h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Well, yeah, Stone Aart, I DO imagine that there's a 50/50 chance in that case. Why not?

Reason: either:

1. I picked the right door the first time.
2. I picked the WRONG door the first time, in which case the presenter is required to leave the right door out of his selection.

But I don't know which of those possibilities is true. 50/50, right?

That's wxactly what I think, too, Dmitri.

Okay now imagine there are 10,000 guns. 9999 are loaded with bullets and one is a blank. You are held by a terrorist.

The terrorists asks you to select one gun. You select a gun. His accomplice then puts the gun to your head.

The terrorist tells you that he knows which gun doesn't contain a bullet. He looks at the 9999 remaining unselected guns & says he
will choose out of these 9998 guns containing bullets. He then proceed to select 9998 of these guns and fires them into the wall.
A bullet crashes into the wall from each of the guns. Out of the 9999 guns that he had remaining 9998 guns have been fired and these
contained bullets.

He then takes the remaining gun and points it to your head.

He says you can stick with your original choice and his accomplice will fire the gun at your head ... or you can take the chance with the
gun that the terrorist is holding.

By your above reasoning you are saying you might as well stick with your original choice because there is a 50 - 50 choice that your
original selection will be empty.

.... 50 - 50 chance ....

Exactly.

Anyway that is as far as I can take it in the absence of a practical demonstration involving 10,000 guns and 9999 bullets (with the test repeated many times) - so I take my bow from this thread.

Yeah, SA. It's a math problem.

Trying to intimidate us with your terrorist story willnot change our minds...

In the good old days maths teachers were allowed to intimated their students with a heavy stick across the knuckles - but alas no more

Anyway, this is not a maths problems - it's a real life problem with 10,000 guns 9999 bullets and a set of rules. Some can use their intuition to decide what to do while others can resort to "maths" to work it out. You saying it is only a 50 - 50 chance suggests you are only considering it as a "maths problem". Anyway I apologise for resorting to terrorism

intimidate (not intimated) - [wow I average one misspelling / mis word per comment].

I still say the statistical probabiity is not predictive in this form of Russian roulette.

Would I feel more sanguine changing my choice of guns? I would not.

Now, if I had 10,000 lives, sort of like a horde of cats, and I wanted to better my chances, then changing would make sense, non?

OK the only thing to do is to show how this works in practice. Let's use Stone Aart's example of 1000 doors and 1 Lamborghini.

You make your choice. Free choice.

THEN the host tells you:

You can keep your door OR you can switch to ALL 999 other doors.

What do you do - stay with your first choice, or take ALL 999 other doors? What is likeliest: that the car is behind the door you chose first, or that it is behind one of the other doors?

Another example. I have in my hand a six-sided die and I'll stake my life on it that it's a fair die. I offer you a chance: You an bet on me throwing a 6 OR you can bet on me throwing any number OTHER than a 6. In either case you win a thousand dollars. What do you choose?

But that's not at all the same. In your first example it was one door or one other door. That's a huge difference to one door or 999 other doors.
Of course there is an equal probability of the car being in any of the doors, so it could be in my own door just as well as in any other. But with 999 doors you have many more *chances* than with one door.

Or, why people buy more than one raffle ticket? Good point, Tav.

Hi again Tav and Dmitri (and anyone else still hanging around)! Now what I want to show is that it comes to exactly the same thing.

Let us still stay with the 1000 doors one. Let me be the game show host!

So exactly what happens is this:

1. The car is hidden behind any one of the doors, the goats behind all the others.

2. You choose a door - your first choice which is free.

3. I note the door you picked. *Knowing* which door the car is behind, I now open *all* the other doors, showing where the goats are - except for one door. This door might then either have a car behind it, but if your first choice hit the car, then this door will also have a goat behind it, but it's a 'decoy' and I'm not telling you! You have to decide whether to switch or not.

Can you visualise this game? If you have difficulty - how about we actually play a few games? There's an online random number generator. What I can do is, as your host, first let the number generator generate a random number between 1 and 1000. That will be the door with the car; I won't tell you which it is. Then *you* choose a number. Then I as your host, tell you which doors I've opened, showing goats, and which door I'm keeping closed - and you can decide whether to switch or not. We can keep a score. How about it, anyone up for that?