Riddles

[JP Dogberry]

Nazis are WHITE though.

[Archon]

I know, but if I said it was the anarchist calling the Nazi white, then it wouldn't be as direct a parody of "the pot calling the kettle black."

[Legion]

Ok I am still not convinced about the other riddle..I still see the problem as this

X = your pick
Y = not your pick
Z = host eliminates
W = guaranty win

X+(Y+Z) = W

Now at this point X is only at a 1/3 chance of being right while Y+Z has 2/3 chance of being right.  However we know that one of 2Y is going to be wrong and the host will show one of them.

Ok so now Z has been eliminated because the host revealed it

X+(Y+0) = W

Ok now you are only left with X which had a /13 chance of wining and Y+0 which had a 2/3 chance of winning.  Since half of Y+Z have been eliminated it stands to reason that it loses half of its chances of winning…lowering it down to 1/3 chance of being right.  So now you can see that the two boxes are equal.  So really there are only 2 scenarios that are valid.  

1)      You pick the correct box and one of the other two are eliminated
2)      You pick the incorrect box and the other wrong box is eliminated.

The third scenario that you stated is as useless as my 4th scenario that I posted.  Since all boxes are equal in the beginning, there is no real difference if box A has it or if box B has it because at the time when the host allows you to switch there still is only 2 boxes to choose from.

[Firemeboy]

Since half of Y+Z have been eliminated it stands to reason that it loses half of its chances of winning…lowering it down to 1/3 chance of being right.

 No, because then you are only left with 66 percent total.  Just because you know it's wrong, doesn't change the fact that there was a 1 in 3 chance that it was correct.

Get a buddy, try it with cards, check the percentages...

And Archon, your guess is not correct.

[Legion]

 No, because then you are only left with 66 percent total.  Just because you know it's wrong, doesn't change the fact that there was a 1 in 3 chance that it was correct.

Get a buddy, try it with cards, check the percentages...

And Archon, your guess is not correct.

its 33% of the original 100%, because we just found out that 33% that could have been right is now wrong.  So that means that at this point in the game each box is at a 50% chance of winning, because we know now that 1 of the 3 is not it.

And your right there was a 1 in 3 chance, but we are not talking about at the start of the problem we are talking about once 1 box is eliminated is it any better to switch from your box to the only other box still not eliminated...

So if you want to do it with cards what you do is not pick 1 card out of 3 and see if you are righ,t but instead pick 1 card out of 3 then eliminate one of the wrong ones ( but no matter what can you eliminate yours)  and see if you have any better chance of winning if you switch or if you stay with your origninal guess.

[The Jade Knight]

Legion, you have agreed that there is a [33%] chance the correct answer will be each box (because it is randomly distributed among the three).

You have still provided nothing which accounts for 3 [33%] chance scenarios dealing with this fact which would support your current hypothesis.

You cannot say "well, let's erase the old scenarios and voilà, we have two new ones!" or something to that effect.  You have to account for probability of placement.  How does your solution take into account the fact that 1/3 of the time the box will be in box A, 1/3 in box B, and 1/3 in box C?


As for the new riddle, Firemeboy, I'm stumped.  Is this going to turn out to be a political joke?

Each of the other two people are going to see two blue hats, or a blue and a white.  You see two blue hats.  For you, your hat could equally be blue or white.  For them, their hats could equally be blue or white, and whether yours was blue or white wouldn't make a difference to them figure out their own.

[Legion]

I an not trying to erase my other post just trying to show you what I was talking about in a different way...I will try to clarify what I was saying

SO that me start over and try this and see if this helps you understand what I am getting at.

Formula
A = you pick
B = a different box
C = The last box
So then the formula would be A + B + C = 100% chance of winning...right?

Now the host eliminates box C...now you have more information to add into the formula right?


Formula is now

A = still your pick
B=another box
C = been eliminated so its zero now

A + (B + 0) = 100% chance of winning...is that right?

So then it's really A + B = 100% chance of winning...2 choices both equal it’s a 50/50 chance that either is true...

Can you show me anything that is wrong with this logic?

Since it's been more than 24 hours, again...  Here is another one.

You and two other very smart people were brought before the President.  He want's to see which of you is the smartest, so you can figure out how to lower gas prices.  

You are led blind-folded into a small room, and seated around a table.  The president describes the test.

"Upon each of your heads I have placed a hat. You are either wearing a blue hat or a white hat. You don't know which, but I will tell you this; at least one of you is wearing a blue hat. There may be only one blue hat and two white hats, there may be two blue hats and one white hat, or there may be three blue hats. But you may be certain that there are not three white hats."

The president explains that when the blindfolds are taken off, the first to correctly announce the color of his hat shall be his advisor.

With that, the president uncovers your eyes and you see that your two competitors are each wearing blue hats. You see from the look in their eyes that they are thinking, "What is the color of my hat?"

For hours nobody speaks, then finally you stand up and say, "The color of the hat I am wearing is..."

What color is your hat?  And how do you come to that conclusiong?

It never states that you had a hat put on your head....so do you not have one?

[The Holy Saint, Grand High Poobah, Master of Monkeys, Ehlers]

no, you have one, because it says that they put one on you.

[Tink]

Can you just look at what hat you're wearing? The President doesn't say you have to guess without looking, does he? Although if this were true, it wouldn't take hours to figure out.

And Legion, the President says he has placed a hat on each of their heads. So it was done before the test was explained. Plus you should be able to feel it, so you'd know if there wasn't one, and you'd say, "There isn't a hat on my head," not "The color of the hat I'm wearing is. . ."

[Eric James Stone]

Formula is now

A = still your pick
B=another box
C = been eliminated so its zero now

A + (B + 0) = 100% chance of winning...is that right?

So then it's really A + B = 100% chance of winning...2 choices both equal it’s a 50/50 chance that either is true...

Can you show me anything that is wrong with this logic?

Yes, your logic makes the false assumption that the two choices are equal. That is not the case, because your choice was made randomly, while the host's choice is made knowingly.

When you picked A, there was a 1 in 3 chance that it was correct.  Since you have no way of knowing which box contains the prize, your choice is random.

Now, when the host eliminates one of the two remaining boxes, he is not acting randomly.  If he were acting randomly, then there would be a 1/3 chance he would eliminate the box with the prize.  But since he can only eliminate a box that does not have the prize, that means he does not affect the original probability of 2/3 that the prize is not in box A.  Which means there is a 2/3 chance the prize is in the box the host did not eliminate.

[The Jade Knight]

To recap my point:

When you start the game, if you played 30 times and things were perfectly distributed statistically, the answer would be:

A = 10 times
B = 10 times
C = 10 times

Okay?  When you start, there's a 33% chance it'll be in each.  So that's how many it will be in each.

According to my analysis, if you keep your box, you should win 10/30 times.  According to your analysis, you should win 15/30 times.

Let's try it out:

Answer in A (10 times)
Choose A, host kills B [50% - 5 times] or C [50% - 5 times]
You're going to win 10/10 times if you keep A.

Answer in B (10 times)
Choose A, host kills C [100%]
You're going to win 0/10 times if you keep A.

Answer in C (10 times)
Choose A, host kills B [100%]
You're going to win 0/10 times if you keep A.

So, if you keep A, you'll win 10/30 times - 33%.


Let's try reverse engineering:
According to your analysis, you're going to win 50% of the time.  Let's run your scenario 36 times (distributing answers as you expect them to be), and see what happens.

Okay, Legion's scenario:

You choose your box.  You keep your box.  You win 50% of the time, and lose 50% of the time.

50% of the time the answer is in your box (because you kept it and won 50% of the time):
A = 18
50% of the time [18] the answer is not in your box (divided equally among the other two)
B = 9
C = 9

So, according to your plan, legion, the answer is in A 50% of the time, B 25% of the time, and C 25% of the time.  This is, assuming you start with A, and keep A the entire game.

In other words, in order for you to be correct, Legion, the prize would have to be placed in your box twice as often as either of the others.


JadeKnight's scenario:

You choose your box.  You keep your box.  You win 33% of the time and lose 66% of the time.

33% of the time the answer is in your box (because you chose it and kept it and won 33% of the time):
A =12
66% of the time [24] the answer is not in your box (divided equally among the other two)
B = 12
C = 12

So, according to this plan the answer is in A 33% of the time, B 33% of the time, and C 33% of the time.  And it doesn't matter which box you choose and keep the entire game (it works as well for A as it would for C).

[Legion]

According to my analysis, if you keep your box, you should win 10/30 times.  According to your analysis, you should win 15/30 times.

No you are missing my point...we are not dealing with 1 quess to pick the correct box out of 3 choices.  Instead you are given another chance to guess from 1 of 2 boxes.  Your question was after the Host eliminated 1 of the 3 boxes is it better to switch to the other one, and at this point no it is not because even though the host knows where the price is, the price has the same chance of being in either of the 2 boxes.  Thats why I chanced it to a math formula with variables.  Which box is eliminated first does not really matter since it will always not be yours and it is always wrong.  So you are left with did you pick the correct one or does the host know that you are wrong.  So now the host says to you you can switch to the other box are keep yours.....it's like he is letting you choice all over again....so a choose between 2 boxes is a 50% chance with either one.  The only reason why you guys are saying its not is because the host knows where the price is and he knowningly eliminates the only box that does not have the price or is your choice.  This does not chance the fact that the question still leaves you with only 2 choices.

[Archon]

Look at it as a system of equations Legion. Each variable represents the probability of one of the boxes.

X+(Y+Z) = 100
(Y+Z) = 66
Z= 0

You know that, together, all of the boxes represent a 100% chance. You know that together Y and Z have a 66% chance. When the chances of Z are reduced to 0, Y must equal 66.

[Eric James Stone]

This does not chance the fact that the question still leaves you with only 2 choices.

The fact that there are two choices does not automatically mean the probability of each choice being a winner is the same.

Assume that after you have chosen box A, the host gives you the following 2 choices:

1. Stick with box A.
2. Choose both B and C.  If a prize is in either one, you win.

So you have only two choices, but that doesn't mean the probability of the prize being in A has suddenly jumped from 33% to 50%.  With choice 1, your probability of winning is only 33%; with choice 2, it is 67%.

When the host eliminates one box, he is really still giving you choice 2.  Since at least one box out of B and C must be empty, the fact that the host shows you that one of them is empty does not change the fact that there was a 67% chance the prize would be in either B or C.

Let's make it even more clear.

The host shows you 1,000,000 boxes, and tells you that there is a prize hidden in just one of the boxes.  You choose box 234,567.  What are the chances you selected the box with the prize? One in a million (0.0001%).

Now the host has his assistants open up 999,998 of the boxes, showing them as empty.  Only your box and box 327,649 are left.

You have only two choices, now.  Do you seriously believe that the probability that you picked the right box has jumped from 0.0001% to 50%, merely because all the other boxes have been eliminated?

[The Jade Knight]

Legion, you start with 3 boxes.  Either you have random distribution of prizes (between three boxes, or you don't.  If you look at my above analysis, it is clear that random distribution is not possible given your scenario.

You start with 3 boxes, not 2, and so you must divide probability between the three of them.