Riddles

[The Jade Knight]

Wrong.

You're not taking into account the fact that the box which is being revealed is not random.  Were it random, there would be a 50/50 chance.

However, they will never reveal a) the correct answer, and b) your box.

This plays with the equation.  It's not just a matter of randomly picking between two boxes.


As I have stated, in the initial group of 3 boxes, the right answer is NOT your box 66% of the time.  Seeing as they will NEVER reveal your box, whether it's right or wrong, that means that 66% of the time, the other WRONG answer will be eliminated, leaving the other RIGHT answer.  Which means that 66% of the time, switching boxes will make you get the RIGHT one.

[Peter Ahlstrom]

I have to admit that Jade Knight (and those who agree with him) is right. Didn't make sense to me for the longest time, but I bow to some good explanations...one from my brother in law...

You're really playing one of two different games. One game, in which you win 100% of the time by sticking with your initial door, is what you're playing a third of the time. The other game, in which you win 100% of the time by switching your door, comes up two thirds of the time. (Just do the simulation I linked above, for experimental evidence.)

You don't know which game you're playing, but the guy who eliminates a wrong answer does know, and THAT keeps it from being random.

[Firemeboy]

I thought that the fact that the host would know where the item is was implied.  I should have probably stated it better.

[The Jade Knight]

It's not that the host knows where the item is.  It's that the host never reveals the correct answer, and (very important), never reveals your own box.  This as a trend, not just a one-off deal.

[Legion]

yeah I was talking to a Math major about your statistics to see if you were right about all that and she told me that your logic wasn't right and that in fact it wouldn't matter if you switch or not...This girl was the Valedictorian of her college so I am assuming that she knows what she is talking about

[Firemeboy]

She was wrong.  Or at least, didn't bother to get all the facts.  Get her online and maybe we can convince her, and then she can convince you.  

After you pick your door, the host, as he always does, picks another door and opens it up.

Cecil Adams, from the article, states three criteria must be met.  

(1) The host knows which door conceals the prize; (2) he only opens doors that do NOT conceal the prize; and (3) he always opens a door.

The second and third criteria are met.   The first was implied, but I should have been more clear.  

You can argue that maybe the way I stated the puzzle implies that it wouldn't matter, but in the spirit of the puzzle, (what was implied), when those criteria are met, you are foolish not to switch because you have a greater chance of winning.

[Firemeboy]

It's that the host never reveals the correct answer, and (very important), never reveals your own box.

 

This is kind of irrelevent.  Because if the host revealed the correct answer, or your box, there is no longer any guess work involved. You know.  The whole idea is that after the other door is revealed, you don't know where the prize is.  But if the host knows, and always reveals the other door that does not have the prize, you have a great chance if you switch.

[The Jade Knight]

No, but see, it's statistical.  Knowing that it's not random (as opposed to you just being lucky/unlucky) is statistically important for determining your odds.


And no offense, Legion, but she obviously didn't think about my post long enough (did she actually read this thread, or did she just hear your explanation of my post?), or she's not as statistically inclined as you make out.  If you're having trouble grasping how it works this way, I recommend you click on Ookla's link and think it out.


Once again:

If there's a chance they'll open your door, the odds are 50/50, no matter what you do, you'll win.  If there's a chance they'll reveal the correct answer, your odds jump to 66% automatically.  If they will definitely never do either, your odds are 66% only if you switch.

[Firemeboy]

No, but see, it's statistical.  Knowing that it's not random (as opposed to you just being lucky/unlucky) is statistically important for determining your odds.

 Sorry, I misundestood you.  I thought you were saying that in defense of not switching.  

[scAri]

So people just need to get the 50/50 thing out of their heads. You make a 1/3 choice and nothing that happens after that can change the fact that your choice has a 1/3 chance. We're not dealing with that sort of probability/statistics because there's no randomness.

[The Jade Knight]

There may be some randomness, but there is much less randomness, and it affects the equation.

[Legion]

And no offense, Legion, but she obviously didn't think about my post long enough (did she actually read this thread, or did she just hear your explanation of my post?), or she's not as statistically inclined as you make out.  If you're having trouble grasping how it works this way, I recommend you click on Ookla's link and think it out.

She did read your post and that is where she stated that your logic is wrong.  She said that after the host eliminates one of the wrong boxes that is not yours, it is, statistically no better if you switch boxes.  This is due to the fact that the host will always revealed the wrong door no matter if you pick a correct one or not.  Therefore it comes down to you picking one box with a 100% chance of not having your box eliminated...then you are given a chance to switch between 2 boxes which gives you a 50/50 chance of picking the right box.

So having 3 boxes is really not important since you will always have a different box (that is wrong) eliminated first.  So then you only have to really choose between the two boxes that are left.  Now this is just like calling a coin heads or tails in the air.  Statically you have a 50% chance of being right no matter what way you go.

[The Jade Knight]

Wrong.

You didn't read my post carefully enough, and it seems that she didn't, either.

Do you agree, or disagree, that, initially, the box you choose has a [33%] chance of being right, and the other two boxes, together, have a [66%] chance of containing the right answer?

Additionally, do you agree or disagree that there are three possible scenarios, each with a [33%] chance of occuring, and that they are:
a) the correct choice is in box A
b) the correct choice is in box B
c) the correct choice is in box C


Remember, this is a math puzzle, so the math has to add up.

[Legion]

I agree with both of those statements.....but it is really 1/3 not 33%
1/3+1/3+1/3=1
33%+33%+33%=99%

1/3 does not equal 33%, but for this we will assume that 1/3 does =33%

but yes I know what you are saying and I agree that you, at the begining only have a 33% chance of being right, and yes there are only 3 possible scenarios

[The Jade Knight]

I'm rounding down, for the sake of convenience.

Okay, let's run this down for you on a possibility by possibility basis:

Given:  You choose box A.

Possibility 1 [33%]:
The correct answer is box A.  Host reveals box B or C (statistically, it doesn't make a difference).
Keep = [win]  [33%]
Trade (for either) = [lose]  [0%]

Possibility 2 [33%]:
The correct answer is box B.  Host reveals box C.
Keep = [lose] [0%]
Trade for box B = [win] [33%]

Possibility 3 [33%]
The correct answer is box C.  Host reveals box B.
Keep = [lose] [0%]
Trade for box C = [win] [33%]


Those are the only 3 possibilities.

You'll note:
Keep = [33%+0%+0%] = 33% chance of winning
Trade = [0%+33%+33%] = 66% chance of winning.


The rules of the game limit the possibilities, and those are the only three possibilities, and I hope this makes it easier for you to see how the precentages add up.