Game Show

[HeyWayne]

What if it's a revolving door?

[OllieMartin]

This is fun.

Yes, but it alters the odds from 1/3:1/3:1/3 to 1/2:1/2

I agree before a door is shown to you, there is a 1 in 3 chance you have picked the right door. But I still maintain that when one has been eliminated, that elimination does not make your choice anything other than 1 in 2 of the 2 remaining doors. You're left with 2 doors and it's an equal chance the prize is behind either, therefore that must be 50:50.

I think the best arguement is that:
Stephen Fry said you should swap doors.
And the 11th commandment is "Thou shalt not question Stephen Fry".

[Caro]

What if it's a revolving door?

Then I best watch it doesn't slap me on the ass as I leave

[HeyWayne]

Then I best watch it doesn't slap me on the ass as I leave

Now what are the odds of that happening!?

[HeyWayne]

By the way, are we talking probability here, or possibility?

[Sheepish]

There's a detailed explanation of this on Wikipedia
Monty Hall problem - Wikipedia, the free encyclopedia
One point to be clear about - the host doesn't open a door at random.

So there was a 1/3 chance you picked right, in which case changing looses.
There was a 2/3 chance you picked wrong - in which case the host always opens the other loosing door and the other unpicked door contains the prize, so changing always wins.

[FionaH]

As in Numb3rs the cop show? I've never seen it, is it good?

It's ok if there is nowt else on, and there usually isn't

[SueA]

I can see the reasoning behind both explanations but tend to lean towards Caro's 50/50, although if it ever happens to me I'll make sure I swap doors just in case!
It's a bit like the National Lottery isn't it, you'd think if your numbers came up one week there'd be absolutely no chance of the same ones coming up again the next week (& they wouldn't!) but the odds are exactly the same aren't they because each draw is new & independent & the numbers all have the same probability each time?

[OllieMartin]

I think the only way to prove this is to run a practical experiment.
100 instances of sticking with the same door, 100 instances of switching doors.
I'd predict that sticking with the same door would return 33% success and switching would return 66% success, however if it returned 50% success then our theory would be proven wrong.

[HeyWayne]

I'd go with whichever one I am fated to choose.

[SueA]

I think the only way to prove this is to run a practical experiment.
100 instances of sticking with the same door, 100 instances of switching doors.
I'd predict that sticking with the same door would return 33% success and switching would return 66% success, however if it returned 50% success then our theory would be proven wrong.

O.K., now who's got 3 doors, a prize, a patient friend to hide the prize ( & open the doors) & lots of time on their hands?

[HeyWayne]

Question is, which door would Shakey choose?

[Pete C]

All you are doing is changing the odds. That the host can eliminate one incorrect option does not affect this. All that happens is that the odds change with the circumstances. With two doors the chances are Even, which is better than they were, but they're still Even odds, so there's no answer.

To think that by the odds being reduced, the probability of the object being somewhere changes is nonsensical. Whilst the fall in odds might indicate to someone that doesn't know a door has been opened that the other door is the best option, the reality is that neither door has an advantage.

Simplest example is this: whenever I go to the dogs, if I'm up I always go 'Sh*t or Bust' on the last race. I also follow a time based system. I was at a small track and a friend and I were about £200 up. In the last race, the time-based system indicated that a rat of a dog would win. We went for it. Because of that (two drunk Londoners in the Midlands putting a few hundred an a hopeless dog at a penny-hapenny track), the odds fell from 6-1 to 2-1 joint favourite. Now, usually that symbolises a significant punt from someone in the know. Because it was a rainy Tuesday night at a track were most bets are a few pounds, it affected the book. In betting shops across the country, people reacted and in seconds the dog was an odds on favourite. Statistically, the fall from odds of 6-1 to 1-3 in less than a minute indicates that the dog's chances of winning are greatly increased. The reality, however, was that the dog was still clowns' shoes, and it came last.

Odds aside, the reality is that the object could equally be behind either door.

[kernowyon]

Question is, which door would Shakey choose?

yeah thanks for that! any guesses what I'm singing now?

[OllieMartin]

After the great mathematical minds of our time have worked on this problem and have universally agreed that switching doors is the best solution, i'm inclined to agree with them.
That said, it goes against one of my lifetime philosophies:
"It makes no sense to agree with the majority, by definition there are already enough people doing that"!