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After talking about the film '21' in the pub last night, much argument ensued about the 'Monty Hall Problem'. Not sure if its been done before on here but I thought it would be fun for you all to have a go at this problem. Two things though, if you've already done it then don't try to be a hero by posting your prior knowledge answer here and don't go straight to Google/Wikipedia - you're only cheating yourself.
Here it is:
[b]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 No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?[/b]
Yes
I already have a car, but I don't have a goat. Does this change things?
I rememvber watching some horizon thing about this with Alan Davies a while back, but he did it with toy cars and cups....
....i think the outcome was that probability says that when you get a choice like that, if you change your mind you win more often than if you don't
I could set my calendar with the regularity that this thread comes up.
What are the chances of that, though?
Yes.
No mmmmm curried goat 🙂
I made a fantastic curry goat last week. The spices I used were from a tin of Tesco Ras El Hanout rub which I ground up a bit more. It was flipping lovely.
And yes - in the 3-box game, if you do it 100 times you will win more if you switch. But for a single instance there is no difference. The car stays where it is.
And yes - in the 3-box game, if you do it 100 times you will win more if you switch. But for a single instance there is no difference. The car stays where it is.
??? So your probability of winning the car is increased, but only if you play it more than once? That sounds a little dubious 😕
I don't think it's a particular spoiler to give the answer, because a percentage of people won't believe you anyway. We discussed this at length here a few weeks back.
You double your chances (from one in three to two in three) by changing your mind when offered the choice.
Fox everyone by picking again but picking the same door as the first time.
Goats are nice anyhow.
Fox everyone with a goat? *blinks*
Anyway, about this plane on a conveyor belt....
Which door is the fox behind?
switching is clearly better but Monty - who knows where the car is - might be a bit of a ****t and be trying to trick you into losing the car, so he can make more money from the show and sleep with impressionable women, so it's not quite so straightforward...
Who is Monty Hall? Why does he have a car full of goats?
well whenever i've seen deal or no deal (ahem) if you change your mind you always loose more then if you don't.
Doesnt it depend if the host knows where the car is? If it does then you should swap. If he's just randomly picking a door then not.
And yes - in the 3-box game, if you do it 100 times you will win more if you switch. But for a single instance there is no difference. The car stays where it is.
Wrong.
Noel has the Fox, he keeps it in a box.
He does not like green eggs and ham....
when you first pick you have a 1/3 chance of choosing door with car.
the other two doors have 2/3 chance.
i think if the host now picks a door and opens it to reveal a goat then the door that is left now has a 2/3 chance of the car being there.
so you are better off swapping.
Does this apply to switching boxes at the end of [b]Deal or no Deal[/b] then too? *scratches head* 😯
This hurt my head the last time it came up here and I'm not getting involved again.
If 1 & 2 stay as they are then surely it wouldnt make any difference.
This is messing with my head.....
surely, regardless of how you got there, you now have a choice of two doors, behind one of which is a car.
The chance of it being behind either door is equal.
So how does swapping your choice, from one 50% choice to another, increase your odds of winning?
Like I say, it's playing with my mind...
This is the explanation I posted on the [url= http://www.singletrackworld.com/forum/topic/the-monty-hall-problem-or-2-goats-and-a-car ]original thread[/url].
On the first choice you have a one in three chance of making a correct choice, and a two in three chance of goating out.
When the host asks if you want to change your mind, what's essentially happening is that you're being offered to swap the one door you have for [i]both of the other doors.[/i]
You both know one of the remaining two doors contains a goat, by opening a door (with his insider knowledge) the host simply confirms something you knew anyway. Changing your mind nets you both doors, the opening of one of them is a bit of very effective misdirection.
Does this apply to switching boxes at the end of Deal or no Deal then too? *scratches head*
Not really as the game is played differently and in some versions of the show the "banker" knows the value of each box and in others he/she doesn't. There probably is some sort of mathematical model but it will be horrifically complicated.
Incidently you should probably deal at anything over £1000 as there is a only a 50% chance that your selected box will contain more than this amount, despite the fact that the mean prize is £26000ish
This principil only works if the host knows where the car is and takes away one of the goat options.
Basiclly you have a 66% chance of picking a goat in the first place. The host has to show you a goat so the remaining door has to be a car.
This only doesn't work if you picked the car in the first instance, but as you only had a 33% chance of doing that then the numbers are in your favour.
Rustler, it makes a difference as the host knows and is showing you a goat. That is not random it is calulated.
The OP states that the host knows where the car is.
Which, TBH, is pretty much a prerequisite, otherwise he could open door number three and accidentally find that the car is there, buggering the entire premise.
Cougar. well put, good explanation
Thank you (-:
+1 TREKEX8
I think?
I understand the maths/theory behind this and i understand Cougars explanation and it makes sense. I also believe it to be correct but my head just refuses to accept it!
OK, say there were 1000 doors, 999 had goats behind and 1 had the car.
You choose one of the doors - so a 1 in 1000 chance of being a car. Host then opens 998 doors with goats behind leaving one unopened door.
Would you change doors now ? 🙂
I also believe it to be correct but my head just refuses to accept it!
... which is the entire point of the puzzle. It's to show that our intuition isn't to be trusted.
There's a number of other explanations on the earlier thread I linked to ^^ there, if it helps.
all the pies, same principal but even more important to switch
Your 1st choice had a 0.1% chance of being correct
swapping gives you a 99.9% of winning.
As Cougar said your swapping your 1 door for 999 doors but knowing which of those 999 doors are goats.
Sorry to the people have seen this on here before - I come on here a bit but it was the first time I'd heard about it last night in the pub
surely, regardless of how you got there, you now have a choice of two doors, behind one of which is a car.
Yes.
The chance of it being behind either door is equal.
No.
Just because there are two possbible outcomes doesn't mean that they are equally likely. If you were to by a lottery ticket there are two possible outcomes; you will either win the jack pot, or you won't. The odds of you winning the jackpot aren't 50:50.
So with the 1 in 1000 example, after he has showed you the goats are behind the other 998 doors so youre left with youre door and 1 onther door the odds of the car being behind the other door and not your door are 99.9% despite the fact you know you have 2 options, 1 being a goat and 1 being a car?
poop example gonefishin.... there are millions of combinations on the lottery to simplfy it to win or lose isn't really the same
Crispo your basing it on the fact of just looking at the two doors.
Think of it this way, the host has to show you 998 goats. Before he show's you them either he has the car in the 999 doors he has or you do in your 1 door. What is more likely?
all the pies, same principal but even more important to switch
No need to tell me 🙂 I understand the need to swap, my extrapolation example was an attempt to convince the errr, unconvinced 😉
If anyone doesn't believe you should switch, I've a tenner here and three cups. Bring your own tenner and we'll play.
One tenner three cups, it's the next Internet sensation!
gravity sucks - I don't know, which is more likely??
Am I missing the point here? Surely there are now 2 doors, and there is no more likelihood that the car is behind one or the other? So, how do you increase your chances of winning by switching to the other door with equal probability of having a car behind?
Go from 2:40
Am I missing the point here?
Yes?
Surely there are now 2 doors, and there is no more likelihood that the car is behind one or the other? So, how do you increase your chances of winning by switching to the other door with equal probability of having a car behind?
There's a 2/3 chance that there will be a car behind one of the two doors that you didn't choose first. Effectively, you're being given the chance to choose [i]both[/i] those doors instead of your first chance.
So, you're twice as likely to win the car if you switch.
Am I missing the point here? Surely there are now 2 doors, and there is no more likelihood that the car is behind one or the other?
Well if that messes with your head then try [url= http://www.singletrackworld.com/forum/topic/the-boy-girl-puzzle ]this old thread for a probability reality check[/url] 😀
Am I missing the point here? Surely there are now 2 doors, and there is no more likelihood that the car is behind one or the other?
If the premise was simply "one of these doors has a car behind it, which one?" then you'd be right. But that's not what's been set up.
You're presented with two doors. However, the odds of winning have been stacked by previous events; ie, the two doors are not equal.
At the start of the game, you pick a door and have a 33% (rounded) chance of being a winner, yes? Think about that. You can now do what you like with the other doors, add more, remove one, paint them pink, set them on fire, write "here be goats" on them; it does not, [i]can not[/i] change the likelyhood that your door contains a car or a goat - it will always be 33%, that was the choice you originally made.
In the second round, you're given another option. You can keep your existing choice (33% chance of being right, remember) or you can trade that choice for another door which, now that we've removed an obvious wrong choice, represents all other odds other than the one you already hold. 100% - 33% = 67%.
Thanks Cougar, however.....at the beginning, the chance of winning was1/3. If the host opened two doors, to reveal goats, would the odds still remain at 1/3 that opening the remaining door would reveal a car?
Simplest to illustrate by going through all the options.
Assume you pick the one in bold in each of the 3 options below. Whatever you choose, Monty opens a goat* door (the one crossed out). So, sticking means you win 1/3, switching means you win 2/3.
G G [b]C[/b] becomes [s]G[/s] G [b]C[/b]. Stick = Win. Switch = Lose.
G [b]G[/b] C becomes [s]G[/s] [b]G[/b] C. Stick = Lose. Switch = Win.
[b]G[/b] G C becomes [b]G[/b] [s]G[/s] C. Stick = Lose. Switch = Win.
As far as i know this should work with sheep, cows or anything else. I'm surprised questions weren't asked about Monty's interest in goats.
this was demonstrated on BBC Horizon with toy cars and toy goats. One player always changed the other always stuck with the original choice. After 100 goes the pile of cars for the always changes was much bigger than the always stays in line with the probability theory...