OT: Statistical question

[Roger Mills]

The owner of a genealogical website posed a statistical problem a little while ago, and has now given a solution with which I don't agree. What are the views of the esteemed mathematical gurus who frequent this NG?

The question was this:

"Imagine that you're a contestant on a TV game show and have to pick one of three doors, in the knowledge that only one of them conceals a prize. When you've made your selection the host (who knows where the prize is hidden) opens one of the other two doors to show that there's nothing behind it - and then gives you the option of staying with your original choice, or switching to the remaining closed door.

What should you do, and why?"

What do *you* think, and why? I'll tell you the answer given, later!

[Tim+]

The correct answer is that you should change your choice. I must admit the stats boggle my head a bit but it's to do with the remaining doors not having the same odds. Originally you had a one in three chance of being right. Because the game show host *doesn't* choose randomly, the remaining doors have a 1/3 and 2/3rds odds.

Tim

[Andy Burns]

Your initial pick had a 1/3 chance of being correct, in which case the host can show you either empty door, switching guarantees you loose.

Your initial pick had a 2/3 chance of being an empty door, the host must show you the *other* empty door, switching guarantees you win.

So pick a door, then always switch as that has 2/3 chance of winning, it won't stock you kicking yourself if you switch and lose though.

[Huge]

This is a well known probability puzzle;

Bu, naq gur nafjre vf lbh fubhyq nyjnlf fjvgpu.

[Dave Liquorice]

Switch.

Some ones been watch James May's Man Lab. They did this the other week except with beer cans, one of the three was primed to fizz up when opened.

Can't remember the maths but it made sense at the time. It's to do with the cumulative odds, you start at 1 from 3 but the next choice is 1 from

1. It hinges on the fact the host always removes a safe can.

[Brian Gaff]

Well, if you believe that the show host is always going to show the other empty one, then surely its not worth changing horses only if that is true. If he has on occasion shown the prize door and said you lost then stick where you are.

Brian

[newshound]

+1.

Very nicely explained in Mark Haddon's "Curious incident of the dog in the night-time" (which is well worth reading anyway)

[David.WE.Roberts]

I have never really been able to fully accept this.

Especially with past performance being no guarantee of future performance.

You take part in a contest where you have a 1/3 chance of picking the winning door.

You win that contest - and then you are in a new contest where you have a

50:50 chance of picking the winning door.

I fail to see where you have an advantage over someone who replaces you in the contest, not knowing which door you selected, and then selects between the two doors.

Then again, if you toss a coin and call 'heads' and win then I suppose that there is a little bit of statistical pressure to call 'tails' next time (and more pressure each time the subsequent toss is 'heads' not 'tails') but this doesn't seem to weight the individual contest - just gives a weighting to a choice as part of a series.

Remember that the coin has no memory.

So possibly if you treat this as a very short series of coin tosses then 'statistically' if you have called 'heads' then your next call should be 'tails'.

However maths (not arithmetic) is just a specialised language for creating and perpetrating abstract arguments about esoteric things.

A mathematical proof does not mean that something is true - just that it conforms to the current view of mathematical truth.

You can tell that I'm not a mathematician!!

Cheers

Dave R

[Andy Burns]

At that point *you* don't know whether you've won or not, the host does.

[Malcolm G]

I am loathed to get into a debate on the monty hall problem but I think the problem with this argument is simply how do you know it is a 50:50 chance of picking the winning door?

[GB]

I always have trouble explaining this, but the simplest and most convincing explanation (I could find) is as follows:

Say the game host offered you the chance to give up your one can and open *both* the others, you would obviously take that chance - provided he always offered you that chance! You'd obviously be doubling your chances of winning.

Now, ask yourself the question whether it makes any difference whether you open both cans or the game host opens one for you and you open the other?

By the game host opening one of the cans and then offering you the other, you are still effectively getting the benefit of both cans.

HTH

[GB]

I've just checked Wikipedia, and that's the Adams and Devlin solution (although found by GB independently!)

[Huge]

Wrong.

Read the Wikipedia article.

[Roger Mills]

Indeed - and that lines up with the answer given on the genealogical website. But I still struggle to agree with it!

My take is this: When the first choice is made the probability is 1/3 that you will choose the correct door and 2/3 that you won't.

*However* once the host has opened a non-winning door, the game has changed - and you are starting from scratch. You're now in the same situation as you would be in if there had only been two doors to start with.

You know that the prize is behind one of the remaining doors, but you don't know which one. It may be the one you've already chosen or it may be the other one. Each is equally likely - so the probability is now 1/2 whichever way you jump. If you switch, you've got just as much chance of moving from the right to the wrong answer as the other way round.

Remember that the host will open a non-winning door regardless of whether or not your original choice was correct.

I can't for the life of me see how it can be claimed that the 1/3 chance which door opened by the host originally had will automatically be transferred to the other door which you *didn't* choose rather than the one which you *did*. Eliminating one door elevates the probability of

*each* of the other two doors to 1/2 - *not* to 1/3 vs 2/3.

Where am I going wrong?

[Dave Liquorice]

Because the host *always* opens a non-winning door. It's that fact that most people overlook. Curiously when I watched the James May prog I picked up on that straight away as crucial to what one should do. Didn't bother thinking it through to the conclusion that switching is the best option though. B-)

[Roger Mills]

That's my view too!

I have read it - and it makes no sense!

[Andy Burns]

OK so far.

It's not from scratch, it involves the host telling you additional information about the doors.

No, you're not.

True

At the point you initially choose, you had 1/3 chance. Together the two other doors had 2/3, but after the host eliminates a door *that* door is known to be 0/3, therefore the door he didn't open is 2/3 by itself, your door remains at 1/3

[Davey]

I agree, and I have read the Wiki page. Following on your question, if there were two contestants instead of one, and each chose a different door, then the host would still open up the other non-prize door. So does this switching theory mean that both contestants must each change their minds? But only one of them will win, and their odds are surely equal at this point. I, also, am confused. But I am not a probability mathematician.

[polygonum]

Not sure you are going wrong.

Where I keep coming across statistical issues which seem utterly mad is medicine.

We see that 1 in 200 people (or thereabouts - let us not allow any facts to get in the way :-) ) have haemochromatosis. So when you see a doctor he assumes that it is fairly unlikely that you have it. But that is purely a matter of perception - you were either born with it - or not. The selection took place way back at conception - or before. The box was already opened but no-one looked inside!

The individual's chances of having haemochromatosis are way out of line with 1:200 - they are either much higher or much lower, depending on their heredity.

The doctor is looking a a population distribution and assuming an even spread.

And all of that is probably confused, wrong and adds little to the discussion.

[dennis]

Statistics tends to be different from intuition.

The show host does, and that's why the odds change. If he didn't know the answer might be different.

Statistically you can't tell what the next throw will be.

However you can say that there will be a 50% chance it will be heads (well very close, I have flipped a coin and had it land on its edge once). You can also give the chance of the next 3 being heads, etc.