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Top Hat
16th September 2005, 12:15
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

Jayne
16th September 2005, 13:07
No i'd take the goat :lol: Good for keeping grass short, never have to mow again, well at least until it snuffs it.

Jayne

daveashton
16th September 2005, 14:24
No its a 50 % chance of a goat or a car.

SillyJokes
16th September 2005, 21:35
Both wrong - you'd be a fool not to switch.

MinuWeb
17th September 2005, 05:26
Behind one door is a car, behind the others, goats. You pick a door, say number 1,

Of course you don't switch

Behind one door is a car
Behind 3 door is a goat
so, there must be a goat behind 2 door as well :D

SillyJokes
17th September 2005, 05:33
No, with the first choise you had a one in three chance of getting the car.

now the game is squewed.

You know that one of the two remaining doors has a goat but the chances are better that it is in the one you didn't pick.

MinuWeb
17th September 2005, 07:12
but the chances are better that it is in the one you didn't pick.

And in real life we know this still remains true after you change your mind, it will still be behind the door that you didn't choose......

daveashton
17th September 2005, 07:50
You know that one of the two remaining doors has a goat but the chances are better that it is in the one you didn't pick.

Sorry how do you work that out.

There are 2 doors left

1 has a goat
1 has a car

Odds 50 : 50 so it make do difference is you keep the door or swop it.

:?

Top Hat
17th September 2005, 11:13
Odds 50 : 50 so it make do difference is you keep the door or swop it.


When you pick the 1st door the odds are 1 : 3 you are right.
And 2 : 3 that its one of the other doors

The host then opens one of the other doors, remember he knows where the prize is and will always show you a loosing door.

Your door still has odds of 1 : 3, the other unopened door has odds of 2 : 3, so you're best to switch.

Its a lovely counterintuitive problem.

Marina Stone
17th September 2005, 11:17
Sorry how do you work that out.

There are 2 doors left

1 has a goat
1 has a car

Odds 50 : 50 so it make do difference is you keep the door or swop it.


I'm with you on this one, as long as the goat doesn't have green eyes! :D
Marina

MinuWeb
17th September 2005, 12:21
But wouldn't you be p****d off if you switched and was wrong :D

Always stick to your original decision in cases like this

daveashton
17th September 2005, 13:37
I would be pissed off to be on the game show in the first place. These people always have naff jobs an earn very little money.

epiphany
17th September 2005, 13:38
I don't get your maths Top hat :D

Before you pick each door has a 1/3 chance of being the car

After the goats are revealed behind door three; door 1 and 2's probability alters to a chance of 50/50 each of having the car behind them.

Top Hat
17th September 2005, 17:00
Odds at the begining is 1:3 behind each door, agree?

You pick a door, the odds of it being one of the other doors is 2 : 3.

So the host open a loosing other door, the prize does not move.

The odds of it being one of the other doors is still 2 : 3, but now there is only 1 other door.

So its best to switch.

Becomes a bit clearer if there are 100 doors, and the host opens 98 loosing doors.

daveashton
17th September 2005, 17:12
this sorry is getting daft.

Top click your odds are wrong

if there are 2 doors left and 1 is a car and the the other is goat the odds are not 2: 3 but evens

i.e. 50 50 chance, remember to take the other door off you sum as well as the goat!

Top Hat
17th September 2005, 17:43
this sorry is getting daft.

Top click your odds are wrong

if there are 2 doors left and 1 is a car and the the other is goat the odds are not 2: 3 but evens

i.e. 50 50 chance, remember to take the other door off you sum as well as the goat!

This is a great puzzle.

always causes problems.

When the host opens the door he is giving you a clue, he knows it is a loosing door. The odds for the other reaming door, therefore equals the odds for the 2 original doors, so switching you have a 2:3 chance of winning. (interestingly, if he opens a random door, and its a loosing door, the odds are even)

Have a google for Monty hall dilemma, and you'll find lots of simulators.

daveashton
17th September 2005, 17:49
sorry your maths is wrong and all you can go on is facts not conjecture

2 doors left

1 prize=

50:50

Rob Holmes
17th September 2005, 17:53
Find a good use for a goat (i.e. sell it to a curry house) and at least whatever door you choose you'll be pleased to get something from it.

Top Hat
17th September 2005, 17:57
sorry your maths is wrong and all you can go on is facts not conjecture

2 doors left

1 prize=

50:50

But its your maths that is wrong :)

have a go with some cards and some paper, or write a simulator

Rob Holmes
17th September 2005, 19:18
Both your maths is wrong - theres a prize behind each door. 1 door a car, 1 door a goat.

Rob

Jayne
17th September 2005, 21:52
Find a good use for a goat (i.e. sell it to a curry house) and at least whatever door you choose you'll be pleased to get something from it.

Yuck! Goat curry.

I agree with the above, except the curry bit :D

Jayne

DuaneJackson
17th September 2005, 22:00
Saw this in the book "Curious Incident of the Dog in the night time".

the logic there convinced me it was best to switch. My bro-in-law strongly disagrees though.

Cornish Steve
18th September 2005, 01:24
I'm in the President's Club at Newark airport trying to understand this puzzle. At last, I think I understand.

Let's call the three doors A, B, and C. There's a 1 in 3 chance that the car is behind door A, 1 in 3 for door B, and one in 3 for door C.

The contestant selects door A. There's a 1 in 3 chance that the car is behind this door. There's a 2 in 3 chance, however, that it's behind one of the other doors.

Wanting to help, the game show host will always open a door with a goat. If the car is behind door B, he'll open door C. If the car is behind door C, he'll open door B. If the car is behind door A, it doesn't matter which door he opens.

At this point, there's a 1 in 3 chance that the car is behind door A and a 2 in 3 chance that it's behind door B or door C. However, one of those two doors is open, and there stands a goat! With this new information, there's a 1 in 3 chance that the car is behind door A and a 2 in 3 chance that it's behind the unopened door - B or C.

The contestant should switch!

chris1317
19th September 2005, 19:18
Heres the maths

http://www2.sjsu.edu/faculty/watkins/mhall.htm

Heres a page where you can download a simulator to test it for yourself. I didnt believe it for a start either and thought it was nonsense but understand it now.

http://www.grand-illusions.com/monty.htm

Chris.

MinuWeb
19th September 2005, 20:12
What happens if you already have a car and really want to win a goat ? :D

Top Hat
20th September 2005, 08:57
What happens if you already have a car and really want to win a goat ? :D

If you want the goat, you should stick, as you have a 2 :3 chance of getting it :D