- New Logic/Math Puzzles - BrainDen.com

bonanova · August 2, 2010

Ten prisoners are selected and told by the warden that next morning a hat with a single digit 0-9 will be placed on each of their heads.

The digits can be repeated; not all digits need be used.

Each prisoner can see the other nine numbered hats.

After the hats are placed, each prisoner will be brought to the warden, one by one and out of sight and hearing of the others, to guess his own number.

If at least one prisoner gets it right, all ten will be released.

Otherwise they all will be executed.

A strategy may be agreed upon tonight.

The prisoners then cannot communicate with anyone else until they have all spoken to the warden.

What strategy gives them the best chance of being released?

August 2, 2010

First they'll give a "line number" to each prisoner(1-10). The prisoners must remember it. Number 10 will be the leader. When they have to say their number to the warden they'll look at the leader and if his "hat number" is "6", for example, they will line up in order of line numbers and leave a little gap for him just after number (5) (1)(2)(3)(4)(5)(10,leader)(6)...When the leader is taken he will just say his position he is in starting from (1)

Sorry if you don't get a thing, English is not my first language.

By the way there need to be more rules because right now i can think of at least 5 ways that include secret communication.

Edited August 2, 2010 by kristmark1

August 2, 2010

First they'll give a "line number" to each prisoner(1-10). The prisoners must remember it. Number 10 will be the leader. When they have to say their number to the warden they'll look at the leader and if his "hat number" is "6", for example, they will line up in order of line numbers and leave a little gap for him just after number (5) (1)(2)(3)(4)(5)(10,leader)(6)...When the leader is taken he will just say his position he is in starting from (1)
Sorry if you don't get a thing, English is not my first language.
By the way there need to be more rules because right now i can think of at least 5 ways that include secret communication.

After the hats are placed, each prisoner will be brought to the warden, one by one and out of sight and hearing of the others, to guess his own number....The prisoners then cannot communicate with anyone else until they have all spoken to the warden.

Secret communication is ruled out. The prisoners must form a strategy that does not rely on communication of any information they may gain after they have received their hats.

August 2, 2010

After each prisoner goes to the warden and makes his guess, will the other prisoners know whether the guesser was right or wrong? Or will the result only be given after all guessing is complete?

August 2, 2010

A similar was posted before.

August 2, 2010

...gather around the prisoner with the biggest number,for example if you see that the largest number is (7) you stand behind him. Then somebody else sees it and comes to your side, or if your number is bigger he'll stand behind you and so on till the number of people matches the biggest number seen. After that everybody says that number and eventually someone gets it right

Edited August 2, 2010 by kristmark1

preflop · August 2, 2010

What does no communication mean here, are prisoners allowed to do something like raise their hand? or chose their position in line?

If so i think I have a simple solution

The night before the prisoners assign themselves a number from 0 - 9 (lets call this the prisoner value) - each prisoner gets a different number.

Then come the morning when they see each others hats, they raise their hand if they see their own prisoner value on someone's head.

All of the prisoners then use the lowest prisoner value who raised their hand, as the guess for their hat. One is bound to be correct.

Edited August 2, 2010 by preflop

unreality · August 2, 2010

unless choosing a spot in line or 'standing near' other prisoners counts as other people have suggested, I can't think of any way to convey information seeing as their guess is private.

So the base probability to try to beat:

each prisoner guesses 0 (or really any number he wants)

1 - (9/10)^10 = 0.65132156 chance that at least one prisoner will guess correctly.

So that's the number to beat I suppose

superprismatic · August 2, 2010

unless choosing a spot in line or 'standing near' other prisoners counts as other people have suggested, I can't think of any way to convey information seeing as their guess is private.

So the base probability to try to beat:

each prisoner guesses 0 (or really any number he wants)

1 - (9/10)^10 = 0.65132156 chance that at least one prisoner will guess correctly.

So that's the number to beat I suppose

Well, you can beat the random guess probability
of .651 by having each prisoner guess randomly
from the set of numbers which he doesn't see
worn by any other prisoner (there must be at
least one). In this case the probability jumps
to a whopping .684.

unreality · August 2, 2010

Well, you can beat the random guess probability
of .651 by having each prisoner guess randomly
from the set of numbers which he doesn't see
worn by any other prisoner (there must be at
least one). In this case the probability jumps
to a whopping .684.

unless they have some pattern (but then again the subsets of all patterns just makes it random numbers anyway) this is not true. In the OP bonanova states that:

The digits can be repeated; not all digits need be used.

After a line of bubbling in C on a computer-generated scantron test, many kids start getting suspicious and think that there's an ever decreasing probability that the next one will be C as well. But there's a 1/4 (or 1/5 with 'E') chance that C will be the correct answer every time, independent of the others.

It's called the gambler's fallacy when you flip a coin a bunch of times and get heads every time and as a result you expect the next coin to be more than .50 chance to be tails

superprismatic · August 2, 2010

unless they have some pattern (but then again the subsets of all patterns just makes it random numbers anyway) this is not true. In the OP bonanova states that:

The digits can be repeated; not all digits need be used.

After a line of bubbling in C on a computer-generated scantron test, many kids start getting suspicious and think that there's an ever decreasing probability that the next one will be C as well. But there's a 1/4 (or 1/5 with 'E') chance that C will be the correct answer every time, independent of the others.

It's called the gambler's fallacy when you flip a coin a bunch of times and get heads every time and as a result you expect the next coin to be more than .50 chance to be tails

"unless they have some pattern (but then again the subsets of all patterns just makes it random numbers anyway) this is not true. In the OP bonanova states that:

The digits can be repeated; not all digits need be used."

What exactly isn't true? No matter what the digit distribution is, each player is only looking at 9 players. Because there are 10 possible numbers, each player must be looking at an incomplete set of digits.

August 2, 2010

There's nothing better to do than guess. Why? Its because your number would be independent of all the other numbers. Just because your friend's numbers are all different or same doesn't mean anything.

August 2, 2010

iIf everyone picks the same number, one person has to be correct.

August 2, 2010

iIf everyone picks the same number, one person has to be correct.

Yeah, in a perfect world, the prisoners would all have different numbered hats. Unfortunately, numbers can be repeated and not all numbers will be used.

August 2, 2010

this is similar to the coin toss experiment posted a while ago.
two people go into separate rooms and must make it so both don't guess the other's coin.
the answer is one person guesses his own coin (heads if if heads, tails if tails) after flipping and the other person guesses the opposite coin( heads if tails, tails if heads). so basically we can use a similar strategy here. every person adds up all the numbers, mod 10, and guesses the person's hat number in that position. if he gets his own position then guess 0.
let's take some example numbers.
3,5,7,2,1,1,3,8,2,6
prisoner 1: 5 mod 10, guess 1
prisoner 2: 3 mod 10, guess 2*
prisoner 3: 1 mod 10, guess 5
prisoner 4: 6 mod 10, guess 3
prisoner 5: 7 mod 10, guess 8
prisoner 6: 7 mod 10, guess 8
prisoner 7: 5 mod 10, guess 1
prisoner 8: 0 mod 10, guess 3
prisoner 9: 6 mod 10, guess 3
prisoner10: 2 mod 10, guess 7

superprismatic · August 2, 2010

Here, I assume that each prisoner gets a digit drawn
independently from the uniform digit distribution.
Consider the following strategy: Each prisoner
guesses that the digit on his hat is the same as the
most frequent digit that he sees on the other nine.
Ties are broken randomly. In this case, the probability
that the prisoners survive is .804.

unreality · August 3, 2010

Here, I assume that each prisoner gets a digit drawn
independently from the uniform digit distribution.
Consider the following strategy: Each prisoner
guesses that the digit on his hat is the same as the
most frequent digit that he sees on the other nine.
Ties are broken randomly. In this case, the probability
that the prisoners survive is .804.

each hat is independent. It doesn't matter what you guess it'll be a 1/10 chance... unless you can find some way to convey extra info. Which this method does not do. How did you calculate .804???

August 3, 2010

He's definitely on to something. Have the prisoners pair off. One agrees to guess that they have the same hat number and the other agrees to guess that their hats are different numbers. Chance that they are both wrong: (9/10)*(8/9). This increases their chance of survival to about 67.3%. Not a whole lot better, but it's something.

August 3, 2010

There is a way that one and only one person will guess their hat number every time.

Edited August 3, 2010 by d3k3

superprismatic · August 3, 2010

each hat is independent. It doesn't matter what you guess it'll be a 1/10 chance... unless you can find some way to convey extra info. Which this method does not do. How did you calculate .804???

First, anytime there is one number which appears
on 3 or more hats and every other number appears
on 1 or no hats, my strategy (see post #16) works.
This is because each of the people who are wearing
a 3 see that there are 2 of the 9 wearing a 3
whereas no other number appears more than once.
So, everyone who is wearing a 3 will guess correctly
-- all the others will be wrong, but who cares
because it only takes one to be correct.
Now there are many other scenerios like this one.
For example, 4 of one number and no more than 2
of any other number. These cases happen often
when the numbers are chosen from a uniform
distribution. They, and some just a little bit favorable,
happen often enough to push the probability from .65 to .80.

The full proof is an involved argument that counts
the number of cases where the strategy is sure to
work (favorable cases). Then there are cases where
the strategy isn't perfect but works better than random.
In all the other cases, we use the random probability.

Edited August 3, 2010 by superprismatic

August 3, 2010

First, anytime there is one number which appears
on 3 or more hats and every other number appears
on 1 or no hats, my strategy (see post #16) works.
This is because each of the people who are wearing
a 3 see that there are 2 of the 9 wearing a 3
whereas no other number appears more than once.
So, everyone who is wearing a 3 will guess correctly
-- all the others will be wrong, but who cares
because it only takes one to be correct.
Now there are many other scenerios like this one.
For example, 4 of one number and no more than 2
of any other number. These cases happen often
when the numbers are chosen from a uniform
distribution. They, and some just a little bit favorable,
happen often enough to push the probability from .65 to .80.

The full proof is an involved argument that counts
the number of cases where the strategy is sure to
work (favorable cases). Then there are cases where
the strategy isn't perfect but works better than random.
In all the other cases, we use the random probability.

I'm inclined to agree with unreality here; As you look at your specific cases, you fail to properly consider all possibilities and how your solution would affect their probability. Thus, you are committing gambler's fallacy.

In this specific case, you consider how if there are 3 of a single number, there is a guaranteed win. However, if everyone follows this rule, then in cases where the maximum number is 2, you greatly diminish your chances or guarantee a loss:

If there 9 numbers are used, 1 repeated, then every person that see's the pair will guess incorrectly. The people wearing the pair will see 9 other numbers and guess randomly, bringing your chances of success down to 0.19.

If there is more than one pair (these are the most common distributions), then everyone will see at least 1 or more pairs, all of which will not be their number, bringing your chances of success down to 0.

I'm sure if you were to list out all possible initial cases with their probabilities, then calculate your average probability, it would still come out to be 0.65132...

In fact, any "strategy" used equate to that probability because the initial distributions are still random. What would be interesting is too see a proof of this conjecture, but that's way out of my league. Can anyone disprove this by providing a strategy that will lower the chances?

Edited August 3, 2010 by vinays84

superprismatic · August 3, 2010

I'm inclined to agree with unreality here; As you look at your specific cases, you fail to properly consider all possibilities and how your solution would affect their probability. Thus, you are committing gambler's fallacy.
In this specific case, you consider how if there are 3 of a single number, there is a guaranteed win. However, if everyone follows this rule, then in cases where the maximum number is 2, you greatly diminish your chances or guarantee a loss:

If there 9 numbers are used, 1 repeated, then every person that see's the pair will guess incorrectly. The people wearing the pair will see 9 other numbers and guess randomly, bringing your chances of success down to 0.19.
If there is more than one pair (these are the most common distributions), then everyone will see at least 1 or more pairs, all of which will not be their number, bringing your chances of success down to 0.

I'm sure if you were to list out all possible initial cases with their probabilities, then calculate your average probability, it would still come out to be 0.65132...
In fact, any "strategy" used equate to that probability because the initial distributions are still random. What would be interesting is too see a proof of this conjecture, but that's way out of my league. Can anyone disprove this by providing a strategy that will lower the chances?

You said:
"I'm sure if you were to list out all possible initial cases with their probabilities, then calculate your average probability, it would still come out to be 0.65132... ".
I performed the messy calculations and I came up with .804. It seemed too high, so I genetated a bunch of random hat number assignments. Sure enough, my strategy
saved the prisoners about 80% of the time. All that work had not gone down the drain! I would post my calculations if it wasn't such a daunting task.
So, if you don't believe me, generate a bunch of random assignments and see for yourself. The difference between 65% and 80% is large enough to make itself
evident fairly quickly. I happen to have a soft copy of all 42 possible ways to make how many of each number. Here it is:

* 1 10
* 2 9 1
* 3 8 2
* 4 8 1 1
* 5 7 3
* 6 7 2 1
* 7 7 1 1 1
* 8 6 4
* 9 6 3 1
* 10 6 2 2
* 11 6 2 1 1
* 12 6 1 1 1 1
1 13 5 5
1 14 5 4 1
* 15 5 3 2
* 16 5 3 1 1
* 17 5 2 2 1
* 18 5 2 1 1 1
* 19 5 1 1 1 1 1
1 20 4 4 2
1 21 4 4 1 1
1 22 4 3 3
1 23 4 3 2 1
1 24 4 3 1 1 1
* 25 4 2 2 2
* 26 4 2 2 1 1
* 27 4 2 1 1 1 1
* 28 4 1 1 1 1 1 1
1 29 3 3 3 1
1 30 3 3 2 2
1 31 3 3 2 1 1
1 32 3 3 1 1 1 1
1 33 3 2 2 2 1
1 34 3 2 2 1 1 1
1 35 3 2 1 1 1 1 1
* 36 3 1 1 1 1 1 1 1
X 37 2 2 2 2 2
X 38 2 2 2 2 1 1
X 39 2 2 2 1 1 1 1
X 40 2 2 1 1 1 1 1 1
0 41 2 1 1 1 1 1 1 1 1
0 42 1 1 1 1 1 1 1 1 1 1

* strategy always works
1 strategy sometimes works (better than guessing)
0 strategy works randomly here
X strategy always fails

August 3, 2010

I ran all the strategies given thus far in a computer program.

If everyone guesses the same number (i.e. 3) or guesses randomly: .651

superprismatic's strategy: .289

phillip1882's strategy: .507

Tuckleton's strategy: .672

These are the averages of 1 million iterations.

Here's the code, for anyone who wants to modify/verify it:

// Ten prisoners are selected and told by the warden that next morning a hat with a single digit 0-9 will be placed on each of their heads.

// The digits can be repeated; not all digits need be used.

// Each prisoner can see the other nine numbered hats.

// After the hats are placed, each prisoner will be brought to the warden, one by one and out of sight and hearing of the others, to guess his own number.

// If at least one prisoner gets it right, all ten will be released.

// Otherwise they all will be executed.

//

// A strategy may be agreed upon tonight.

// The prisoners then cannot communicate with anyone else until they have all spoken to the warden.

//

// What strategy gives them the best chance of being released?


#include "time.h"

#include <stdlib.h>

#include <stdio.h>


const int NUM_TESTS_TO_RUN = 1000000;

const int NUM_PRISONERS	=  	10;


int hat_numbers[NUM_PRISONERS] = {0};


/*****************************************************************************/

/*                                                                   		*/

/*  sameNumberStrategy()                                             		*/

/*                                                                   		*/

/*  Strategy: All prisoners guess the same number, i.e. 3.           		*/

/*                                                                   		*/

/*****************************************************************************/

int sameNumberStrategy()

{

  int guess = 3;


  for (int i = 0; i < NUM_PRISONERS; i++) {

//	printf("Prisoner %d guess: %d\n", i+1, guess);

	if (hat_numbers[i] == guess)

  	return 1;

  }

  return 0;

}


/*****************************************************************************/

/*                                                                   		*/

/*  randomGuessStrategy()                                                	*/

/*                                                                   		*/

/*  Strategy: All prisoners guess a random number.                   		*/

/*                                                                   		*/

/*****************************************************************************/

int randomGuessStrategy()

{

  int guess;


  for (int i = 0; i < NUM_PRISONERS; i++) {

	guess = rand() % 10;

//	printf("Prisoner %d guess: %d\n", i+1, guess);

	if (hat_numbers[i] == guess)

  	return 1;

  }

  return 0;

}


/*****************************************************************************/

/*                                                                   		*/

/*  computeModalAverage()                                                	*/

/*                                                                   		*/

/*****************************************************************************/

int computeModalAverage(int prisonerNum)

{

  int numberFreq[10]   = {0};

  int modalAverage 	=  -1;

  int modalAverageFreq =  -1;


  for (int i = 0; i < NUM_PRISONERS; i++) {

	if (i != prisonerNum)

  	numberFreq[hat_numbers[i]]++;

  }


  for (int j = 0; j < 10; j++) {

	if (numberFreq[j] > modalAverageFreq) {

  	modalAverageFreq = numberFreq[j];

  	modalAverage = j;

	}

  }


  return modalAverage;

}


/*****************************************************************************/

/*                                                                   		*/

/*  superprismaticsStrategy()                                            	*/

/*                                                                   		*/

/*  Strategy: Each prisoner says the number of the modal average of the  	*/

/*        	numbers he CAN see.                                        	*/

/*                                                                   		*/

/*****************************************************************************/

int superprismaticsStrategy()

{

  int guess;


  for (int i = 0; i < NUM_PRISONERS; i++) {

	guess = computeModalAverage(i);

//	printf("Prisoner %d guess: %d\n", i+1, guess);

	if (hat_numbers[i] == guess)

  	return 1;

  }

  return 0;

}


/*****************************************************************************/

/*                                                                   		*/

/*  phillipsStrategy()                                               		*/

/*                                                                   		*/

/*  Strategy: Every person adds up all the numbers, mod 10, and guesses the  */

/*        	person's hat number in that position. If he gets his own   	*/

/*        	position, then guess 0.                                    	*/

/*                                                                   		*/

/*****************************************************************************/

int phillipsStrategy()

{

  int guess;

  int sumOfHats;

  int pos;


  for (int i = 0; i < NUM_PRISONERS; i++) {

	sumOfHats = 0;

	for (int j = 0; j < 10; j++) {

  	if (i != j)

    	sumOfHats += hat_numbers[j];

	}

	pos = sumOfHats % 10;

	if (pos == i)

  	guess = 0;

	else

  	guess = hat_numbers[pos];

//	printf("Prisoner %d guess: %d\n", i+1, guess);

	if (hat_numbers[i] == guess)

  	return 1;

  }

  return 0;

}


/*****************************************************************************/

/*                                                                   		*/

/*  tuckletonsStrategy()                                             		*/

/*                                                                   		*/

/*  Strategy: Have the prisoners pair off. One agrees to guess that they have*/

/*        	the same hat number and the other agrees to guess that their   */

/*        	hats are different numbers.                                	*/

/*                                                                   		*/

/*****************************************************************************/

int tuckletonsStrategy()

{

  int guess;


  for (int i = 0; i < NUM_PRISONERS; i++) {

	if (i % 2 == 0)

  	guess = hat_numbers[i+1];

	else {

  	do {

    	guess = rand() % 10;

  	} while (guess == hat_numbers[i-1]);

	}

//	printf("Prisoner %d guess: %d\n", i+1, guess);

	if (hat_numbers[i] == guess)

  	return 1;

  }

  return 0;

}


/*****************************************************************************/

/*                                                                   		*/

/*  main()                                                           		*/

/*                                                                   		*/

/*****************************************************************************/

void main()

{

  // random number seed

  srand( (unsigned)time( NULL ) );


  int sameNumberSuccess  	= 0;

  int randomGuessSuccess 	= 0;

  int superprismaticsSuccess = 0;

  int phillipsSuccess    	= 0;

  int tuckletonsSuccess  	= 0;


  for (int i = 0; i < NUM_TESTS_TO_RUN; i++) {

	// randomize hats

	for (int i = 0; i < NUM_PRISONERS; i++) {

  	hat_numbers[i] = rand() % 10;

//  	printf("%d ", hat_numbers[i]);

	}

//	printf("\n");


	// Same Number Strategy

	sameNumberSuccess += sameNumberStrategy();


	// Random Guess

	randomGuessSuccess += randomGuessStrategy();


	// Modal Average Strategy

	superprismaticsSuccess += superprismaticsStrategy();


	// Phillip1882's Strategy

	phillipsSuccess += phillipsStrategy();


	// Tuckleton's Strategy

	tuckletonsSuccess += tuckletonsStrategy();

  }


  printf("Same Number: .%d\n", (sameNumberSuccess*1000/NUM_TESTS_TO_RUN));

  printf("Random Guess: .%d\n", (randomGuessSuccess*1000/NUM_TESTS_TO_RUN));

  printf("superprismatic's: .%d\n", (superprismaticsSuccess*1000/NUM_TESTS_TO_RUN));

  printf("phillip1882's: .%d\n", (phillipsSuccess*1000/NUM_TESTS_TO_RUN));

  printf("Tuckleton's: .%d\n", (tuckletonsSuccess*1000/NUM_TESTS_TO_RUN));

}

August 3, 2010

You said:
"I'm sure if you were to list out all possible initial cases with their probabilities, then calculate your average probability, it would still come out to be 0.65132... ".
I performed the messy calculations and I came up with .804. It seemed too high, so I genetated a bunch of random hat number assignments. Sure enough, my strategy
saved the prisoners about 80% of the time. All that work had not gone down the drain! I would post my calculations if it wasn't such a daunting task.
So, if you don't believe me, generate a bunch of random assignments and see for yourself. The difference between 65% and 80% is large enough to make itself
evident fairly quickly. I happen to have a soft copy of all 42 possible ways to make how many of each number. Here it is:
*    1 10

*    2  9  1

*    3  8  2

*    4  8  1  1

*    5  7  3

*    6  7  2  1

*    7  7  1  1  1

*    8  6  4

*    9  6  3  1

*   10  6  2  2

*   11  6  2  1  1

*   12  6  1  1  1  1

1   13  5  5

1   14  5  4  1

*   15  5  3  2

*   16  5  3  1  1

*   17  5  2  2  1

*   18  5  2  1  1  1

*   19  5  1  1  1  1  1

1   20  4  4  2

1   21  4  4  1  1

1   22  4  3  3

1   23  4  3  2  1

1   24  4  3  1  1  1

*   25  4  2  2  2

*   26  4  2  2  1  1

*   27  4  2  1  1  1  1

*   28  4  1  1  1  1  1  1

1   29  3  3  3  1

1   30  3  3  2  2

1   31  3  3  2  1  1

1   32  3  3  1  1  1  1

1   33  3  2  2  2  1

1   34  3  2  2  1  1  1

1   35  3  2  1  1  1  1  1

*   36  3  1  1  1  1  1  1  1

X   37  2  2  2  2  2

X   38  2  2  2  2  1  1

X   39  2  2  2  1  1  1  1

X   40  2  2  1  1  1  1  1  1

0   41  2  1  1  1  1  1  1  1  1

0   42  1  1  1  1  1  1  1  1  1  1


* strategy always works

1 strategy sometimes works (better than guessing)

0 strategy works randomly here

X strategy always fails

Any time the most frequent number is tied with another number, your strategy will fail. For example, your "4 4 1 1" example. You have that your strategy will work sometimes. Let's try it:
Prisoner 1 has a 2 on his head.
Prisoner 2 has a 2 on his head.
Prisoner 3 has a 2 on his head.
Prisoner 4 has a 2 on his head.
Prisoner 5 has a 5 on his head.
Prisoner 6 has a 5 on his head.
Prisoner 7 has a 5 on his head.
Prisoner 8 has a 5 on his head.
Prisoner 9 has a 7 on his head.
Prisoner 10 has a 0 on his head.

Prisoners 1-4 will see 5 as the most predominant and guess 5. They will be wrong. The same thing will happen to Prisoners 5-8. And Prisoners 9 and 10 have no chance.

August 3, 2010

LOL. I still claim 100% success. No simulations.

If one out of the ten can correctly "guess" the sum of all ten digits, they win.

superprismatic · August 3, 2010

Any time the most frequent number is tied with another number, your strategy will fail. For example, your "4 4 1 1" example. You have that your strategy will work sometimes. Let's try it:
Prisoner 1 has a 2 on his head.
Prisoner 2 has a 2 on his head.
Prisoner 3 has a 2 on his head.
Prisoner 4 has a 2 on his head.
Prisoner 5 has a 5 on his head.
Prisoner 6 has a 5 on his head.
Prisoner 7 has a 5 on his head.
Prisoner 8 has a 5 on his head.
Prisoner 9 has a 7 on his head.
Prisoner 10 has a 0 on his head.

Prisoners 1-4 will see 5 as the most predominant and guess 5. They will be wrong. The same thing will happen to Prisoners 5-8. And Prisoners 9 and 10 have no chance.

What does computeModalAverage do?

I posted an early (and incorrect) version of my table.
Here is the correct one:

* 1 10
* 2 9 1
* 3 8 2
* 4 8 1 1
* 5 7 3
* 6 7 2 1
* 7 7 1 1 1
* 8 6 4
* 9 6 3 1
* 10 6 2 2
* 11 6 2 1 1
* 12 6 1 1 1 1
X 13 5 5
1 14 5 4 1
* 15 5 3 2
* 16 5 3 1 1
* 17 5 2 2 1
* 18 5 2 1 1 1
* 19 5 1 1 1 1 1
X 20 4 4 2
X 21 4 4 1 1
1 22 4 3 3
1 23 4 3 2 1
1 24 4 3 1 1 1
* 25 4 2 2 2
* 26 4 2 2 1 1
* 27 4 2 1 1 1 1
* 28 4 1 1 1 1 1 1
X 29 3 3 3 1
X 30 3 3 2 2
X 31 3 3 2 1 1
X 32 3 3 1 1 1 1
1 33 3 2 2 2 1
1 34 3 2 2 1 1 1
1 35 3 2 1 1 1 1 1
* 36 3 1 1 1 1 1 1 1
X 37 2 2 2 2 2
X 38 2 2 2 2 1 1
X 39 2 2 2 1 1 1 1
X 40 2 2 1 1 1 1 1 1
0 41 2 1 1 1 1 1 1 1 1
0 42 1 1 1 1 1 1 1 1 1 1

* strategy always works
1 strategy sometimes works (better than guessing)
0 strategy works randomly here
X strategy always fails

Sign In

Question

Link to comment

Share on other sites

32 answers to this question

Recommended Posts

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Guest

Link to comment

Share on other sites

Link to comment

Share on other sites

Join the conversation

Recently Browsing 0 members