[bonanova]

A magician blindfolded seven subjects, then for each of them he rolled a

pair of fair dice and asked them the probability of their dice totaling 7.

He said to the first, here's a hint: truthfully, at least one of your dice shows a 6.

The subject counted 11 cases of at least one 6, two of which, 1-6 and 6-1, total 7.

So he answered, my chances of having 7 are 2/11. Very good, said the magician.

He then said to the second, I've looked at your dice, and at least one of them is a 5.

This subject counted 11 cases of at least one 5, of which 2-5 and 5-2 made 7.

So he answered 2/11. Very good, said the magician.

He told the third subject, I see at least one 4 on your dice.

That subject also found 11 cases of 4, of which 3-4 and 4-3 made 7.

So he answered 2/11. Very good, said the magician.

The next subject was told at least one of his dice was a 3.

Like the others, he found 11 cases, and of them only 4-3 and 3-4 were favorable.

So he answered, my chances of having 7 are 2/11. Very good said the magician.

The next two were told their dice showed at least one 2 and one 1, respectively.

They found 5-2, 2-5 for one, and 6-1, 1-6 for the other, among 11 cases gave 7.

They both answered their chances of 7 were 2/11. Very good said the magician.

The seventh subject had been listening to all of this. And before the magician

could speak, he said, I don't need a hint. I know that you're going to tell me

some number appears on at least one of my dice. And you've already confirmed

what the right answer is in each case. So whatever you were going to say,

I know most certainly what the odds probability of my dice totaling 7 are is.

My answer is 2/11.

But we know his odds are probability is 1/6, right?

Edited December 29, 2008 by bonanova
Scraff's observation

[bonanova]

Thanks Bonanova - I've really enjoyed this topic! I had read the kids problem previously and was convinced the answer was 1/3, but this thread got me thinking again. It's great to be able to challenge your own thoughts and, as ever, this forum continues to make you do just that!!

Y/w.

It is a great place.

You have a pleasant, persuasive way of making a case. A rarity.

[Guest]

Thanks Bonanova - I've really enjoyed this topic! I had read the kids problem previously and was convinced the answer was 1/3, but this thread got me thinking again.

I hadn't joined brainden on time of kids problem, but I guess that formal answer was given as 1/3.

Now, as you I think it is wrong, provided that kids question was nearly same as this one:

You meet a couple, they say they have two children, you see one of the children is girl. And they ask you to know gender of the other child. In this case chances are equal=1/2.

Actually, in kids question, the family says "at least one of our children is girl", I think these are identical, and both have an answer of 1/2.

But if the question was that:

In an exam, there is such a question: Among the couples who have two children and one of them is girl, what is the chance of other child to be girl= 1/3

After a hundred of posts, all I deduce is this. Am I right?

[Guest]

I hadn't joined brainden on time of kids problem, but I guess that formal answer was given as 1/3.

Now, as you I think it is wrong, provided that kids question was nearly same as this one:

You meet a couple, they say they have two children, you see one of the children is girl. And they ask you to know gender of the other child. In this case chances are equal=1/2.

Actually, in kids question, the family says "at least one of our children is girl", I think these are identical, and both have an answer of 1/2.

But if the question was that:

In an exam, there is such a question: Among the couples who have two children and one of them is girl, what is the chance of other child to be girl= 1/3

After a hundred of posts, all I deduce is this. Am I right?

All correct!

The subtlety is in the word "and". "Among the couples who have two children AND one of them is a girl" clearly defines the population of all couples with two children who don't have two boys. If there was a break in the sentence, i.e. "Among the couples who have two children" and separately you are told for a particular family "one of them is a girl" then the population is all couples with two children of any gender.

[Guest]

After looking at all the posts, I'm still dumbfounded as to the exact bit of extra information the first 6 had to increase their odds that the 7th didn't.

Out of curiosity, I ran a small Java app to see a simulation (using 100 million rolls):

	private int roll() {

		return (int) Math.floor(6*Math.random());

	}


	private boolean randBoolean() {

		return Math.random() >= 0.5;

	}


	public void diceTest() {

		final int ROLLS = (int) Math.pow(10, 8);

		final boolean ONE_DIE_ONLY = true;

		int d1, d2;

		int[] contains = new int[6];

		int[] sumsToSeven = new int[6];

		int allSumsToSevens = 0;

		boolean randBool;

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

			d1 = roll();

			d2 = roll();

			randBool = randBoolean();

			if (ONE_DIE_ONLY) {				

				contains[randBool?d1:d2]++;

			} else {

				contains[d1]++;

				if (d1 != d2)

					contains[d2]++;

			}


			if (d1 + d2 == 5) {

				if (ONE_DIE_ONLY) {

					sumsToSeven[randBool?d1:d2]++;

				} else {

					sumsToSeven[d1]++;

					sumsToSeven[d2]++;

				}

				allSumsToSevens++;

			}	

		}		

		System.out.println("1/6 == " + (double)1/6);

		System.out.println("2/11 == " + (double)2/11);

		System.out.println("Chekced one die only? " + ONE_DIE_ONLY);

		for (int i=0;i<6;i++) {

			System.out.println("Contains at least one " + (i+1) + ": " + (double) sumsToSeven[i]/contains[i]);

		}

		System.out.println("All\t\t	   : " + (double) allSumsToSevens/ROLLS);

	}

Output when only one die were looked at only (After a double roll, a die was picked as random and only its results were added):

1/6 == 0.16666666666666666

2/11 == 0.18181818181818182

Checked one (random) die only? true

Contains at least one 1: 0.16674083362946113

Contains at least one 2: 0.16655349378981213

Contains at least one 3: 0.16666807678407305

Contains at least one 4: 0.16661091962227978

Contains at least one 5: 0.16651701024636423

Contains at least one 6: 0.1665549155933424

All : 0.16660754

Output when both dice were looked at (After a double roll, the results for both dice were added):

1/6 == 0.16666666666666666

2/11 == 0.18181818181818182

Checked one (random) die only? false

Contains at least one 1: 0.1818427658505504

Contains at least one 2: 0.1818128729072829

Contains at least one 3: 0.18192065081955375

Contains at least one 4: 0.18190028764613636

Contains at least one 5: 0.18177532062203824

Contains at least one 6: 0.18175545421438394

All : 0.16667847

Weird how when I only added the results for one die, the probabilities came to 1/6, but when I added them from both, the probabilities came to 2/11.

This must somehow be related to that "extra information." Can anyone shed any light?

[Guest]

After looking at all the posts, I'm still dumbfounded as to the exact bit of extra information the first 6 had to increase their odds that the 7th didn't.

They didn't - that's where a lot of people on this post initially made their mistake. Ordinarily, based on what the magician said, as they don't know the rule by which the magician decides which number to tell them, each of 1 to 6 can't actually make an accurate calculation as to their odds. As Bonanova points out in one of his later posts, the fact that they said 2/11 and were told they were correct was actually luck on their part and can only be used to tell us what rule the magician was using in each of the cases (i.e. for first person he would definitely tell them if he saw at least one 6, for second person he would definitely tell them if he saw at least one 5, etc.) As we don't know the rule for the seventh person - and we haven't been told anything, nothing can actually be determined.

Weird how when I only added the results for one die, the probabilities came to 1/6, but when I added them from both, the probabilities came to 2/11.

This must somehow be related to that "extra information." Can anyone shed any light?

What your program is doing is simulating the different rules for how you are told about a number (and actually helps prove that the answer varies depending on the rule).

When you look at the results for one die, this is simulating the rule "The number I say you have at least one of will be chosen at random from the two numbers shown."

When you look at the results for both dice, this is simulating the rule "If the two dice contain at least one of x then I will tell you about it."

I hope that helps clear things up!