[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

[Guest]

The probability of 7 is always 1/6 in the above cases. The calculation which leads to 2/11 is flawed because :

The classical definition of probability is identified with the works of Pierre Simon Laplace. As stated in his Théorie analytique des probabilités,

" The probability of an event is the ratio of the number of cases favorable to it, to the number of all cases possible when nothing leads us to expect that any one of these cases should occur more than any other, which renders them, for us, equally possible."

(source:wikipedia)

When the blindfolded persons calculated 2 favorable outcomes and 11 total possible outcomes, they calculated the probability as 2/11 which is not correct by the very definition of probability because those 11 outcomes are not equally likely possible. e.g the probablity of 1,6 is not the same as 1,1. So, they can't use desired outcome divided by total outcomes to calculate the probability in this case.

An important thing to note : If I ask you what is the probability of a 6,6 and you say 1/36 because there is one favourable outcome and 36 possible outcomes so probability is 1/36, then I'd say this is the wrong way of reaching the right answer by fluke. Because here also the probabilities of all those 36 outcomes is not the same. The correct way will be the probability of 1st dice to have a six is 1/6 (one favourable / six possible which are all equally likely) multiplied by prob of second dice to have a six i.e 1/6 which equals to 1/36.

Similarly the prob of 1 and 6 is not calculated as 2 favorable (1,6 and 6,1) / 36 total but as prob of 1,6 + prob of a 6,1 i.e 1/6*1/6 +1/6*1/6

This the biggest flaw that is always overlooked and that is the reason you will see a number of problems based on a pair of dice.

[Guest]

If I ask you what is the probability of a 6,6 and you say 1/36 because there is one favourable outcome and 36 possible outcomes so probability is 1/36, then I'd say this is the wrong way of reaching the right answer by fluke. Because here also the probabilities of all those 36 outcomes is not the same.

They are exactly the same (assuming a pair of fair dice). All of the following have equal probability:

1 x 1, 1 x 2, 1 x 3, 1 x 4, 1 x 5, 1 x 6

2 x 1, 2 x 2, 2 x 3, 2 x 4, 2 x 5, 2 x 6

3 x 1, 3 x 2, 3 x 3, 3 x 4, 3 x 5, 3 x 6

4 x 1, 4 x 2, 4 x 3, 4 x 4, 4 x 5, 4 x 6

5 x 1, 5 x 2, 5 x 3, 5 x 4, 5 x 5, 5 x 6

6 x 1, 6 x 2, 6 x 3, 6 x 4, 6 x 5, 6 x 6

Out of 36 equal choices, only one of them is 6,6. This is the reason the probability of rolling two sixes is 1/36.

What is the probability of rolling a 6 and a 5 in any order? 1/18. This is simply because all the possible rolls above are equal and there are two of them.

[Guest]

However, I must a say this is a good puzzle to clear the basics except the part where the magician says "Very good" in response to 2/11 but then maybe that plays on us psychologically and we take it for granted that 2/11 is correct.

[Guest]

However, I must a say this is a good puzzle to clear the basics except the part where the magician says "Very good" in response to 2/11 but then maybe that plays on us psychologically and we take it for granted that 2/11 is correct.

2/11 is correct.

[Guest]

They are exactly the same (assuming a pair of fair dice). All of the following have equal probability:

1 x 1, 1 x 2, 1 x 3, 1 x 4, 1 x 5, 1 x 6

2 x 1, 2 x 2, 2 x 3, 2 x 4, 2 x 5, 2 x 6

3 x 1, 3 x 2, 3 x 3, 3 x 4, 3 x 5, 3 x 6

4 x 1, 4 x 2, 4 x 3, 4 x 4, 4 x 5, 4 x 6

5 x 1, 5 x 2, 5 x 3, 5 x 4, 5 x 5, 5 x 6

6 x 1, 6 x 2, 6 x 3, 6 x 4, 6 x 5, 6 x 6

Out of 36 equal choices, only one of them is 6,6. This is the reason the probability of rolling two sixes is 1/36.

What is the probability of rolling a 6 and a 5 in any order? 1/18. This is simply because all the possible rolls above are equal and there are two of them.

Ok. Let me first ask you this. Dou you agree that the probability of 7 in the puzzle is wrongly calculated as 2/11 or do you say that the probability should be 2/11 because 2 favourable / 11 total possible ??

I'll give u the the explanation why the latter calculation is wrong ?

[Guest]

Ok. Let me first ask you this. Dou you agree that the probability of 7 in the puzzle is wrongly calculated as 2/11 or do you say that the probability should be 2/11 because 2 favourable / 11 total possible ??

The first seven subjects had a probability of 2/11 given the information the magician gave. I thought I made my feeling on this clear, especially given not only the title of my last link, but the contents.

[Guest]

Still not a probability of 1/3. Once you know that one is a girl, take that girl and put her out of your mind. What is the probability another child is a boy or a girl. You only have 2 options, boy or girl, so the probability cannot be 1/3. They are 2 distinct statistical events and have to be considered separate. Once you know the gender of one the order no longer matters, I just explained it that way for simplicity. Here's the full picture.

If the older child is a girl, the younger child could be a boy or girl

If the younger child is a girl, the older child could be a boy or girl

Either way you look at it, the probability is 1/2.

Let me return to the dice though. You all have brought up that you should label them A and B. That's fine. Here's how that'd work.

Assume you're the contestant the magician told you had at least one 3.

If A=3

Then B=1, 2, 3, 4, 5, or 6

and your probability of getting 7 total is 1/6

SEPARATE PROBABILITY EVENT

If B=3

Then A=1, 2, 3, 4, 5, or 6

and your probability of getting 7 total is 1/6

It isn't two separate probability events though. There is only one probability of having the dice sum to 7 in each scenario. In your example, if you add the two probability events together, you will see that A=3 AND B=3 is accounted for twice, leaving 11 unique possibilities, 2 of which sum to 7. The whole trick to the riddle is that you are given information, however vague it may be, which alters the possible outcomes. The final contestant, having not been given any information as to what number(s) he may have rolled, then has what we all know to be the overall probability of having 7: 1/6, not 2/11 like the others.

[bonanova]

Two dice [call them A and B] are rolled.

With equal likelihood, A and B can independently be 1, 2, 3, 4, 5 or 6.

There are 36 equally likely combinations for A-B:

1-1, 1-2, 1-3, 1-4, 1-5, 1-6

2-1, 2-2, 2-3, 2-4, 2-5, 2-6

3-1, 3-2, 3-3, 3-4, 3-5, 3-6

4-1, 4-2, 4-3, 4-4, 4-5, 4-6

5-1, 5-2, 5-3, 5-4, 5-5, 5-6

6-1, 6-2, 6-3, 6-4, 6-5, 6-6

Some say [lfac[post #27]] these outcomes are not equally likely.

Not so. For fair dice, they are equally likely, each with p = 1/36.

Now the magician makes a statement [to Subject 1] "At least one of your dice is 6"

This statement [make a truth table to confirm it for yourself] is logically equivalent to

A=6 or B=6, where "or" has the inclusive sense of either or both.

A=6 or B=6 conflicts with 25 of the above cases.

When they are removed, these remain.

.........................1-6

.........................2-6

.........................3-6

.........................4-6

.........................5-6

6-1, 6-2, 6-3, 6-4, 6-5, 6-6.

Some say [vinays84[post #14]] these cases are no longer are equally likely.

Not so. Nothing the magician said changed their likelihood.

One [rossbeamer[post 5]] asserts p[A=6] = p = 1/2, concluding p[7] = 2/12 = 1/6.

Not quite. p[A=6] = p= 6/11. So p[7] = 2/11.

Or just count the favorable cases.

armcie[7] says 6-6 got more likely. Well, yes, from 1/36 to 1/11.

But not more likely than any of the other ten cases; all are 1/11.

Some say [rossbeemer[4] PT[10]] 6-6 should be counted twice.

No. [A=6 B=6] and are not distinct cases.

Some say [Beastmaster[12,17]] once 1-6 is counted, 6-1 should not be counted.

No. "6-1" says [A=6 B=1]. "1-6" says [A=1 B=6]. These are distinct outcomes.

Some say [BM[22]] "6-1" indicates time sequence: first 6 is rolled then 1 is rolled.

Amazingly, BM then concludes that 6-1 and 1-6 are the same!

No. Sequence is nowhere implied. But actually, it doesn't matter.

The magician said: A=6 or B=6. Either could have been "first".

Some say [HH[post 8]; lazboy[9]; BM[12, 17, 23]] the magician said "A=6."

That leaves only the 6 outcomes for B, one of which makes 7, giving p[7] = 1/6.

No. The magician said A=6 or B=6.

Putting it another way, some [listed above] say the magician identified

ONE of the dice as a 6. So the six cases for the OTHER die give p[7] = 1/6.

No. The magician said A=6 or B=6. Same mistake; different wording.

The mistake above is neglecting the inconvenient truth that the ONE die

that is 6 can be EITHER of the dice A and B. The magician said A=6 or B=6.

There is both a full row and a full column remaining in our outcomes table.

They give 11 enumerable, equally likely cases, 2 of which are favorable.

p[7] = 2/11.

Finally, Scraff has lead the charge for truth justice and ... the correct answer for

One Boy One Girl insofar as establishing the 2/11 probability for Subjects 1-6.

And lazboy recanted his early objections and is now a 2/11 believer. Now back to the question.

All well and good.

Those who reject 2/11 for the first 6 subjects have no question to answer:

For them, p[7] = 1/6 for all seven subjects. Good. Let's not hear from them again.

But for those who believe p[7] = 2/11 for Subjects 1-6:

where does the argument Subject 7 gives that his p[7] is also 2/11 break down?

Or does it?

[Guest]

But for those who believe p[7] = 2/11 for Subjects 1-6:

where does the argument Subject 7 gives that his p[7] is also 2/11 break down?

Or does it?

Subject 1-6 has a clue. What is that:

Knowing at least one of his dice is n (any value) gives no clue, because certainly one of his dice would be at a n value, and this seems to give no clue.

But subject x knows that both of his dice are not distinct from n.

Whatever n is, knowing both of his dice are not away from a n value is a clue, because it is not certain that one of his dice would be distinct from a n value.

This is a conctrete knowledge, and now subject x can ignore the possibilities of both dice to be away from n.

This knowledge is given to subject 1-6 but not to subject 7, thus without this knowledge, his chance is 1/6 while others are 2/11.

[Prof. Templeton]

Some say [rossbeemer[4] PT[10]] 6-6 should be counted twice.

No. [A=6 B=6] and are not distinct cases.

I wasn't implying that boxcars should be counted twice. I was just stating the obvious (not to some). On 36 possibles a "six" shows up 12 times, 1/6 as expected. If you were told "at least one of the die shows a six", then there are 11 possible scenarios and of those 11 only 2 can total "seven". I completely agree with Scraf on this one.

[bonanova]

I wasn't implying that boxcars should be counted twice. I was just stating the obvious (not to some). On 36 possibles a "six" shows up 12 times, 1/6 as expected. If you were told "at least one of the die shows a six", then there are 11 possible scenarios and of those 11 only 2 can total "seven". I completely agree with Scraf on this one.

Yeah, I got that on a re-read. Apologies.

That a "6" shows with probability of 1/6, sounded like p[7]=1/6 because of two counts from double 6 on first read.

I should have edited my post to that effect. Let this admission accomplish that.

I compiled the summary in probably too short a time; in haste, to direct things back to the question.

[bonanova]

Subject 1-6 has a clue. What is that:

Knowing at least one of his dice is n (any value) gives no clue, because certainly one of his dice would be at a n value, and this seems to give no clue.

But subject x knows that both of his dice are not distinct from n.

Whatever n is, knowing both of his dice are not away from a n value is a clue, because it is not certain that one of his dice would be distinct from a n value.

This is a conctrete knowledge, and now subject x can ignore the possibilities of both dice to be away from n.

This knowledge is given to subject 1-6 but not to subject 7, thus without this knowledge, his chance is 1/6 while others are 2/11.

Any clue that is imparted by a statement is also imparted by a statement that is its logical equivalent.

At least one of your dice is 6 is logically equivalent to A=6 or B=6: either or both are 6; i.e., not neither are 6.

Also equivalent is both of your dice are not distinct from 6; more precisely: it is not the case that both are distinct from 6.

All these statements do nothing other than to remove the 25 cases where neither of the dice show a 6.

Their common outcome table is the truth table that proves their logical equivalence. For any value of n, not just for 6.

Subject 7 says that p[7] = 2/11 once it is known A=n or B=n for some [any] value of n.

He argues that with certainty his dice permit the statement for some value of n.

Then since, for any n, 25 cases are removed, and among the remaining cases 2 are favorable, his p[7] must be 2/11.

Do you agree that "at least one of your dice is 6" and "both dice are not distinct from 6" [in your words] are logically equivalent?

[Guest]

Bonanova, I know where others stand, but I'm not quite sure if you're asserting that the 2/11 is correct or if that falacy is part of your riddle. I'm somewhat bothered by your dismissal of myself and a couple others who seem to be the only ones who know statistics well enough to catch this falacy, unless of course your intention is to get as many people to chase the white rabbit as possible. You can't dismiss truth just because you don't like it or it doesn't fit into your agenda. Please consider the possibility that your riddle is flawed. I'm not trying to be confontational, I'm just trying to stand up for truth, even if it is a truth as insignificant as this (afterall, this is just a fun site for riddles).

Not that anyone will listen, but here is my final attempt to show you why the 2 children riddle was flawed, and, by extension the die riddle.

If there are 2 children and you don't know either gender, there are 4 EQUALLY possible outcomes. These outcomes are...

Boy THEN Boy

Boy THEN Girl

Girl THEN Boy

Girl THEN Girl

Therefore, the probability of Girl THEN Girl is 1/4

If you know that one is a girl, you can't simply remove the Boy THEN Boy scenario and say that the new probability is 1/3. If you know one is a girl, you have a totally new problem. The new problem is,

If there are 2 children and one of them is a girl, what is the probability that they are both girls.

For this there are only 2 EQUALLY possible outcomes. These outcomes are...

Girl AND Boy

Girl AND Girl

More simply, this question could have been more easily be worded as, "What's the probability ANY child is a girl." Unless you know something I don't, there are only 2 possible outcomes to that question.

You see, in the first problem, the order of events matters in order to attain EQUALLY POSSIBLE statistical events because you have 2 variables. However, in the second problem, order no longer matters because you only have 1 variable (the OTHER child). Whether the OTHER child is second or first is inconsequential. Therefore, the probability is 1/2. If you do care about order still, you would have write that as a probability subset of the first event's probability.

Girl AND Boy = 1/2

Girl THEN Boy = 1/2 of that or 1/4 of total

Boy THEN Girl = 1/2 of that or 1/4 of total

Girl AND Girl = 1/2

This is what others have tried to explain. Once you know that one is a girl, the events are no longer EQUALLY PROBABLE when written as A THEN B. They have to become A AND B. When you do that, things like 1 x 3 and 3 x 1, are obviously repeats and you only account for one.

So...if you know that one die is a 3, then the problem is a totally new problem, unrelated the original 36 outcomes (this is where the false logic comes in). Your new outcome possibilities are...

3 AND 1 (which is comprised of 1 THEN 3 and 3 THEN 1)

3 AND 2 (which is comprised of 2 THEN 3 and 3 THEN 2)

3 AND 3 (which is only comprised of 3 THEN 3)

3 AND 4 (which is comprised of 3 THEN 4 and 4 THEN 3)

3 AND 5 (which is comprised of 3 THEN 5 and 5 THEN 3)

3 AND 6 (which is comprised of 3 THEN 6 and 6 THEN 3)

You now no longer care about order because you only have 1 variable. You want to know what the probability is that you have a 3 AND a 4.

Now, each AND statement has a probability of 1/6 (which should be obvious because there are only 6 EQUALLY PROBABLE events)

However, WITHIN each AND statement, there are 2 EQUALLY PROBABLE subevents, so each of these carries 1/2 of the original event probability.

That means, if you insist on writing ALL 11 events out, you have to associate the correct probability to them. They are not all equally probable.

So

3 THEN 1 = 1/6x1/2 = 1/12

3 THEN 2 = 1/6x1/2 = 1/12

3 THEN 3 = 1/6x1 = 1/6

3 THEN 4 = 1/6x1/2 = 1/12

3 THEN 5 = 1/6x1/2 = 1/12

3 THEN 6 = 1/6x1/2 = 1/12

1 THEN 3 = 1/6x1/2 = 1/12

2 THEN 3 = 1/6x1/2 = 1/12

4 THEN 3 = 1/6x1/2 = 1/12

5 THEN 3 = 1/6x1/2 = 1/12

6 THEN 3 = 1/6x1/2 = 1/12

and the probability of rolling a 7 is the probability of rolling a 3 AND a 4, which is the probability of rolling a 3 THEN a 4 PLUS the probability of rolling a 4 THEN a 3 = 1/12 + 1/12 = 1/6.

This is a proof, not an opinion. It is based on correct logic and statistical methods. Please tell me you all can accept the truth when it's presented, regardless of how many people before you have believed a falacy.

So, Bonanova is correct, there is no problem to solve, other than to answer why so many people get sucked into this blatant misuse of statistics.

[HoustonHokie]

To adapt slightly from another source, "the truth table will set you free". Now I understand. And it appears that the real question is whether subject 7 is removing the 25 cases prematurely.

Take a trip to Vegas and play craps: if you pick up the dice and roll them, all 36 cases are available and only 6 will give you a total of 7 spots showing. But if the dice are already rolled and you're allowed to see one of them (not your choice but someone else's), you're left with 11 cases, 2 of which will give you 7. Which is why you have to place your bets before the dice are rolled - the house is not stupid. [side note: I've never actually been to Vegas - paid just enough attention in statistics class, I guess, to know I'd either lose or be arrested for cheating - so my assumptions on the house's behavior are just that - assumptions]

So a review of the facts is in order. What do we know?

1. The dice have been rolled for all 7 subjects.

"for each of them he rolled a pair of fair dice"

2. The magician supplied hints to the first 6 subjects.

"here's a hint: truthfully, at least one of your dice shows a 6."

3. The magician did not say anything to subject 7.

"before the magician could speak"

4. Subject 7 assumed he knew what the magician would say.

"I know that you're going to tell me some number appears on at least one of my dice."

5. Subject 7 gives an answer without having been supplied a clue.

"So whatever you were going to say, I know most certainly what the probability of my dice totaling 7 is. My answer is 2/11."

So, what can we say now?

1. Subject 7 does not know for certain what the magician will say.

Even if subject 7 is correct and the magician was going to tell him that at least one of his dice shows a n, he does not know the value of n. So there are 6 possible hints the magician could give him, and he doesn't yet know which hint he will receive.

2. All 6 hints are equally likely to be given.

A quick count of the possible values of the dice shows that each value 1-6 appears at least once in 11 of the 36 available cases; therefore, if there is no prejudice on the magician's part, he could give either of the 6 possible hints with equal likelihood.

3. A truth table can be constructed for each of the possible hints.

Here they are:

2-1
3-1
4-1
5-1
[color="#ff0000"]6-1[/color]

1-2
2-1 2-2 2-3 2-4 [color="#ff0000"]2-5[/color] 2-6
3-2
4-2
[color="#ff0000"]5-2[/color]
6-2

1-3
2-3
3-1 3-2 3-3 [color="#ff0000"]3-4[/color] 3-5 3-6
[color="#ff0000"]4-3[/color]
5-3
6-3


1-4
2-4
[color="#ff0000"]3-4[/color]
4-1 4-2 [color="#ff0000"]4-3[/color] 4-4 4-5 4-6
5-4
6-4

1-5
[color="#ff0000"]2-5[/color]
3-5
4-5
5-1 [color="#ff0000"]5-2[/color] 5-3 5-4 5-5 5-6
6-5

[color="#ff0000"]1-6[/color]
2-6
3-6
4-6
5-6
[color="#ff0000"]6-1[/color] 6-2 6-3 6-4 6-5 6-6[/font]

[font="Courier New"]1-1 1-2 1-3 1-4 1-5 [color="#ff0000"]1-6[/color] 

It stands to reason, therefore, that because each hint is equally likely, a new truth table can be constructed by combining the other 6 and keeping track of how many times each combination appears. Take the total number of favorable outcomes divided by the total number of appearing combinations, and you get the probability of subject 7 having a 7 appear on his dice.

2-1(2) 2-2(1) 2-3(2) 2-4(2) [color="#ff0000"]2-5[/color](2) 2-6(2)
3-1(2) 3-2(2) 3-3(1) [color="#ff0000"]3-4[/color](2) 3-5(2) 3-6(2)
4-1(2) 4-2(2) [color="#ff0000"]4-3[/color](2) 4-4(1) 4-5(2) 4-6(2)
5-1(2) [color="#ff0000"]5-2[/color](2) 5-3(2) 5-4(2) 5-5(1) 5-6(2)
[color="#ff0000"]6-1[/color](2) 6-2(2) 6-3(2) 6-4(2) 6-5(2) 6-6(1)[/font]

[font="Courier New"]1-1(1) 1-2(2) 1-3(2) 1-4(2) 1-5(2) [color="#ff0000"]1-6[/color](2) 

In all, we have 66 appearing combinations and 12 favorable outcomes. Dividing by 6 gives 2 favorable outcomes in 11 cases, or a probability of 2/11.

Therefore, the knowledge that a hint is coming is enough to change the probability from 1/6 with no hint to 2/11 with a hint.

I'm sure Paul never had to ask this after his conversion experience, but am I right?

[Guest]

Take a trip to Vegas and play craps: if you pick up the dice and roll them, all 36 cases are available and only 6 will give you a total of 7 spots showing. But if the dice are already rolled and you're allowed to see one of them (not your choice but someone else's), you're left with 11 cases, 2 of which will give you 7. Which is why you have to place your bets before the dice are rolled - the house is not stupid. [side note: I've never actually been to Vegas - paid just enough attention in statistics class, I guess, to know I'd either lose or be arrested for cheating - so my assumptions on the house's behavior are just that - assumptions

The idea that the house doesn't let you see one of the 2 dice because it increases your probability, is correct (unless the total you are seeking is 7). 7 is the one case within dice rolling probability that it doesn't matter whether you have a hint or not, you always have a probability of 1/6.

I'm sure Paul never had to ask this after his conversion experience, but am I right?

I feel like a broken record, and I'm sure if I keep this up I'll get banned, but no, you aren't right, because the probability of getting a 7 without a hint is 1/6 and the probability of getting a 7 with a hint is still, in fact, 1/6, not 2/11.

Edited December 30, 2008 by BeastMaster

[Guest]

Bonanova, I know where others stand, but I'm not quite sure if you're asserting that the 2/11 is correct or if that falacy is part of your riddle.

He asserted 2/11 is correct:

Finally, Scraff has lead the charge for truth justice and ... the correct answer for

One Boy One Girl insofar as establishing the 2/11 probability for Subjects 1-6.

I'm somewhat bothered by your dismissal of myself and a couple others who seem to be the only ones who know statistics well enough to catch this falacy

bonanova explained the answer; why does that bother you? Because he doesn't agree with you? Pretty arrogant of you to claim that only you and a few others "know statistics well enough" when I linked to three math websites that disagree with your answer for this and the Boy/Girl problem, not to mention two moderators on this website which specializes in brain teasers.

You can't dismiss truth just because you don't like it or it doesn't fit into your agenda.

Now your just plain being rude. What agenda are you accusing bonanova of having? Do you really believe he's dismissing your answer for any reason other than he knows it's wrong due to his knowledge of mathematics? Why would he not like the truth regarding the answer to this riddle?

That means, if you insist on writing ALL 11 events out, you have to associate the correct probability to them. They are not all equally probable.

So

3 THEN 1 = 1/6x1/2 = 1/12

3 THEN 2 = 1/6x1/2 = 1/12

3 THEN 3 = 1/6x1 = 1/6

3 THEN 4 = 1/6x1/2 = 1/12

3 THEN 5 = 1/6x1/2 = 1/12

3 THEN 6 = 1/6x1/2 = 1/12

1 THEN 3 = 1/6x1/2 = 1/12

2 THEN 3 = 1/6x1/2 = 1/12

4 THEN 3 = 1/6x1/2 = 1/12

5 THEN 3 = 1/6x1/2 = 1/12

6 THEN 3 = 1/6x1/2 = 1/12

and the probability of rolling a 7 is the probability of rolling a 3 AND a 4, which is the probability of rolling a 3 THEN a 4 PLUS the probability of rolling a 4 THEN a 3 = 1/12 + 1/12 = 1/6.

Totally wrong! Each possibility has an equal probability of what the dice show. You really believe you have twice the probability of rolling "3 THEN 3" than you do of rolling "3 THEN 2"? No way!

The below are the possible rolls of two dice being thrown, all with equal probability.

1 x 1, 1 x 2, 1 x 3, 1 x 4, 1 x 5, 1 x 6

2 x 1, 2 x 2, 2 x 3, 2 x 4, 2 x 5, 2 x 6

3 x 1, 3 x 2, 3 x 3, 3 x 4, 3 x 5, 3 x 6

4 x 1, 4 x 2, 4 x 3, 4 x 4, 4 x 5, 4 x 6

5 x 1, 5 x 2, 5 x 3, 5 x 4, 5 x 5, 5 x 6

6 x 1, 6 x 2, 6 x 3, 6 x 4, 6 x 5, 6 x 6

If you're told at least one of your dice is showing a 3, you have an equal probability of having one of the following dice pairs:

1 x 3

2 x 3

3 x 1, 3 x 2, 3 x 3, 3 x 4, 3 x 5, 3 x 6

4 x 3

5 x 3

6 x 3

There are 11 possible dice rolls.

2 of them add up to 7.

So, Bonanova is correct, there is no problem to solve, other than to answer why so many people get sucked into this blatant misuse of statistics.

You clearly didn't understand what he wrote. There is a problem to solve.

[Guest]

Take a trip to Vegas and play craps: if you pick up the dice and roll them, all 36 cases are available and only 6 will give you a total of 7 spots showing. But if the dice are already rolled and you're allowed to see one of them (not your choice but someone else's), you're left with 11 cases, 2 of which will give you 7.

Wrong. You are left with 6 cases, one of which will add up to seven. Let's say the dice you see is a 5, the six possible combinations are:

5-1, 5-2, 5-3, 5-4, 5-5, 5-6

We don't include 1, 5, 2, 5, etc. because we already see a particular die.

If both dice are rolled and we're told at least one is a 5, then there are 11 equal possibilities:

5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 5-1, 5-2, 5-3, 5-4, 5-6

5,5 is not counted twice as there's only one possible way to roll a 5, 5. The first die must be a 5 (1 out of 6) and so must the second (1 out of 6). 6x6=36

There are two possible ways to roll a 5 and a 1. The first die must be a 1 or a 5 (1 out of 3) and the second must be a 1 if a 5 has been revealed or a 5 if a 1 has been revealed ((1 out of 6). 3x6=18

As you can see in the probabilities below, there is only on 5, 5:

1 x 1, 1 x 2, 1 x 3, 1 x 4, 1 x 5, 1 x 6

2 x 1, 2 x 2, 2 x 3, 2 x 4, 2 x 5, 2 x 6

3 x 1, 3 x 2, 3 x 3, 3 x 4, 3 x 5, 3 x 6

4 x 1, 4 x 2, 4 x 3, 4 x 4, 4 x 5, 4 x 6

5 x 1, 5 x 2, 5 x 3, 5 x 4, 5 x 5, 5 x 6

6 x 1, 6 x 2, 6 x 3, 6 x 4, 6 x 5, 6 x 6

Again, this is similar to the Boy/Girl problem. If one child randomly walks out of the house and is a girl, there is a 1/2 probability that the other is a girl. You already see one, so the other has an equal probability of being a boy or a girl. If you're told at least one of two children is a girl, there are three possible combinations:

GG

GB

BG

The probability that both children are girls is 1/3.

Anyone that doesn't believe it can try Martini's challenge on post #31 in the Boy/Girl thread.

[HoustonHokie]

Wrong. You are left with 6 cases, one of which will add up to seven. Let's say the dice you see is a 5, the six possible combinations are:

5-1, 5-2, 5-3, 5-4, 5-5, 5-6

We don't include 1, 5, 2, 5, etc. because we already see a particular die.

If both dice are rolled and we're told at least one is a 5, then there are 11 equal possibilities:

5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 5-1, 5-2, 5-3, 5-4, 5-6

5,5 is not counted twice as there's only one possible way to roll a 5, 5. The first die must be a 5 (1 out of 6) and so must the second (1 out of 6). 6x6=36

There are two possible ways to roll a 5 and a 1. The first die must be a 1 or a 5 (1 out of 3) and the second must be a 1 if a 5 has been revealed or a 5 if a 1 has been revealed ((1 out of 6). 3x6=18

As you can see in the probabilities below, there is only on 5, 5:

1 x 1, 1 x 2, 1 x 3, 1 x 4, 1 x 5, 1 x 6

2 x 1, 2 x 2, 2 x 3, 2 x 4, 2 x 5, 2 x 6

3 x 1, 3 x 2, 3 x 3, 3 x 4, 3 x 5, 3 x 6

4 x 1, 4 x 2, 4 x 3, 4 x 4, 4 x 5, 4 x 6

5 x 1, 5 x 2, 5 x 3, 5 x 4, 5 x 5, 5 x 6

6 x 1, 6 x 2, 6 x 3, 6 x 4, 6 x 5, 6 x 6

Again, this is similar to the Boy/Girl problem. If one child randomly walks out of the house and is a girl, there is a 1/2 probability that the other is a girl. You already see one, so the other has an equal probability of being a boy or a girl. If you're told at least one of two children is a girl, there are three possible combinations:

GG

GB

BG

The probability that both children are girls is 1/3.

Anyone that doesn't believe it can try Martini's challenge on post #31 in the Boy/Girl thread.

So it's the being told, but not actually seeing, that makes the difference...

[Guest]

So it's the being told, but not actually seeing, that makes the difference...

It's actually the difference between knowing specifics.

If you throw two coins in the air, there are four possible ways they can land:

HH

TT

HT

TH

If I tell you the one further left is showing heads, there are now two equal possible combinations:

HH

HT

There is a 1/2 probability of having two heads.

If I tell you at least one landed showing heads, there are three possible combinations:

HH

HT

TH

There is a 1/3 probability of having two heads.

In your dice example, you said, "you're allowed to see one of them". In that case, it's a particular die we have knowledge of.

Again, anyone who disagrees can try the coin experiment. Lay out 100 pairs of pennies. If I tell you I flipped two coins and the one to the left is showing heads, then to simulate the probability of what the other is, remove all coin pairs that are showing a tails on the left (you will eliminate about 50 coin pairs). The coin to the right will be showing heads about 25 times (1/2 probability of having HH).

If I tell you I flipped two coins and the one of the two is showing heads, then to simulate the probability of what the other is, remove all coin pairs that are showing two tails (you will eliminate about 25 coin pairs). Of the 75 coin pairs left, about 25 of them will be HH (about 25 will be HT and about 25 will be TH) and you'll see there is a 1/3 probability of having HH.

[bonanova]

Bonanova, I know where others stand, but I'm not quite sure if you're asserting that the 2/11 is correct or if that falacy is part of your riddle. I'm somewhat bothered by your dismissal of myself and a couple others who seem to be the only ones who know statistics well enough to catch this falacy, unless of course your intention is to get as many people to chase the white rabbit as possible. You can't dismiss truth just because you don't like it or it doesn't fit into your agenda. Please consider the possibility that your riddle is flawed. I'm not trying to be confontational, I'm just trying to stand up for truth, even if it is a truth as insignificant as this (afterall, this is just a fun site for riddles).

Not that anyone will listen, but here is my final attempt to show you why the 2 children riddle was flawed, and, by extension the die riddle.

If there are 2 children and you don't know either gender, there are 4 EQUALLY possible outcomes. These outcomes are...

Boy THEN Boy

Boy THEN Girl

Girl THEN Boy

Girl THEN Girl

Therefore, the probability of Girl THEN Girl is 1/4

If you know that one is a girl, you can't simply remove the Boy THEN Boy scenario and say that the new probability is 1/3. If you know one is a girl, you have a totally new problem. The new problem is,

If there are 2 children and one of them is a girl, what is the probability that they are both girls.

For this there are only 2 EQUALLY possible outcomes. These outcomes are...

Girl AND Boy

Girl AND Girl

More simply, this question could have been more easily be worded as, "What's the probability ANY child is a girl." Unless you know something I don't, there are only 2 possible outcomes to that question.

You see, in the first problem, the order of events matters in order to attain EQUALLY POSSIBLE statistical events because you have 2 variables. However, in the second problem, order no longer matters because you only have 1 variable (the OTHER child). Whether the OTHER child is second or first is inconsequential. Therefore, the probability is 1/2. If you do care about order still, you would have write that as a probability subset of the first event's probability.

Girl AND Boy = 1/2

Girl THEN Boy = 1/2 of that or 1/4 of total

Boy THEN Girl = 1/2 of that or 1/4 of total

Girl AND Girl = 1/2

This is what others have tried to explain. Once you know that one is a girl, the events are no longer EQUALLY PROBABLE when written as A THEN B. They have to become A AND B. When you do that, things like 1 x 3 and 3 x 1, are obviously repeats and you only account for one.

So...if you know that one die is a 3, then the problem is a totally new problem, unrelated the original 36 outcomes (this is where the false logic comes in). Your new outcome possibilities are...

3 AND 1 (which is comprised of 1 THEN 3 and 3 THEN 1)

3 AND 2 (which is comprised of 2 THEN 3 and 3 THEN 2)

3 AND 3 (which is only comprised of 3 THEN 3)

3 AND 4 (which is comprised of 3 THEN 4 and 4 THEN 3)

3 AND 5 (which is comprised of 3 THEN 5 and 5 THEN 3)

3 AND 6 (which is comprised of 3 THEN 6 and 6 THEN 3)

You now no longer care about order because you only have 1 variable. You want to know what the probability is that you have a 3 AND a 4.

Now, each AND statement has a probability of 1/6 (which should be obvious because there are only 6 EQUALLY PROBABLE events)

However, WITHIN each AND statement, there are 2 EQUALLY PROBABLE subevents, so each of these carries 1/2 of the original event probability.

That means, if you insist on writing ALL 11 events out, you have to associate the correct probability to them. They are not all equally probable.

So

3 THEN 1 = 1/6x1/2 = 1/12

3 THEN 2 = 1/6x1/2 = 1/12

3 THEN 3 = 1/6x1 = 1/6

3 THEN 4 = 1/6x1/2 = 1/12

3 THEN 5 = 1/6x1/2 = 1/12

3 THEN 6 = 1/6x1/2 = 1/12

1 THEN 3 = 1/6x1/2 = 1/12

2 THEN 3 = 1/6x1/2 = 1/12

4 THEN 3 = 1/6x1/2 = 1/12

5 THEN 3 = 1/6x1/2 = 1/12

6 THEN 3 = 1/6x1/2 = 1/12

and the probability of rolling a 7 is the probability of rolling a 3 AND a 4, which is the probability of rolling a 3 THEN a 4 PLUS the probability of rolling a 4 THEN a 3 = 1/12 + 1/12 = 1/6.

This is a proof, not an opinion. It is based on correct logic and statistical methods. Please tell me you all can accept the truth when it's presented, regardless of how many people before you have believed a falacy.

So, Bonanova is correct, there is no problem to solve, other than to answer why so many people get sucked into this blatant misuse of statistics.

You ask why I don't say whether 2/11 is correct for subjects 1-6. I did make a case; but it's in a spoiler, since it's part of the puzzle.

The puzzle is solved either of two ways:

[1] Prove that p[7] = 1/6 for all seven subjects.

[2] Either

(a) Give up the notion that Subject 7's p[7] has to be 1/6, or

(b) Debunk S7's argument.

You took approach [1], giving a proof [one that satisfies you] that p[7] = 1/6 for all seven subjects.

You've finished the puzzle.

If I was perceived to put you [or anyone else] down, that's not ever my intent; I apologize.

I believe it's within the bounds of respect for a person to point out what might be

a weak point in an argument, but certainly I will not impugn anyone's intelligence.

I reserve for myself the right to make a mistake; and, frankly, I exercise that right fairly regularly.

So I this give that same latitude to others.

[Guest]

The puzzle is solved either of two ways:

[1] Prove that p[7] = 1/6 for all seven subjects.

[2] Either

(a) Give up the notion that Subject 7's p[7] has to be 1/6, or

(b) Debunk S7's argument.

I would like to then solve the puzzle by taking approach 2(b).

Subjects 1-6 were actually given hints that changed the possible outcomes of their dice, and thus changed each subject's probability of having 7 to 2/11. Subject 7, however, having NOT received a hint (yet, or at all, it doesn't matter), still has the original 1/6 probability of having 7. If he is also given a hint as to "at least one of your dice is X," his probability would then change to 2/11 just like all the others.

Can this now be laid to rest? <_<

[bonanova]

I would like to then solve the puzzle by taking approach 2(b).

Subjects 1-6 were actually given hints that changed the possible outcomes of their dice, and thus changed each subject's probability of having 7 to 2/11. Subject 7, however, having NOT received a hint (yet, or at all, it doesn't matter), still has the original 1/6 probability of having 7. If he is also given a hint as to "at least one of your dice is X," his probability would then change to 2/11 just like all the others.

Can this now be laid to rest? <_<

Sure. We can stop whenever.

But did we debunk his argument?

I'm not just trolling for more posts: I find it persuasive. I don't have a tightly-worded rebuttal.

I paraphrase your answer as [he was given no clue] therefore [1/6 cannot change.]

That ignores his argument. It says, since my argument is right and we differ, you must be wrong.

If [he had heard nothing] then clearly [he was given no clue.] But [he heard the first six conversations].

I am trying to prove, to my satisfaction, that [hearing them] does [not lead to a clue].

I haven't succeeded yet.

It's interesting, probably pivotal, that had he [not heard anything] he could nevertheless reason exactly the same!

[HoustonHokie]

If [he had heard nothing] then clearly [he was given no clue.] But [he heard the first six conversations].

I am trying to prove, to my satisfaction, that [hearing them] does [not lead to a clue].

I haven't succeeded yet.

It's interesting, probably pivotal, that had he [not heard anything] he could nevertheless reason exactly the same!

Question: if he [expects to receive a clue] is that the same as [receiving a clue]?

I think my truth tables showed that as long as his expectation pans out and he receives a clue of the same form as the others, he can say his probability is 2/11 without even knowing the specific value to be revealed. Otherwise - either because you can argue that the form of or future presence of a clue is not guaranteed until actually provided (or at least promised) - he would have to say 1/6 until receiving the actual clue (or the promise of one). In other words, if the magician had stated up front that everyone would be told one of the two values of their dice, they would all be able to say 2/11 and we'd only be arguing the point from the boy/girl problem. No such promise was made, so the argument we must also wrestle with is whether it is reasonable for subject 7 to assume that the pattern of clue-giving would continue.

I think should be obvious that if he had [not heard anything] he would have had no expectation of receiving a clue and could not venture any guess except 1/6. Similarly, if the magician had told each of the previous 6 subjects that the value of a particular die was n, he (like they) would be forced to say the probability is 1/6.

I realize that I am probably arguing out of both sides of my mouth but that's because I think we've left the realm of probability/statistics and moved into human behavior with the subleties of whether and how a clue would be provided, which, in my opinion, is the determining factor in choosing between 1/6 and 2/11 as the appropriate answer for subject 7. That depends on a prediction of the behavior of the magician, and we are given limited data upon which to base our assessment of his future behavior.

[Guest]

It is quite the quandary, for while the 7th subject has not been given a specific clue, he has heard what the magician told the other 6 subjects, and can reasonably assume that out of his two dice, "at least one of them is X," where the value of X could very well be any value 1-6, even if he is not told what X is, he can still reason the probability to be 2/11 for all X. But, for every different value of X, there is a different set of 11 possibilities, with two possibilities overlapping any for any 2 values of X, so I think that without knowing exactly what X is, he can only logically reason that the probability of having 7 is 1/6 from the entire set of 36 possibilities. I have no more definitive proof of this, however, as this puzzle has cramped my brain.

[Guest]

It is quite the quandary, for while the 7th subject has not been given a specific clue, he has heard what the magician told the other 6 subjects, and can reasonably assume that out of his two dice, "at least one of them is X," where the value of X could very well be any value 1-6, even if he is not told what X is, he can still reason the probability to be 2/11 for all X. But, for every different value of X, there is a different set of 11 possibilities, with two possibilities overlapping any for any 2 values of X, so I think that without knowing exactly what X is, he can only logically reason that the probability of having 7 is 1/6 from the entire set of 36 possibilities. I have no more definitive proof of this, however, as this puzzle has cramped my brain.

I think that nails it, actually.

To restate lazyboy (accurately, I hope): It's knowing specifically which possibilities can be eliminated that changes the game. Merely knowing that possibilities can be eliminated if given specific information doesn't change anything by itself. In fact, any potential advantage each permutation would give is canceled by the others when taken as a whole, which it must be, until the information is given.

I'm sure I'll re-read that in the morning and find tons wrong with it, but for now it seems to work.