While that's fine for those who know the notation, it's not very accessible for those who don't. I'm hoping for intuitive solutions that make the head-scratching readers say "Ah. I get it now."

And ... partly because I spent an inordinate amount of time - at an uncivilized hour of the night - producing intuitive visualizations that don't require more than a very basic understanding of probability.

Or this??

To clarify: an astonishing number of people - myself included initially - assumed that the problem reduces to this logic:

You are told one die is a six. Therefore its chance of being six is 1. They are independent events, therefore that die's result is done. The only calculation left to do is the other die. They are independent events, so the 2nd die is simply 1:6. So: the chances of a double 6 are 1:6.

Or, even more simply, this:

Colour the dice red and blue.
If the red turns up 6, then there are 6 options for the blue die.
If the blue turns up 6, then there are 6 options for the red die.
Thus: 12 possible states. 12:36 = 1:6.

(You and I know they contain flawed logic, but my goodness it's hard to point out the flaw to others!)

Like this??

Or this??

Like this??

Or this??

For example, if the instructions to the reporter are as follows:

If the roll includes at least one "1", then report that "at least one of them is a 1".
Otherwise, if the roll includes at least one "2", then report that "at least one of them is a 2".
Otherwise, if the roll includes at least one "3", then report that "at least one of them is a 3".
Otherwise, if the roll includes at least one "4", then report that "at least one of them is a 4".
Otherwise, if the roll includes at least one "5", then report that "at least one of them is a 5".
Otherwise, if the roll includes at least one "6", then report that "at least one of them is a 6".

No. @Ibix understood.
The phrasing of the report is not the issue.
The issue is under what conditions the report is made.
For example, if the instructions to the reporter are as follows:

If the roll includes at least one "1", then report that "at least one of them is a 1".
Otherwise, if the roll includes at least one "2", then report that "at least one of them is a 2".
Otherwise, if the roll includes at least one "3", then report that "at least one of them is a 3".
Otherwise, if the roll includes at least one "4", then report that "at least one of them is a 4".
Otherwise, if the roll includes at least one "5", then report that "at least one of them is a 5".
Otherwise, if the roll includes at least one "6", then report that "at least one of them is a 6".

The most straight forward way I can think of explaining this is as follow:

The die are rolled yielding one of 36 results:
A) In 25 cases, both dice show 1 to 5, and no report is made.
B) In 5 cases, this first dice is 1 to 5, the second dice is 6, and the report is made.
C) In 5 cases, this second dice is 1 to 5, the first dice is 6, and the report is made.
D) In 1 case, both die are 6 and the report is made.

If a report is made, then:
the 25 "A" cases are eliminated.
In the remaining 11 cases, only 1 has both die as 6.

Hence 1/11

(but I am probably a bad jusge of "simple".)

One roll of two dice. Full stop.

Yes. We take them to be honest.

They have seen the result. They tell you enough that you can deduce two things:
1] One is a six.
2] The other one might be a six.

Your job is to determine - first how many possibilities are valid, of the total 36 possible configurations - and subsequently, the odds of a double six, (but that's the easy part).

1 6
2 6
3 6
4 6
5 6
6 6
6 5
6 4
6 3
6 2
6 1

This is the only way I know to solve problems like this. Trying to do it operationally or intuitively is too tricky.

I've done calculations of odds with bridge hands which are similar but with larger numbers. Something you see all the time is the opponents have an even number of cards split between two hands. Let's say six. Then a 4-2 split is about twice as likely as 3-3 since there are two ways it can happen vs. one.

Or there's craps. The dice total 4. Rolling 3-1 or 1-3 is twice as likely as 2-2. 2-2 is called "the hard way."

The whole thrust of probability is to get a static timeless situation instead of thinking of it as a process. Get get a set of all possibilities. That set has measure one. Then find the measures of subsets.

Another way too put it: Additional information reduces the set of possibilities, and there is no reason to assume that the probability distributions in the reduced set are the same as in the original full set.

I think the problem is not accepting / understanding conditional probability:
https://en.wikipedia.org/wiki/Conditional_probability
https://en.wikipedia.org/wiki/Bayes'_theorem

More specifically the fallacy, that just because some outcomes are independent of each other, their probability distributions remain independent, no matter what additional information about their outcomes becomes avaible. See also the Monty Hall problem:
https://en.wikipedia.org/wiki/Monty_Hall_problem

If the dice land 1 and 6, say, then there is a 50-50 chance you are told "at least one die is 1" and a 50-50 chance you are told "at least one die is a 6". The same with 2 and 6 etc. From these 10 outcomes, you will be told "at least on is a 6" 5 times.

If the dice land double 6, then you are always told "at least one is a 6".

Therefore, 6 times out of 36 you are given this information and only once is it a double six.

Alternatively, the answer could be 0. "At least" could be used in the sense "at least I didn't get double 1". I.e. in the sense that things could be worse. In this case, you would be told if both were six. As in, "at last I got double 6".

HH
HT
TH

Now suppose after every such trial one is told either "at least one of them is heads" or "at least one of them is tails". The answer depends on whatever algorithm the announcer uses. Looking back at the first problem we implicitly assumed that if the result was TT that the trial would be thrown out or something like that.

You already have alternate scenarios from @Ibix , myself, and now @PeroK with this:

In this case, it's not clear why someone would tell you there is at least one six. On what basis is this information being supplied? Why didn't they say something like "neither die is a one"? Or, "both dice show six"?

Instead, if you ask a direct question, such as "do you have at least one six", then you know why they gave this information and you can proceed to the 1/11 solution. Otherwise, you're misapplying probability theory. In this case, by making an assumption that appears to be inferred from the question, but is actually not justified.

Note: not trusting the information is a separate issue.

Suppose you repeat the experiment and they say "this time I have two sixes". That throws out your assumption and breaks your solution.

That being said, one could easily avoid the "you are told..." part by simply letting just one die roll out of view, while the still visible one shows a six. If that was the intended meaning.