Impeccable logic - part 2

[pna]

The Monty Hall scenario is quite simple, and the odds can easily be followed......

1) Their are 3 doors, your chance of picking the correct door with your first choice is...1 in 3 Therefore you will only win the car 33 times out of 99 attempts, if you stick with your original choice
( if you think carefully about the above sentence, its all you need to know )

2) So what is the advantage in switching, well we already know that in 2 out of 3 Attempts, the car will be behind the other 2 doors, that you did not choose.
Monty now removes the goat for you........
Thus the remaining door contains the car 66 times out of 99.

Its a must switch, thanks to Monty removing the goat

Point 1 is correct. At that time there are 3 boxes.

Point 2 ('thus the remaining door contains the car 66 times out of 99') is not correct. This is because each 'probability event' is separate, and you have to calculate the odds separately each time. Just because you have tossed a coin 100 times previously does not influence the next throw, and so it is with this scenario.
You might think you have learnt something from seeing the goat, but you havn't, because you ALWAYS see a goat, no matter whether you picked the car or the other goat on your first choice. There are only 2 boxes left - its a straight 50:50. Wondering if you should swap makes for good TV though...

 

[megawatt]

Didn't I say exactly the same in the previous post PNA

 

[blondini]

The idea that swapping makes no difference and the odds of the decision are 50:50 may seem absolutely blindingly obvious, but the correct solution is counter-intuitive and I believe Softus is right.

 

[joe-90]

I must admit that I've lost the plot in this thread a bit, but it would seem to me that:

If Monty ALWAYS shows a goat, then in reality he is cancelling that door.

Let's suppose that Monty ( who knows which door the goats are behind) says " I'm cancelling this door" and paints it over with a big roller. Then he says "We only have two doors. One has a goat behind it and one has a car behind it. Take your pick. Now being as there are only two possible permutations

Door one = Goat, Door two = car

or

Door one = car, Door two = goat.

How can it be anything other than 50-50?

Let's start with ten doors. Monty knows the car is behind door two so he gets his paint roller out and paints over doors three to ten. They have been cancelled and removed from the show.

He then asks the contestant to choose from doors one and two. One has a goat - and one has a car. That equals 50-50. How can it be otherwise?

 

[megawatt]

Blondini wrote:

The idea that swapping makes no difference and the odds of the decision are 50:50 may seem absolutely blindingly obvious, but the correct solution is counter-intuitive and I believe Softus is right.

Then you are as deluded as the Soft one ... Presumably, you also believe in Peter Pan and the Tooth Fairy

I'm not interested in the MH scenario as the original OP's question was about DoND (and I can't remember how we ever started banging on about MH ).

The earlier posts from me and PNA outline the probability for DoND ... Final two boxes, choice is 50:50.

MW

 

[Kes]

Whenever I see multiple explanations all coming to one conclusion, and I come to another, then I concede that I might need to look again. (It's a good thing that Galelio, and countless others who have bucked the trend and forged our view of the world, didn't follow my example, but then I'm not Galelio.)

If you Google 'Monty Hall' you will get over 250,000 hits. Those I've looked at, including Wikipedia, and the biography of mathemetician Paul Erdos which discusses this, come to the conclusion that the MH odds are 1/3 if you stay, 2/3 if you switch.

There's a good site at http://www.grand-illusions.com/monty.htm which discusses the example that Joe states, only with 100 doors. There's also a link to a simulator which you can run for 1000 passes. Go for it!

PS Did Monty really use two goats?

 

[jackpot]

Exactly.

To summarise. Montys theory. 2 in 3 chance. Better swapping.
Deal or no deal. 1 in 2 chance (50:50). Makes no difference.

Montys theory is totally irrelevant to the ops scenario. Hence its a total different experiment

 

[megawatt]

You've hit the jackpot ... Jackpot

 

[blondini]

Ok, let's play the game.

I'm convinced that swapping works in my favour so I'll always swap and will take one of three choices.

Door 1 then swap.
Door 2 then swap.
Door 3 then swap.

Whichever door the car is behind, two of the three choices will win the car.


Do the opposite and stick, you will have to take one of three alternative choices.

Door 1 and stick.
Door 2 and stick.
Door 3 and stick.

Whichever door the car is behind, two of the three choices will lose.

 

[blondini]

Assuming that the MH swap gives 2/3 odds.

If having picked 'your' door you then chose to open one of Monty's two doors instead of Monty doing it, would it make a difference to the subsequent 2/3 odds on swapping to win if that door had revealed a goat? I think not.

So what is the difference between this scenario and DOND? It looks similar to me. You have your one door/box chosen earlier, and Monty/Noel has two. Only one of the three contains the big prize. You choose to open one of Montys/Noels two doors/boxes. If it doesn't contain the big prize you then stick or swap,

 

[joe-90]

If we change the terminology of what Monty is saying then that shouldn't have any bearing on the outcome - it's just words, after all.

In the original version Monty says "Choose a door".
Contestant chooses door 1,2 or 3. (let's say he chose door 2)
Monty then goes over to a door that has a goat behind it and says, "Look it's a goat". (Let's say it was door 3)
Then he goes back to the contestant and says "Would you like to swap door 2 for door 1?"

.....................................

So is there any benefit in swapping or not?

Let's run the same game but Monty will use different words.

Monty says to the contestant, "Look at that, there are three doors. One has a car behind it whilst the other two have goats. You get to keep whatever is behind the door you choose."

As the contestant is just about to choose, Monty says "Stop! I'm in a good mood so I'll make it easier for you." Monty then walks across and opens a door (door 3) to reveal a goat, and says "There you go (contestant), there are just two doors now. One has a car and one has a goat. Which door would you like ? One or two?

...............................................

It's the same game with the same moves - just Monty saying different words.

Looks like 50-50 to me.

 

[blondini]

If we change the terminology of what Monty is saying then that shouldn't have any bearing on the outcome - it's just words, after all.

In the original version Monty says "Choose a door".
Contestant chooses door 1,2 or 3. (let's say he chose door 2)
Monty then goes over to a door that has a goat behind it and says, "Look it's a goat". (Let's say it was door 3)
Then he goes back to the contestant and says "Would you like to swap door 2 for door 1?"

.....................................

So is there any benefit in swapping or not?

Let's run the same game but Monty will use different words.

Monty says to the contestant, "Look at that, there are three doors. One has a car behind it whilst the other two have goats. You get to keep whatever is behind the door you choose."

As the contestant is just about to choose, Monty says "Stop! I'm in a good mood so I'll make it easier for you." Monty then walks across and opens a door (door 3) to reveal a goat, and says "There you go (contestant), there are just two doors now. One has a car and one has a goat. Which door would you like ? One or two?

...............................................

It's the same game with the same moves - just Monty saying different words.

Looks like 50-50 to me.

But it's not the same order of moves Joe and it's the order of moves that are important not the words.

In your 'same as' version Monty opens and discards one dud door first, and leaves a straight genuine 50:50 choice. He can only choose between the two duds though.

In the real version the contestant chooses one door first and Monty can't touch it. Monty then opens one of two remaining doors, but he only has a choice if the contestants first door conceals the car. It's 2/3 odds that one of Montys doors conceals the car and if it does (2/3 remember) then he has to reveal his goat.

 

[joe-90]

OK I can see what you are saying but (if I understand it right) Monty ALWAYS uncovers a goat. A goat is a goat, a known value, but it doesn't matter which one it is. By exposing a goat he removes the goat from the game as if it had never been there in the first instance. He plays the game with 'the car' and (either) 'goat A or Goat B' but it really doesn't matter which one it is. At the point where the 'swap' takes place it is as if the other goat had never existed.

 

[Trazor]

Point 1 is correct. At that time there are 3 boxes.

Point 2 ('thus the remaining door contains the car 66 times out of 99') is not correct. This is because each 'probability event' is separate, and you have to calculate the odds separately each time. Just because you have tossed a coin 100 times previously does not influence the next throw, and so it is with this scenario.
You might think you have learnt something from seeing the goat, but you havn't, because you ALWAYS see a goat, no matter whether you picked the car or the other goat on your first choice. There are only 2 boxes left - its a straight 50:50. Wondering if you should swap makes for good TV though...

If you believe point 1 is correct, then you can not believe what you followed up with.

joe-90..... Quite interesting how you CHANGED the way the game is played, in order to make it 50-50


Softus....... the whole essence of the Monty Hall game, is that the host always reveals the goat. Thus 2 doors, each with a 1 in 3 chance becomes 1 door with a 2 in 3 chance.
Monty,s knowledge is irrelevant during the first pick, but his knowledge is the only factor which turns the game in your favour, when given the option to change.

 

[blondini]

OK I can see what you are saying but (if I understand it right) Monty ALWAYS uncovers a goat. A goat is a goat, a known value, but it doesn't matter which one it is. By exposing a goat he removes the goat from the game as if it had never been there in the first instance. He plays the game with 'the car' and (either) 'goat A or Goat B' but it really doesn't matter which one it is. At the point where the 'swap' takes place it is as if the other goat had never existed.

I'm not trying to get your goat Joe, honest, but Montys exposed goat was there in the first place and that makes a difference because there were two hidden goats that the contestant could have chosen. Thus the odds of the contestant picking a hidden goat first try are 2/3. The contestant is therefore more likely than not to be holding a goat. So when Monty reveals another goat, Montys other door is more likely than not to be hiding the car.