Cue PH brainiacs - logic puzzle!

[gasblaster]

apparently you should always switch and this is the reason:

3 doors:

door 1 = goat. Stick with door 1 = get a goat. Change = get a car
door 2 = goat. Stick with door 2 = get a goat. Change = get a car
door 3 = car. Stick with door 3 = get a car. Change = get a goat

Note that if you chose a goat door (1&2), if you change you get a car because the other goat has already been shown by the host.

So, you stick = 2/3 chance of getting a goat
you change = 2/3 chance of getting a car



>> Edited by gasblaster on Friday 9th December 08:57

[deckster]

JonRB said:

So returning to the original problem, changing your mind and choosing the other door is an independant probability. You have a 50% chance of getting the right door regardless of your previous choice, so it makes little difference whether you stay with your original choice or choose the other as you are either right or wrong - the probability of success is independant of the previous choice.

This is the obvious - and wrong - answer. The subtle point is that your final choice is *not* an independent random event - it is influenced by the fact that the host knows which door the car is behind. The stats quoted above and in the linked solution are quite correct.

I don't remember much from my university stats course, but the one thing that stays with me is that the obvious solution is almost always wrong

[JonRB]

deckster said:

This is the obvious - and wrong - answer.

Oh well. At least I know that the aircraft won't take off and the slinky will keep going, which is obviously far more important.

[minicity]

joust said:

moleamol said: With a less known solution: www.sover.net/~nichael/puzzles/monty/solution.html Ah - the wonders of assuming that what the stats say therefore proves it. You are falling into the trick that if the numbers 10, 12, 27, 33, 48 and 49 come up on the lottery, should you buy a ticket the next weekend with the same numbers on it? Logic says no, status say "yes" (or, to put it in LB terms - "computer says no...." Your "solution" that you quote assumes that you would choose the same thing more than once (that's how the stats work in that example), that's how it proves its better to switch. It is indeed statistically correct, as it assumes the person sticking the prize in the boxes in the first place is a perfect random machine. However, people are not perfect random machines, and so the answer is, for normal human beings, go with the one you feel "lucky" on. J >> Edited by joust on Friday 9th December 00:16

Well there's a good piece of advice - ignore the facts and take a punt at it!


The second choice is not 50:50. It's a choice between the originally selected door (which has only a one in two chance of being correct) and the remaining door from the other two. As the total of the two options must add up to one whole, the odds for going with this option must be two to one on. So you change. You could be wrong but there is no rational reason to stick with the first option unless you think there is something underhand going on.

[jesusbuiltmycar]

gasblaster said:

door 3 = car. Stick with door 3 = get a car. Change = get a goat

Then give the host a slap for tempting you to change your mind.....

[Mr Whippy]

I watched QuizTV last night with a really random "how many chips?" question, with several piles of gaming chips.

Some were obscured, but their value denoted elsewhere on the tops of other piles. Anyway, the answer was 234 or something totally obscure. There were 13 physical chips, 4 different value chips, and the total value of the chips was 370.

How they even begin to say 234 is an answer to the puzzle I have no idea. I laughed as people said things like 230, then when they said 370 or 13 or 4 was astounded they got it wrong! WTF!?


TV puzzles are bollox, they hide the answer so it's down to random guess work. Kinda like answering a question there isn't an answer to until they have enough money to "let" someone win I guess.

Dave

[speedchick]

JonRB said:

deckster said: This is the obvious - and wrong - answer. Oh well. At least I know that the aircraft won't take off and the slinky will keep going, which is obviously far more important.

Life's sorted then

[BliarOut]

speedchick said:

JonRB said: deckster said: This is the obvious - and wrong - answer. Oh well. At least I know that the aircraft won't take off and the slinky will keep going, which is obviously far more important. Life's sorted then

No it's not, I just bumped the aircraft thread with a twist

[speedchick]

Yea I saw!

This question was asked on a science forum, the overwhelming answer was yes, it would take to the air.

It was asked on an aviation forum, everyone said no it wouldn't, except one person that got quite p1ssed at the thickness of everyone that said it wouldn't.

Question has now been asked to aerospace engineers..... lets see what happens there!

[BliarOut]

speedchick said:

Yea I saw! This question was asked on a science forum, the overwhelming answer was yes, it would take to the air. It was asked on an aviation forum, everyone said no it wouldn't, except one person that got quite p1ssed at the thickness of everyone that said it wouldn't. Question has now been asked to aerospace engineers..... lets see what happens there!

We need a url

[speedchick]

BliarOut said:

We need a url

Can provide one to the aviation one....
www.aviationforum.org/forums/showthread.php?s=&threadid=6025

[axeman30]

speedchick said:

Yea I saw! This question was asked on a science forum, the overwhelming answer was yes, it would take to the air. It was asked on an aviation forum, everyone said no it wouldn't, except one person that got quite p1ssed at the thickness of everyone that said it wouldn't. Question has now been asked to aerospace engineers..... lets see what happens there!

Right, there's only one way to settle this, we need to build a test rig.

ps, I still think the key is the weight of the plane relative to the thrust of the jets, in my mind I am thinking of a large commercial airliner, not a missile.

>> Edited by axeman30 on Friday 9th December 09:59

[v8nrg]

WOW, WOW, WOW...... slow down a minute.

Forget this chance and percentage mumbojumbo.

Do you get to keep the goat if you get the right door ?

[bigandclever]

gasblaster said:

You are on a TV game show. The main prize is a car. You have to choose between three doors. Behind one of the doors is the car. Behind the other two doors are goats. You are asked to choose your door. You choose your door, but the host does not open it. The host then opens one of the other doors (that you did not pick) to reveal a goat. The host then gives you one final chance to change your mind about which door you want. Your choice is to stick with your original choice, or to pick the other unopened door. What should you do? And why?

Can I not just take the goat? What if the car is a Nissan Almera Tino?! Or, God forbid, an MR2 with a F40 kit?!?!

[ATG]

deckster said:

JonRB said: So returning to the original problem, changing your mind and choosing the other door is an independant probability. You have a 50% chance of getting the right door regardless of your previous choice, so it makes little difference whether you stay with your original choice or choose the other as you are either right or wrong - the probability of success is independant of the previous choice. This is the obvious - and wrong - answer. The subtle point is that your final choice is *not* an independent random event - it is influenced by the fact that the host knows which door the car is behind. The stats quoted above and in the linked solution are quite correct. I don't remember much from my university stats course, but the one thing that stays with me is that the obvious solution is almost always wrong

If you read the original post it doesn't suggest tha host knows where the car is .. so JonRB is correct. If the host knows where the car is then you're solving a different problem.

(If you check the link to the other website's statement of the puzzle, you'll see they make it clear that the host knows where the prize is.)

>> Edited by ATG on Friday 9th December 10:12

[Jinx]

Its always 50:50 - the cars either behind the door or it isn't.

[minicity]

Jinx said:

Its always 50:50 - the cars either behind the door or it isn't.

Perhaps it is not clear that the host knows that he is getting rid of an option that does not lead to the car.

When making the second choice your are being asked the following question:

"Is it more likely that the car is behind the door that I orginally chose? Or is it more likely that the car is behind one of the two doors that I did not originally chose?" The odds of the first option are 2 to 1 against, and for the second option 2 to 1 on.

[Dakkon]

joust said:

Old old old.... The best option statistically is to change your choice. If you want to know why, type "Monty Hall Problem" into google and read, and read, and read, and read, and read. You'll then realise that it makes sod all difference in practice. If everyone followed the statistical view, then everyone would stick 1,2,3,4,5,6 on their lottery card. Statistics only describe perfect worlds. The world isn't perfect. If you are ever in that position, go with whatever you feel at the time, and pray you are lucky! J

I am always reminded of this when I see Roulette tables and they show the previous numbers, who cares, it makes no odds to what the next number will be.

[blindswelledrat]

v8nrg said:

WOW, WOW, WOW...... slow down a minute. Forget this chance and percentage mumbojumbo. Do you get to keep the goat if you get the right door ?

good point- you can't shag a car can you?

[ATG]

blindswelledrat said:

v8nrg said: WOW, WOW, WOW...... slow down a minute. Forget this chance and percentage mumbojumbo. Do you get to keep the goat if you get the right door ? good point- you can't shag a car can you?

My mate parked his 635CSi backwards in a hedge last week. I'd say both he and it were pretty well ed. Does that count?

>> Edited by ATG on Friday 9th December 11:18