Monty Hall Problem

[ACE2M]

So he shows you the door with the goats - if they then mix up the remaining 2 doors so you cant remember which one you picked originally and ask you to pick what are the odds of you finding the car?

50/50

But they are not doing this. It is based on the fact that your first pick is 2/1 against.

[Decider]

So he shows you the door with the goats - if they then mix up the remaining 2 doors so you cant remember which one you picked originally and ask you to pick what are the odds of you finding the car?

Yes, if you change the problem like that then it does become 50/50

[Decider]

So he shows you the door with the goats - if they then mix up the remaining 2 doors so you cant remember which one you picked originally and ask you to pick what are the odds of you finding the car?

50/50

But they are not doing this. It is based on the fact that your first pick is 2/1 against.

Correct.. Here's my explanation:

Basically at the start of the show you know you can pick one door which has a 1 in 3 chance of being the car.

This leaves a 2 in 3 chance that the car is behind one of the other doors.

You know that he will always give you the option to stick with your original choice or switch.

The exposure of the door with a goat is a red herring, it's completely irrelevant other than to confuse you into thinking the problem is now 50/50.

He may as well have said at the start,choose a door, and then decide if you want to stick with your original choice or switch to the other 2 doors.

[ACE2M]

http://en.wikipedia.org/wiki/Monty_Hall_problem

For a very detailed explanation, it also points out that the monty hall advantage is lost in the deal or no deal game.

[Decider]

Another explanation I love this problem, when I first saw it it was clearly 50/50...

It's equivalent of having 1 million doors and having a car behind 1 of them and goats behind the others.

You choose 1 door a 1 in 1 million chance of getting the car. The host (who knows where the car and goats are) then shows 999,998 doors with goats and leaves just your original choice, and 1 other door. He offers you the chance to stick or switch.

[matt674]

Correct.. Here's my explanation:

Basically at the start of the show you know you can pick one door which has a 1 in 3 chance of being the car.

This leaves a 2 in 3 chance that the car is behind one of the other doors.

You know that he will always give you the option to stick with your original choice or switch.

The exposure of the door with a goat is a red herring, it's completely irrelevant other than to confuse you into thinking the problem is now 50/50.

He may as well have said at the start,choose a door, and then decide if you want to stick with your original choice or switch to the other 2 doors.

If you are playing cards and you are up against two opponents and you all had the same hand AK o/s.

You are a 33.3% favourite to win therefore making the chances of either of your opponents winning as 66.6% - if you get one of your opponents to fold your chances of winning the pot dont stay at 33.3% - they rise to 50% as you now have an equal chance against just the one opponent.

Ok then by the same principle that you explain - how come in the example i have given your odds of winning the hand dont stay at 33.3%?

[Graham C]

The odds are only increased because of the way they choose to explain it, it's 50/50 once one door has gone.

Take 3 from 6 counting backwards, 6 - 5 - 4.  6-3 isn't 4 but the way you do it makes it 4.

[bolt pp]

Its all so mathematical with you internet players!!!!!

If its not poker tracker its something else.......sigh!!

This is a LIVE situation!!! you have to incorporate whether or not the gameshow host has a well trimmed beard or is clean shaven, this is integral to making the right choice.

Needless to say now Ive highlighted the beard aspect your all able to very easily work out whether or not the correct move is to stick or change, i dont have to tell you that if he has a well trimmed beard........well, its just soooooooo obvious

[AceofWands]

A simple and easy way of looking at it is that the ONLY way of losing by swapping is if you picked the winning door in the first place which was a 1 in 3 chance. So the chance of NOT losing by swapping is 2 in 3.

[Wardonkey]

What if you want the goat?

[The_nun]

What if you want the goat?

Pervert..

[ACE2M]

The example you give matt, bares no resemblence to this problem.

As pointed out earlier this is a problem of prior action and knowledge affecting the choice.

If you choose 1 from 3 there is twice as much chance that you did not choose correctly than you did choose correctly.

The host is always going to expose a goat and offer you a change.

The intial odds now affect the choice as the chance of your intial choice being correct remain at 1/3 whilst the section with 2/3 chance of containing the car still exists but with one of the options exposed.

A simple and easy way of looking at it is that the ONLY way of losing by swapping is if you picked the winning door in the first place which was a 1 in 3 chance. So the chance of NOT losing by swapping is 2 in 3.

or that which is very clear

[matt674]

but then i'd take the goats over a car every day of the week!! 

What if you want the goat?

Pervert..

Oi!! Where i come from a goat is far more practical than a car!! 

[bolt pp]

Goats + boxes usually = indignant Sark!!

They're stuffing goats in boxes for the purpose of entertainment over here Sark, get over here!

http://blondepoker.com/forum/index.php?topic=11985.0 

[ACE2M]

Goats + boxes usually = indignant Sark!!

They're stuffing goats in boxes for the purpose of entertainment over here Sark, get over here man!!!

http://blondepoker.com/forum/index.php?topic=11985.0