OT: Statistical question

[Roger Mills]

And how exactly does that help you?

[Roger Mills]

Why does that make any difference? You don't know whether your door was the winner - only that the opened door wasn't. It's now reduced to a two horse race, with equal odds on each!

[David.WE.Roberts]

Your strategy is to pick a goat in the first round, and then change in the second round.

Do this, and you win because the other goat is eliminated so you must change to the car.

There are two goats and one car.

So you have two chances of picking a goat against one of picking the car.

So if you follow the strategy of "pick and change" you have two chances out of 3 of winning.

Simples :-)

Although if I try and treat the whole thing as two seperate contests my brain metls.

Cheers

Dave R

P.S. the elimination of the goat is the essential part of this problem, making it goat shit statistics.

[Andy Burns]

No, there are still three doors.

A) The one you originally chose remains closed

B) One that has been opened so is definitely not a winner (it's irrelevant whether the stranger can see this still on stage or not).

C) The other the host has left closed.

chance of 1/3 your door is (unbeknown to you) the winner, it's irrelevant which of the other two the host left closed. But chance of

2/3 your door was a loser, the host deliberately chose to leave the winner closed, so you should switch to take advantage of what you know.

The stranger doesn't know which door the host left closed, so he doesn't have your advantage.

[Martin Brown]

But it does, and it is absolutely critical to the calculation. The point is that you chose your door as one out of three free choices so that your chance of winning was 1/3 at the outset and the host has a 2/3 chance of having the winning door. Since he is obliged to only choose a losing door after his move you should take his position.

IOW there are two choices and the one now belonging to the host is twice as likely to win!

No it isn't. You fail to take into account the additional information provided by the host. His *remaining* door is twice as likely to be a winner as the one that you originally chose.

It is actually a very good test of understanding. It should be used as a comprehension filter for anyone asked to serve on complex fraud cases.

[Andy Burns]

You *also* know the host deliberately chose which door to leave closed, two times out of three that is significant.

The host *selected* one of the doors to remain, you chose one of the doors to remain, the stranger doesn't know which is which.

I suggest you stop trying to answer something that wasn't the question, once you see the answer to the original question, you'll see why it's

*not* 50:50 for you to change knowing what you know, but it *is* 50:50 for the stranger to make a selection starting halfway through the game, knowing nothing.

[Clive George]

(on Monty Hall)

I had a discussion with a colleague who was of the same opinion as you. We made a small bet, with the proceeds going to an appropriate charity. He ran a simulation (I think in Excel), and paid up :-)

Try doing the same - it's not too hard. For some people, once they've seen the answer demonstrated, they can then free their mind from the idea that it must be wrong, and get on with understanding why it isn't.

No, it's not a new contest. What happened previously has relevance.

Nope, you're right about coins. One coin toss doesn't affect the next.

Unlike science, maths doesn't change "truth" with time. That's why maths has proofs but science has theories.

[Andy Burns]

And that's crucial, because if the stranger knew which one you originally picked, by elimination he'd know which one the host deliberately picked, and therefore to pick the one you didn't pick.

True, but if he knew which one the host had left, he'd know that one had HIGHER chances of being the winner, nobody is saying there's any CERTAINTY of winning by picking the host's door, only that the odds are better.

Only if you disregard information the host has given to you.

[Andy Burns]

That's the same as saying your strategy is to win.

If can you can be sure of picking a goat, why not just pick the car instead and refuse to change?

[Clive George]

If they both choose a non-prize door, which will happen 1/3 of the time, how does the host open up another one?

[Roger Mills]

That's the best explanation I've seen so far in support of the "official" correct answer.

But I'm afraid that I still don't buy it!

The host knows which door is which and *always* eliminates a goat door. Once he's done that, there are two doors left - one with a car and one with a goat. The host knows which has the car (in order to avoid opening it) but *you* don't. The elimination of a goat door has raised the odds of each of the other two doors having the car to 50% - *not* to 67%/33% against your first choice, as is being claimed.

You are in the same position as a substitute contestant coming in after the first door has been opened, as mooted by someone else.

[Andrew May]

But which door is the 'other' door is itself dependent on which door you picked first.

[Malcolm G]

It is now a two horse race, but one of the horses is the favourite.

[Clive George]

Correct.

You know which door you picked originally, he doesn't.

[Roger Mills]

Agreed

As it always will, because he knows where the car is

No. There are now only two possible locations for the car. You first choice now has a 1:2 chance

Agreed that the odds are now 1:2. That was achieved by eliminating a door.

No. they both have a 1:2 chance. You original choice was irrelevant.

[Andy Burns]

Just because there are two doors remaining does *not* mean they're equal, because two times out of three the host discarded a loser, knowing the your door was also a loser, in which case his remaining door is definitely a winner.

The stranger doesn't know which door was the host's and which was yours, so for him it's 50:50. But you *do* know which door was yours, so the odds are better if you switch to the host's door ...

You need to understand this is about knowledge as well as probability!

[Roger Mills]

Good analogy! Of course it doesn't! It now becomes 1:1 - just as your original door goes from 1:3 to 1:2 when one door is eliminated.

[Scion]

Roger Mills put finger to keyboard:

No, it hasn't.

IF you choose to switch, then if you originally picked the car, you'll lose; if you originally picked a goat, you'll win.

Statistically you'll originally pick a goat 2/3 of the time.

Nope. If you pick a goat the first time, you are forcing the host to open the only other door with a goat. You're in a different position to a substitute contestant because you know which door you chose originally - with a 1/3 chance of a car behind it.

[Roger Mills]

is the winner, thus the odds are now 0.5 for each door. Thus it makes no difference whether one switches doors or not. Statistical probability figures are nothing more than reflections of the data they're based on, and opening one door changes the available data. Missing that point is where so many are going wrong.

Well said!

[Roger Mills]

Sorry and all that, but it's made for a lively discussion!