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Excellent Math Riddle

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naturalplastic
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11 May 2013, 10:27 am

b9 wrote:
Shatbat wrote:
Not really, there are 1001 hats and 1000 wizards. So the first wizard will have to choose between two numbers, and if we keep the iterative process as you described it, every wizard afterwards will have to choose between two hats as well. That's where I got that, on average, half of the wizards will guess their number correctly, but as the OP said that is not the best angle to approach this problem.


since the first wizard can see every other hat, they will successfully eliminate every hat except their own from the list of possibilities of the number on their hat. so will everyone else if they are equally intelligent as the sequence progresses.
.


Its not clear what you are saying here.

If there were an equal number of hats and wizards then yes.the first wizard can see every other hat BUT HIS OWN- and yes he will successfully eliminate every number as possible number for his hat except the one number that IS the correct number of his hat and he will correctly state that one remaining number as his 'guess', and he will win.

But the number of hats is one GREATER than the number of wizards. There is one extra hat that (presumably) none of the wizards is allowed to see that none of them is wearing.

So that means that the first wizard sees all of the other wizards' hats and knows all of the numbers to eliminate-except the number on his own unseen hat- AND the number on then one extra unworn AND unseen hat.

So he has not ONE, but TWO possible numbers for the number on his own hat to pick from as a guess. So its not a one-out-of-one sure thing guess, but a one-out-of-two (fifty-fifty) cointoss.

Capice?

( that means 'do you understand' in Italian).



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11 May 2013, 10:48 am

b9 wrote:
you have tried to complicate a puzzle to make it harder to work out, but in the process, you have made the puzzle easy to calculate.

wizard 1000 sees every hat in front of him and the number between 1 and 1001 that is not seen on a hat by him must be the number on his hat.
the person in front of him who also has the ability to see every hat ahead of him can ignore the number on the hat behind him as a possibility to be his own answer considering the rearmost person will be correct. if he can see all the all the hats in front of him, he can calculate what number will be on his hat since all the hats he sees will not contain 2 numbers (since one is already taken)
and the remaining unseen number must be the one on his cap. the process inevitably resolves itself so that every person correctly identifies what their hat number is.


Are you certain?

Reassess the data provided in the riddle: 1000 wizards, 1001 hat values and a total of 999 wizards/hat values encountered initially by the first wizard. He can use the process of elimination as you've argued but he nonetheless encounters two missing hat values and discerning the one placed on his own head is impossible with respect to total certainty. This means that the probability after using the process of elimination by the first wizard aided by the x-ray function will result in a .5 probability of correctness should the strategy be an iteration of blind guessing. If your strategy is the iteration of blind guessing then the first and subsequent wizards are enrolled in a continual inheritance of a probability of .5. which proves with certainty neither 500 or 999 wizards.



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11 May 2013, 11:36 am

I think I have a way for 999 to win with certainty, with a possibility for 1000 to win. The only problem is that it requires the first wizard to intentionally guess the wrong answer, sacrificing his chance of becoming an archmage.


So the first wizard sees 999 numbers in front of him, leaving 2 numbers to choose from. Instead of guessing either of these numbers, he adds them together and guesses that number e.g. if he knows his number is either 312 or 115 he will guess 427

The next wizard sees 998 numbers in front, and so knows that his number must be one of three numbers. He then finds which two of these three numbers add together to get the first wizard's guess, and then guesses the third number. e.g. if he knows his number must be 312, 115 or 512, then he can work out that the two numbers that the first wizard couldn't see must have been 312 and 115, as they add up to 427. He then knows that his number must be 512.

Similarly, the third wizard can immediately narrow his number down to four possibilities. He also knows that as long as everyone follows the plan, the second wizard must have guessed correctly, so he can eliminate that number, leaving three possibilities. He then applies the same technique the second wizard did to eliminate two more, leaving his number. e.g. If he sees that his possibilities are 312, 115, 512 and 65, he can immediately eliminate 512 as that was the second wizard's number. He then uses the same thinking that the second wizard did to eliminate 312 and 115, and so guesses 65 correctly.

Each wizard in succesion uses this same method of first eliminating all numbers in front of him, then all guesses made before him (except for the first guess) to leave him with three possibilities. He then finds which two of these three numbers add together to get the first wizard's guess, and then guesses the remaining number.

The only 'hiccup' comes for the wizard who has the number that the first wizard guessed. He would use the technique to narrow his possibilities down to three, so in our example he would be left with 312, 155 and 427. He knows that his number is 427, but he cannot guess this as it has already been guessed. Instead he guesses an obviously wrong number, any number over 1001. This tells all the wizards who come after him that his number was the same as that first guess, allowing them to succeed even if he cannot.

So this is how 999 can win with certainty. It is also possible to one more to win, if the first wizard sees that his two possibilities add together to make a number greater than 1001, he can guess this number and then every wizard after him can guess right. There is no 'hiccup' as no wizard will have a hat with that number on it.



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11 May 2013, 11:38 am

Are guesses of invalid hat numbers permitted?

If so, then the first wizard to guess just guesses the number of the hat in front of him plus 10,000. The second wizard then subtracts 10,000 from the previous wizard's guess and uses that as his guess and gets it correct.

The third wizard sees the hat number in front of him and adds 10,000. The fourth wizard subtracts 10,000 from that guess and gets it right.

Thus, every even numbered wizard becomes an archmage.



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11 May 2013, 11:41 am

eric76 wrote:
Are guesses of invalid hat numbers permitted?

If so, then the first wizard to guess just guesses the number of the hat in front of him plus 10,000. The second wizard then subtracts 10,000 from the previous wizard's guess and uses that as his guess and gets it correct.

The third wizard sees the hat number in front of him and adds 10,000. The fourth wizard subtracts 10,000 from that guess and gets it right.

Thus, every even numbered wizard becomes an archmage.


You're right but that would be too easy. This is why I stipulated the additional assumption that no wizard can guess a value greater than 1001.



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11 May 2013, 11:43 am

Ellingtonia wrote:
I think I have a way for 999 to win with certainty, with a possibility for 1000 to win. The only problem is that it requires the first wizard to intentionally guess the wrong answer, sacrificing his chance of becoming an archmage.


So the first wizard sees 999 numbers in front of him, leaving 2 numbers to choose from. Instead of guessing either of these numbers, he adds them together and guesses that number e.g. if he knows his number is either 312 or 115 he will guess 427

The next wizard sees 998 numbers in front, and so knows that his number must be one of three numbers. He then finds which two of these three numbers add together to get the first wizard's guess, and then guesses the third number. e.g. if he knows his number must be 312, 115 or 512, then he can work out that the two numbers that the first wizard couldn't see must have been 312 and 115, as they add up to 427. He then knows that his number must be 512.

Similarly, the third wizard can immediately narrow his number down to four possibilities. He also knows that as long as everyone follows the plan, the second wizard must have guessed correctly, so he can eliminate that number, leaving three possibilities. He then applies the same technique the second wizard did to eliminate two more, leaving his number. e.g. If he sees that his possibilities are 312, 115, 512 and 65, he can immediately eliminate 512 as that was the second wizard's number. He then uses the same thinking that the second wizard did to eliminate 312 and 115, and so guesses 65 correctly.

Each wizard in succesion uses this same method of first eliminating all numbers in front of him, then all guesses made before him (except for the first guess) to leave him with three possibilities. He then finds which two of these three numbers add together to get the first wizard's guess, and then guesses the remaining number.

The only 'hiccup' comes for the wizard who has the number that the first wizard guessed. He would use the technique to narrow his possibilities down to three, so in our example he would be left with 312, 155 and 427. He knows that his number is 427, but he cannot guess this as it has already been guessed. Instead he guesses an obviously wrong number, any number over 1001. This tells all the wizards who come after him that his number was the same as that first guess, allowing them to succeed even if he cannot.

So this is how 999 can win with certainty. It is also possible to one more to win, if the first wizard sees that his two possibilities add together to make a number greater than 1001, he can guess this number and then every wizard after him can guess right. There is no 'hiccup' as no wizard will have a hat with that number on it.


I have to think about this approach a bit more, but there is an obvious discrepancy. If the first wizard to guess guesses the wrong one than he not only doesn't become archmage, but neither does the wizard who's hat he guessed if that number is a valid wizard hat number. So you can only guarantee that 998 can win.

However, if you set the guess as 10,000 plus the sum of the two hats he doesn't see, then the wizard with the hat can still get his hat correct.



Last edited by eric76 on 11 May 2013, 12:01 pm, edited 1 time in total.

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11 May 2013, 11:45 am

aequitas1 wrote:
eric76 wrote:
Are guesses of invalid hat numbers permitted?

If so, then the first wizard to guess just guesses the number of the hat in front of him plus 10,000. The second wizard then subtracts 10,000 from the previous wizard's guess and uses that as his guess and gets it correct.

The third wizard sees the hat number in front of him and adds 10,000. The fourth wizard subtracts 10,000 from that guess and gets it right.

Thus, every even numbered wizard becomes an archmage.


You're right but that would be too easy. This is why I stipulated the additional assumption that no wizard can guess a value greater than 1001.


Okay. Can he guess negative hat numbers?



naturalplastic
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11 May 2013, 12:02 pm

Guessing the sum of the two missing numbers is rather clever.

Trouble is that the OP stipulated 'no guesses over 1001' allowed. So if the sum of the two missing numbers is over 1001 it would be illegile to guess that sum.



Last edited by naturalplastic on 11 May 2013, 12:10 pm, edited 1 time in total.

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11 May 2013, 12:06 pm

a) For 500: Odd wizards can give clue to even wizards by guessing closer numbers to their actual number.

Also subtracting numbers can eliminate ''over 1001'' problem for b).


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11 May 2013, 12:08 pm

How about if odd numbered wizards were to pick the number that was closer to the number of the hat in front of him?

If he sees that 715 and 400 are missing and the next hat is number 975, he would pick 715 and have a 50% chance of getting it right.

The next wizard would see that 975 and 400 are missing. Since 975 is closer to 715 than to 400, he would know his hat was 975 and guess it.

The only time this would break down is if the two numbers were equidistant from the one in front of him. For example, if he sees that 400 and 600 are missing and that the wizard in front of him as hat number 500.

Edit number 1: Meant to be odd numbered.

Edit number 2: Noticed that the post above mine has the same method.



Last edited by eric76 on 11 May 2013, 12:10 pm, edited 2 times in total.

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11 May 2013, 12:08 pm

eric76 wrote:
I have to think about this approach a bit more, but there is an obvious discrepancy. If the first wizard to guess guesses the wrong one than he not only doesn't become archmage, but neither does the wizard who's hat he guessed if that number is a valid wizard hat number. So you can only guarantee that 998 can win.

However, if you set the guess as 10,000 plus the sum of the two hats he doesn't see, then the wizard with the hat can still get his hat correct.


Yeah you're right, that's what I get for doing maths at 2:30am. I should have said that I can guarantee 998, with a possibility for 999. I also missed that you can't guess over 1001 so my method is further ruined :(



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11 May 2013, 12:09 pm

Ellingtonia wrote:
I think I have a way for 999 to win with certainty, with a possibility for 1000 to win. The only problem is that it requires the first wizard to intentionally guess the wrong answer, sacrificing his chance of becoming an archmage.


If the first wizard intentionally guesses an incorrect answer then it leaves no possibility for the promotion of 1000 wizards. This is argumentation is paradoxical.

Ellingtonia wrote:
So the first wizard sees 999 numbers in front of him, leaving 2 numbers to choose from. Instead of guessing either of these numbers, he adds them together and guesses that number e.g. if he knows his number is either 312 or 115 he will guess 427


I stipulated that no wizard can guess a value greater than 1001. Resultant from this are limitations to your solution that makes the matter of proving with certainty a difficult task. Also, in doing so the first wizard not only eliminates any chance of successfully guessing his number but creates an inability for the wizard with the 427 hat value unable to successfully guess his own number. This problem creates structural difficulty. What might this imply for another subsequent wizard if the wizard with 427 must guess a number no greater than 1001 thus having to draw his guess from the value of another? I'm not so concerned with the 500 because that answer is provable and we know from the framing of the question itself that the implication is that it has an answer.

Ellingtonia wrote:
The next wizard sees 998 numbers in front, and so knows that his number must be one of three numbers. He then finds which two of these three numbers add together to get the first wizard's guess, and then guesses the third number. e.g. if he knows his number must be 312, 115 or 512, then he can work out that the two numbers that the first wizard couldn't see must have been 312 and 115, as they add up to 427. He then knows that his number must be 512.

Similarly, the third wizard can immediately narrow his number down to four possibilities. He also knows that as long as everyone follows the plan, the second wizard must have guessed correctly, so he can eliminate that number, leaving three possibilities. He then applies the same technique the second wizard did to eliminate two more, leaving his number. e.g. If he sees that his possibilities are 312, 115, 512 and 65, he can immediately eliminate 512 as that was the second wizard's number. He then uses the same thinking that the second wizard did to eliminate 312 and 115, and so guesses 65 correctly.

Each wizard in succesion uses this same method of first eliminating all numbers in front of him, then all guesses made before him (except for the first guess) to leave him with three possibilities. He then finds which two of these three numbers add together to get the first wizard's guess, and then guesses the remaining number.

The only 'hiccup' comes for the wizard who has the number that the first wizard guessed. He would use the technique to narrow his possibilities down to three, so in our example he would be left with 312, 155 and 427. He knows that his number is 427, but he cannot guess this as it has already been guessed. Instead he guesses an obviously wrong number, any number over 1001. This tells all the wizards who come after him that his number was the same as that first guess, allowing them to succeed even if he cannot.


No wizard can guess a number greater than 1001. I added this because it would make the promotion of 999 wizards without such a limitation easily solvable. I figured this limitation was a natural implication of the riddle because of the ease with which it can be solved under such a system. It wouldn't be a riddle if the wizards could simply construct an infinite set of values.

Ellingtonia wrote:
So this is how 999 can win with certainty. It is also possible to one more to win, if the first wizard sees that his two possibilities add together to make a number greater than 1001, he can guess this number and then every wizard after him can guess right. There is no 'hiccup' as no wizard will have a hat with that number on it.


I disagree that there is any probability whatsoever that all 1000 wizards can be promoted with certainty and the consequence of your strategy requires incorrectness on the part of one wizard which results (notwithstanding the limitations of that which can be expressed for sum totals greater than 1001) in another wizard being unable to guess his own number because of his value being guessed by the first wizard. This has implications under your line of reasoning that could similarly leave the wizard with 427 with no choice but to guess the number of someone in front of him.

If it's any consolation, you're close enough to taste it. But that's all I'm going to say.



Last edited by aequitas1 on 11 May 2013, 12:12 pm, edited 1 time in total.

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11 May 2013, 12:11 pm

eric76 wrote:
How about if odd numbered wizards were to pick the number that was closer to the number of the hat in front of him?

If he sees that 715 and 400 are missing and the next hat is number 975, he would pick 715 and have a 50% chance of getting it right.

The next wizard would see that 975 and 400 are missing. Since 975 is closer to 715 than to 400, he would know his hat was 975 and guess it.

The only time this would break down is if the two numbers were equidistant from the one in front of him. For example, if he sees that 400 and 600 are missing and that the wizard in front of him as hat number 500.


That equidistant number problem can be eliminated by a ''Smaller guesses cover equidistant ones'' statement before the exam.

For b): Why addition, not subtraction? First wizard can subtract the smaller unknown from the greater one, solving the 1001 problem. However, that number will surely be some other wizards number. Poor him.


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Last edited by Somberlain on 11 May 2013, 12:18 pm, edited 2 times in total.

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11 May 2013, 12:13 pm

eric76 wrote:
aequitas1 wrote:
eric76 wrote:
Are guesses of invalid hat numbers permitted?

If so, then the first wizard to guess just guesses the number of the hat in front of him plus 10,000. The second wizard then subtracts 10,000 from the previous wizard's guess and uses that as his guess and gets it correct.

The third wizard sees the hat number in front of him and adds 10,000. The fourth wizard subtracts 10,000 from that guess and gets it right.

Thus, every even numbered wizard becomes an archmage.


You're right but that would be too easy. This is why I stipulated the additional assumption that no wizard can guess a value greater than 1001.


Okay. Can he guess negative hat numbers?


All wizards are limited to guesses between 1 and 1001.



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11 May 2013, 12:17 pm

aequitas1 wrote:
eric76 wrote:
aequitas1 wrote:
eric76 wrote:
Are guesses of invalid hat numbers permitted?

If so, then the first wizard to guess just guesses the number of the hat in front of him plus 10,000. The second wizard then subtracts 10,000 from the previous wizard's guess and uses that as his guess and gets it correct.

The third wizard sees the hat number in front of him and adds 10,000. The fourth wizard subtracts 10,000 from that guess and gets it right.

Thus, every even numbered wizard becomes an archmage.


You're right but that would be too easy. This is why I stipulated the additional assumption that no wizard can guess a value greater than 1001.


Okay. Can he guess negative hat numbers?


All wizards are limited to guesses between 1 and 1001.


I would ask if they could guess complex numbers, but obviously 512+715i is not between 1 and 1001.

How about numbers that aren't integers? :)



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11 May 2013, 12:22 pm

Somberlain wrote:
eric76 wrote:
How about if odd numbered wizards were to pick the number that was closer to the number of the hat in front of him?

If he sees that 715 and 400 are missing and the next hat is number 975, he would pick 715 and have a 50% chance of getting it right.

The next wizard would see that 975 and 400 are missing. Since 975 is closer to 715 than to 400, he would know his hat was 975 and guess it.

The only time this would break down is if the two numbers were equidistant from the one in front of him. For example, if he sees that 400 and 600 are missing and that the wizard in front of him as hat number 500.


That equidistant number problem can be eliminated by a ''Smaller guesses cover equidistant ones'' statement before the exam.


So if the wizard sees numbers 2 and 4 are missing and the number in front of him is 3, he guesses 2. The next wizard sees that 3 and 4 are missing and guesses 3 because that is closer to 2.

I think you have it.