Three Doors with one prize

Hi,

There are three doors and one prize behind one door.
The host of the show knows which one has the prize.
A person fron the audiance has to pick a door (at random), he is not allowed to open it

He has 1/3 chance of winning, so far so good.

Now the host opens one of the remaining empty doors.

Show the man be better staying with his original selection or move to the other door?

How does it affect the probabilities?

Not off topic, I found this in an astronomy book. What do you think?

John

With no other information, the probability is initially distributed 33/33/33 behind each door.

With the disclosure that one door has a 0, the probability between the other two doors has moved to 100, but there's still no information to pick which, so it's 50/50.

It will neither increase nor decrease the chances of the participant winning to select the other door.

At least that's what logic would dictate. But these questions usually have a twist in them that I've probably failed to account for.... :oops:

Cheers,
-- Jeff.

He had a 1-in-3 chance.
Eliminating the empty door changed his chances to 1-in-2.
Changing his choice of door is pointless because he is still only guessing.
Same as flipping a coin.

P.S. You might have found it in an Astronomy book but it IS "off topic" for Astronomy.

We are assuming that the prize actually exists. Until it is revealed, it may exist, or it may not exist. It may even exist beyond both doors.
Read: en.wikipedia.org/w/index.php?title=Schr%C3%B6dingers_Cat

All of this logic puzzles have a twist to them. The best way to go about them is to make a logic table. So that's what I'm going to do.

And so my logic would be that you should change your decision after he opens one of the 2 doors and shows you that nothing is in it. Because the gameshow host is aware of what is behind the doors, he will 100% open a door that has nothing behind it. The probability that the last remaining door hides the prize will now be 2/3, and so your decision should change.

Some people at first glance might think no, the probability is 1/3 for each door. But if you think of all the options available it begins to make sense. Lets see what you can do...

Logic Table (remember host always reveals empty)

It turns out that there are 6 possible combinations


Original Choice :: Behind Hosts door :: Last door hides

Prize :: Empty :: Empty
Prize :: Empty :: Empty
Empty :: Empty :: Prize
Empty :: Empty :: Prize
Empty :: Empty :: Prize
Empty :: Empty :: Prize


You can see from this table, the probability that the 3rd remaining door is not 50/50 as one might think, but in fact it is 2/3 or 66% probabilty that there will be a prize.

Of course, this is all assuming that the host KNOWS and is trusted to open an empty door. And of course as Michael says, that the prize actually exists. But this is the premise the thread poster gave so I believe this is the solution. If he didn't know where the prize was and just opened an empty door, the probability would still just be 1/3 and you would have no reason to change your choice.

My brain hurts now