Apolitical Intelligence Tests With Important Point to Follow

Old English Riddle

As I was walking to St. Ives, I met a man with seven wives
Each wife had seven sacks, each sack had seven cats
Each cat had seven kits.
Kits, cats, sacks, wives, how many were going to St. Ives?

Second Test

A prison was overcrowded, so the warden decided to release a prisoner and wanted to reward intelligence.
He had three prisoners brought into his office and said:
"The guard behind you will draw a hat and put it on each of your heads from behind you. He will take it at random out of a bag containing two red hats and three black hats. If you can tell me the color of the hat on your own head, you will be released. But if you guess and guess wrong, you agree to be shot immediately or you may not play. Agreed?" All agreed.
Hats are placed on the prisoners and they look at each other.
Prisoner #1: "I don't know."
Prisoner #2: "I don't know."
Prisoner #3, who is blind: "I know."

Does he? Explain, and please don't look either puzzle up or confer with others. Solve them on your own.

After a sufficient number of attempts or passes, I will provide the answers and the Important Point implicit in these tests.

 

Fourteen views so far and nobody has even made an attempt to solve either puzzle. Nobody.
What is the important lesson to be gleaned, and can it be gleaned whether or not anyone so much as offers up one, or two answers?
Stay tuned and learn, if you still do that sort of thing.

 

Answer: Zero. "Kits, cats, sacks, wives, how many (of THOSE) were going to St. Ives?"
I "met" them, going the other way. I was going to St. Ives. They were coming from St. Ives.

Answer: Prisoner #1 does not see two red hats so he doesn't know if his is red or black, so he passes.
Prisoner #2 does not see two red hats, but in addition, he does not see a red hat on #3, for if he did, he would know that his own hat was black because Prisoner #1 said he "didn't know."
Therefore Prisoner #3 knows that the hat on his head is black and he is immediately set free.

At least twenty views and nobody has so much as attempted to solve either intelligence test. What important point follows?
The fact that although everyone reading them has the same information, still many can't figure out the answer. Others arrive at incorrect conclusions when everyone has exactly the same information!!! This widespread failure is applicable to the Holy Bible, Darwinism, Climate Change, and politics, to name but a few controversial domains that are divisive.

 

Both of your answers are incorrect. You never said anything about the wives "going the other way."

You failed your own test.

 

What point is it that you were trying make. As I read the OP the answers came to me. Your puzzles were quite easy. But you had to go and jump the gun.

I'm not sure what point you are making.


How about. A rooster was sitting on the ridge of a roof. If the ridge runs north to south and the wind is blowing from west to east, which way will latest eggs roll.

Or a little harder;

Two cars are traveling from two different points and in opposite directions in a circuit race at a constant speed. The cars cross for the first time at point A. The second time at point B. The third time is at point C and the fourth time at point A. How much faster is one car than the other?

 

"As I was going to St. Ives
I MET a man with seven wives".

1. It's not "my" test. It's an old English riddle.
2. "Met" infers a face to face meeting, which statistically is far more likely to happen when the riddler is walking towards the group, than if he had overtaken them.
3. Had they all been going in the same direction, the riddle would be "I passed a man with seven wives." He did not say anything remotely like that.

Since you're in California:

http://PeoplesRepublicOfCalifornia.wordpress.com

 

The bag contains 2 red hats and 3 black hats. Supposes the two sighted prisoners have black hats on - and the blind one a red hat.

Each of the sighted prisoners would then see a red had and a black hat on the other two. There is no way for either of the sighted prisoners to know what hat is on their head from the information ... and no way to for the blind prisoner to know that his hat is red and not black.

 

Twice as fast

 

Jolly good!

 

I had the exact same thought as Giftedone.

Since you say that this was easy and that you had it figured out, can you respond to his retort as well as my added possibility?..

-"The bag contains 2 red hats and 3 black hats. Supposes the two sighted prisoners have black hats on - and the blind one a red hat.

Each of the sighted prisoners would then see a red had and a black hat on the other two. There is no way for either of the sighted prisoners to know what hat is on their head from the information ... and no way to for the blind prisoner to know that his hat is red and not black."

-Additionally, what if they all 3 had on black hats? The first 2 people would only see 2 black hats which would leave them no way of knowing whether their hat was red or black, and their not knowing would not be enough to indicate the blind mans hat color



It seems to me that this logic only applies if there were 2 red and 2 black hats in the bag. What am I missing?

 

OP is correct. (assuming that each prisoner understands the logic and properly analyzes the situation and doesn't lie or make a mistake)

If anyone sees 2 reds, then that person must be in black. The necessary logic comes from the fact that P2 has access to P1's answer, and P3 has access to both P1's and P2's answer.

Possibilities:

1. B B B
2. B B R
3. B R B
4. B R R
5. R B B
6. R B R
7. R R B

The key lies in the 3 possibilities that have P3 wearing red. Let's examine the P3 red possibilities.

2. B B R P1 says unknown. P2 now knows he is wearing black and would declare this.

4. B R R P1 knows he is wearing black and would declare this. P2 can safely declare that he is in red.

6. R B R P1 says unknown. P2 knows he is wearing black, and would declare this.

All of the possibilities with P3 wearing red can be eliminated because both P1 and P2 declared "unknown." This leaves nothing but possibilities with P3 wearing black.

 

The puzzle only works if there are only two red hats and three or more black hats, and the red hats are on the sighted prisoners.

 

Thank you for the response but it does not address all possibilities facing number 3....

Lets say for instance that all 3 are wearing black. Number 1 and 2 would see 2 blacks, and hence not know what they are wearing because remaining there are 2 reds and one black, and hence would subsequently say "I dont know". Number 3 does not know that they both saw black, or whether one or both saw red and black. Even if he did know that, he still would not know if his own hat was black or red. The only thing that the blind number 3 knows for sure is that both 1 and 2 did not see 2 reds.

 

I woulda said one. You were still going to st. Ives.

 

Geeze my math skills tell me roosters dont lay eggs...

 

Then please tell me what I am missing in the following logic....


Lets say for instance that all 3 are wearing black. Number 1 and 2 would see 2 blacks, and hence 1 and 2 would not know what they are wearing because remaining there are 2 reds and one black, and hence would subsequently say "I dont know". Number 3 does not know that they both saw black, or whether one or both saw red and black. The only thing that the blind number 3 knows for sure is that both 1 and 2 did not see 2 reds.

 

"Kits, cats, sacks, wives, how many (OF THOSE SUBJECTS OF THE SENTENCE) were going to St. Ives?"

 

Prisoner #1 does NOT see two red hats on #2 and #3. So he said "I don't know."
Prisoner #2 does NOT see a red hat on #3 for if he did, he would say "My hat is black. Bye!"
Prisoner #3 understands the logic of the first two prisoners and knows his own hat must be black.
Think that through.

 

The first one is:

2 men (possibly 1 man and 1 woman), 7 wives, 49 sacks, 343 cats, 2401 kits

"How many" is the question.

How many what? People? Items? Living creatures?

"Kits, cats, sacks, wives" is not the answer, since there were also two men going with them.

 

Appreciated.

When I first read GiftedOnes reply (which coincides with your thinking), I thought that he/she was correct. But then I began thinking.... this logic puzzle and answer most likely wasn't thought up by the OP - it is probably a standard logic puzzle. And therefore, the answer given must be correct.

So, I thought about it some more. When considering the BBB possibility, we are actually considering 2 possibilities, the BBB and the BBR. Both P1 and P2 are wearing black, but P3 doesn't know if he is wearing red or black. Let's examine this:

B B R : P1 sees a black and a red and says "unknown." P2 uses logic and says "I am wearing black." P2 knows this because if P2 was wearing red, P1 would have known he was wearing black.

B B B : P1 says "unknown." P2 says "unknown". P3 says "I am in black, because if I was in red, P2 would have known he was wearing black.

The point is, in the 3 possibilities where P3 is in red, one or both of P1 and P2 would know what they were wearing. And in the 4 possibilities with P3 wearing black, P1 and P2 do not know their colors. So it is easy for P3 to decide black or red.

Go through the possibilities I presented earlier and you will see.

If P1 and P2 say "unknown," P3 can say black.
If P1 and/or P2 say they know their color, then P3 can say red.

 

1 was going to St. Ives

I'm not sure about the hat question. If prisoner 1 sees 2 black hats he wouldnt know. However, if he saw a black and a red, he also wouldn't know so there is no information that is gained by the "i don't know". I'm sure there's a trick to it, but I don't see it.

 

Following up - possibilities concerning P1 and P2

..................P1...........P2
1. B B B unknown unknown
2. B B R unknown black
3. B R B unknown unknown
4. B R R black ...... red
5. R B B unknown unknown
6. R B R unknown black
7. R R B unknown unknown

In the possibilities with P3 in black, neither P1 or P2 know their color. In the possibilities with P3 in red, either P1 or P2 or both will know their color. And that is what tells P3 whether he is black or red.

 

I HAVE thought it through.

Lets say all 3 prisoners are wearing black hats.

-Prisoner number one sees 2 black hats and says I dont know.
-Prisoner number 2 sees 2 black hats and also says I dont know because the only thing he knows is that number 3 and 1 are both wearing black hats and that number 1 either saw one black and one red OR two blacks
- Prisoner number 3 sees nothing and can only deduce that both number 1 and number 2 did not see 2 red hats. He has no basis for knowing whether they saw 2 blacks or 1 of each.



After this debate, I decided to google this riddle. I found a link on wikipedia that lists a number of various hat riddles. It says....

Three-Hat Variant[edit]
In this variant there are 3 prisoners and 3 hats. Each prisoner is assigned a random hat, either red or blue. In all, there are three red hats and two blue. Each person can see the hats of two others, but not their own. On a cue, they each have to guess their own hat color or pass. They win release if at least one person guessed correctly and none guessed incorrectly (passing is neither correct nor incorrect).

This puzzle doesn't have a 100% winning strategy, so the question is: What is the best strategy? Which strategy has the highest probability of winning?

If you think of colors of hats as bits, this problem has some important applications in coding theory.

The solution and the discussion of this puzzle can be found here (also a solution to the analogous 7-hat puzzle) and other 3 variants are available on this Logic Puzzles page (they are called Masters of Logic I-IV).
https://en.wikipedia.org/wiki/Hat_puzzle


According to my logic, AND this link, there is not enough information in this particular riddle to 100% deduce the correct answer ( and in this one the last prisoner isnt even blind). The only thing you can do is improve your odds with logic. In order to know with certainty, you have to stipulate that at least one prisoner will have a red hat. It is the one out of 7 chance that all 3 have black hats that cannot be deduced with certainty. You could improve your odds to 7 out of 8 with logic.

 

Ahh, then it's a statistical problem. That changes the problem.

 

Your analysis for BBB is incorrect. P2 would not have known he was wearing black if P3 was in black. All P2 would know in that situation is that both 1 and 3 are wearing black which with there being one more black hat out there as well as 2 red ones, his answer would be "I dont know"

I think that you are starting with the assumption that the OP is correct, and are trying to work backwards for proof of it being correct. In truth, the OP is incorrectly stated. There is not enough information provided to give a definitive answer. I have since googled the hat puzzles and found the following link that states that the 3 hat variant does not provide certainty. It is only a means of improving ones odds.

https://en.wikipedia.org/wiki/Hat_puzzle
Three-Hat Variant[edit]
In this variant there are 3 prisoners and 3 hats. Each prisoner is assigned a random hat, either red or blue. In all, there are three red hats and two blue. Each person can see the hats of two others, but not their own. On a cue, they each have to guess their own hat color or pass. They win release if at least one person guessed correctly and none guessed incorrectly (passing is neither correct nor incorrect).

This puzzle doesn't have a 100% winning strategy, so the question is: What is the best strategy? Which strategy has the highest probability of winning?

If you think of colors of hats as bits, this problem has some important applications in coding theory.

The solution and the discussion of this puzzle can be found here (also a solution to the analogous 7-hat puzzle) and other 3 variants are available on this Logic Puzzles page (they are called Masters of Logic I-IV).