

If you know how to program you can see this in action. If you don't, I'm sure I could create a php webpage that played it for you

---

Update: Here is a good one: http://www.nytimes.com/2008/04/08/scienc....html?_r=1

So play the game, and do 10 rounds where you always switch. Then do 10 rounds where you always stick.

I got 60% wins with switching, 30% wins with sticking.
After 20 rounds each, 65% wins with switching, 35% wins with sticking.
After 30 rounds each, 63% wins with switching, only 27% wins with sticking.
After 40 rounds each, 65% wins with switching, 33% wins with sticking.

The pattern is emerging

[Dotard]

They aren't though, because your chances of being on the correct door are much lower than you being on the correct door.

Ahhhh! I see! Your EXISTING chance of choosing the winning door in the first round are lower than your EXISTING chance of having chose the losing door.

Therefore............... (*Dotard spends a long time scratching his head*)..............

So switching doors would yield a higher possibility of choosing the winning door!

I get it now.


Stop that. You make my ears bleed from the pressure in my brain.

[fr0d0]

It makes perfect sense straight away but is essentially deception.

[Tiberius]

Well at least this is easier to understand than the 0.9r thing

[fr0d0]

Yes definitely. i love the 0.9r thang

[Amphora]

It makes perfect sense!

If you tell it to always switch, it wins 2/3 of the time. If you tell it to always stick, it wins 1/3 of the time.

But more than likely murphys law would come into play.

Amp

[Edwardo Piet]

Makes sense to me. You're more likely to have picked a goat, and if you have, switching always wins.

I didn't get it originally, but that may only because I didn't really understand the OP properly, it confused me. I love probabilities, it's by far my favourite kind of maths. Other than that I'm not really bothered.

EvF

[Golbez]

For those who don't get it, here is an explanation using the three options you can choose from:

1) Pick a door with a goat. Host reveals other goat. Switching wins.
2) Pick a door with a goat (the other one). Host reveals other goat. Switching wins.
3) Pick a door with a car. Host reveals one of the other doors (they both have goats). Switching loses.

The idea is that you have a 2/3 chance of picking a door with a goat at the start, so you are more to have picked a goat. When you have picked a goat, switching always wins. If you picked a car at the beginning, switching will always lose, but since you only have a 1/3 chance of picking a car at the start, the chances are against you. On the other hand, if you stay with your choice, 2/3 times you will lose, and 1/3 times you will win.

So it is more advantageous to switch

Sorry to resurrect a thread, but I have just recently come across this problem with a Sam Harris discussion (actually, came across it a couple weeks ago, but never thought to seek it out on a forum). So rather than start a new thread, I did a search which brought me here.
(Link to Sam Harris discussion of it:
http://youtu.be/3c4F4tW7u_A?t=57m37s)
Starts at about 57:37. Not sure if the embedded video held on to the time stamp. Doesn't seem like it.

Now, I've been struggling greatly in trying to accept this, and even with this best explanation, I feel like it's not holding up to scrutiny.

If the idea was a 2/3 or 1/3 chance of picking (in)correctly at the start, these are no longer your options the moment the goat is revealed. The instant it is revealed, you've dodged one of the 2 incorrect bullets and it is no longer a liability of odds stacking against you. There is no memory to chance. And this seems like an easily scientific problem whereby you can actually test it out in many simulations.

Hmm, actually, I almost discarded this post as I thought I had worked out the very clear understanding of this problem. But either my understanding was wrong (as I've worked that analysis to be incorrect again, so I hope this is the case or everyone's been duped), or it is indeed the one everyone holds of this problem. So let me see if I have it here:

The problem is there are three options each with the equal likelihood of being the right answer from the start, of which you can only choose one. So a 33.3% even odds for each choice. So here's my understanding of how people view there being a right answer in switching: your initial selection has a 33% chance of being right, and the remaining two options have a 66% chance of being right. One of the two options of the 66% odds was revealed, and so you know it's not that. But if you switch, you'll have displayed two options, whose combined chances had a 66% chance of being right. So statistically, that means you've played the better odds every time. 66% is always better odds than 33%. Here's a crude visual representation of this (sorry I had to use underscores; the forum format removed my spacing):

1_______2_______3
|_33%|__x____66%|

My (your/whose ever) choice was door number 1, at 33% odds. Doors no.2 and 3 combine for 66% odds. So if door no.2 is revealed not to be it, and I choose door number 3, I've covered 66% of the terrain, so I should always switch to cover this terrain, rather than sticking with my initial 33% terrain instead.

The problem with this analysis is one of poor and arbitrary framing. I could just have easily lumped in door number two with my initial lot and given myself 66% odds instead, knowing that third and isolated door was the one with 33% odds. It reshifts the crude representation thusly:

1_______2_______3
|_66%__x__|__33%|

Now with X'ing out door no. 2, and sticking with door no. 1, I still have 66% odds going against 33% of the door I didn't pick. So this gives me the exact same odds as you think you have for switching, with me not switching. And I think what this means is, we are wrong in assigning that 33% door number 2 odds to either door, which means that the odds are really just 33% against 33%, or even at 50-50. Which again makes sense because random chance has no memory. We have 2 different results from 2 different choices, each of which are equally likely. Why ought it be anything other than 50-50?

So again, hopefully I am misunderstanding the reason for this and that my interpretation above is wrong. But if so, I don't yet understand the rationale for switching. I'll continue to read through the rest of the thread, but otherwise please help point out where I went wrong.

Edit - Nevermind. Went through that NYT link and had it spelled out for me. That's trickery. =P

[Rev. Rye]

Well, since this thread's been resurrected and I haven't heard any mention of "The Curious Incident of the Dog in the Night-Time", I'll explain it as Christopher Boone explained it.

Let the doors be called X, Y and Z.
Let Cx be the event that the car is behind door X and so on.
Let Hx be the event that the host opens door X and so on.
Supposing that you choose door X, the possibility that you win a car if you then switch your choice is given by the following formula
P(Hz ^ Cy) + P(Hy ^ Cz)
= P(Cy)·P (Hz ¦ Cy) + P(Cz)·P(Hy ¦ Cz)
= (1/3 · 1) + (1/3 · 1) = 2/3