Two dice are rolled...

[DaveC426913]

I'm putting this is the Fun 'n Games forum because if I put it in the Math forum I'd get answers in the form of formal probability notation, with elements and sets and unions and intersections.

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.

 

[.Scott]

I think you also need to say that if either of them are 6, you would be told that. Otherwise, there is not enough information.
You haven't asked for spoiler's, but I'll do that anyway.

Spoiler

When the die are tossed, there are 36 possible combinations. 11 of them will result in the "one of them is a 6" report. Of those 11 reports, only one of them is a double 6. So the chance that both are sizes is 1/11.

 

[DaveC426913]

I think you also need to say that if either of them are 6, you would be told that. Otherwise, there is not enough information.

Like this??

Or this??

You haven't asked for spoiler's, but I'll do that anyway.

You got the answer of course, but I was hoping for answers that show, as clearly and concisely as possible, how to get there (for those not familiar with probability notation).



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!)

 

[Ibix]

Like this??

Or this??

I think @.Scott's point is that you must always be told if there's a six, or at least you have to be able to assume no bias. If there's bias in it (e.g., the die-thrower only says there's at least one six if it was on the red die not the blue die, or only if it's actually a double six, or something) your restriction is still satisfied but the odds might shift.

Spoiler

There are 36 equiprobable outcomes of rolling two dice:$$\begin{array}{|c|c|c|c|c|c|}
\hline
1,1&1,2&1,3&1,4&1,5&1,6\\
\hline
2,1&2,2&2,3&2,4&2,5&2,6\\
\hline
3,1&3,2&3,3&3,4&3,5&3,6\\
\hline
4,1&4,2&4,3&4,4&4,5&4,6\\
\hline
5,1&5,2&5,3&5,4&5,5&5,6\\
\hline
6,1&6,2&6,3&6,4&6,5&6,6\\
\hline
\end{array}$$The constraint that "at least one was a six" tells you that the outcome was one of the eleven cells with a six (the right hand column or bottom row). Only one of them is a double six, so the chance of the outcome is 1/11 (given the "no bias" assumption).

 

[A.T.]

12:36 = 1:6

?

 

[DaveC426913]

?

Indeed. That was a mistake on my part. Plz ignore.

 

[A.T.]

I was hoping for answers that show, as clearly and concisely as possible

These are somewhat opposed goals. The clearest way to show it is the least concise brute force method of listing all outcomes and counting (see post #4). People playing board games know this table well.

 

[.Scott]

Like this??

Or this??

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".

In that above case, the answer would be different.
The most straight forward way I can think of explaining this is as follow:

Spoiler

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".)

 

[Ibix]

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".

That's much better than my slightly malicious ideas - it's systematic, and gives a 100% chance of a double six given that the reporter says "at least one six", because a six and anything else would be reported as "at least one of the 'anything else'".

 

[Hornbein]

I'd say it is one in eleven. That's the number of possibilities and all are equally likely.

 

[DaveC426913]

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".

I don't really get how you interpret it that way; it seems overly pedantic, but I allow the possibility that what I take for granted is not what you take for granted.

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

Spoiler

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".)

Yes. That is a third way. Eliminate all the invalid possibilities and you are left with an unequivocal answer. This solution has the advantage of competely side-stepping a "double-dipping" pitfall.

 

[DaveC426913]

I'd say it is one in eleven. That's the number of possibilities and all are equally likely.

Correct. But can you explain it to someone who might struggle with complex probabilities?

 

[.Scott]

I don't really get how you interpret it that way; it seems overly pedantic, but I allow the possibility that what I take for granted is not what you take for granted.

You're presuming it's someone who will say "at least one of them is a six" each time a die roll shows a six.
But, I suspect it's a salesman too busy showing shoes to a potential customer to pay any attention to the die roll.

 

[DaveC426913]

You're presuming it's someone who will say "at least one of them is a six" each time a die roll shows a six.

No! There's no "each time".

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).

 

[Hornbein]

Correct. But can you explain it to someone who might struggle with complex probabilities?

There are eleven possibilities, each equally likely.

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.

 

[A.T.]

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

In this particular case it is easy to show that, given the additional information "at least one 6", the probability distributions of the two dice are not independent of each other. For example, if you are shown one of the dice and it's not a 6, then you know that the other one must be a 6. So you can infer the outcome of one die from the outcome of the other, demonstrating that their probability distributions are not independent.

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

 

[PeroK]

The answer could be 1/6. There are 36 equally likely outcomes with two dice, 11 of which include at least one 6.

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".

 

[A.T.]

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"

Yes, "At least one of them is a six" (emphasis on "one"), could also mean that only one is a six. But that would make the question pointless.

 

[Hornbein]

A simpler example is flipping two coins. If they tell you "at least one of them is heads" then there is a 1/3 chance that both are heads.

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.

 

[.Scott]

No! There's no "each time".

You don't have to use the notion of "each time", but it isn't overly pedantic to describe exactly what conditions will result in the "at least one is a six" report .. it's part of the problem statement and it is normal to be that "pedantic" when stating logic problems.

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

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 one is a 6" 5 times.

When I used the term "each time", I was not saying that there was more than one time. I was using the term to describe how the reporter decided to make a report. And I was emphasizing that his intention to make the report was preexisting (ie, before the die had been rolled). This is very important. If his decision on how to make a report was not based exclusively on things that happened before he looked at the dice, then there is probably no pragmatic way of determining whether he would have reacted to a "6,1" in the same way as a "6,2" or a "2,6", or a "6,6".

 

[A.T.]

... how the reporter decided to make a report...

The idea that the reporter gets to decide anything didn't even occur to me, because it obviously makes the whole question unanswerable, as he could just as well decide to lie to you.

 

[DaveC426913]

The idea that the reporter gets to decide anything didn't even occur to me, because it obviously makes the whole question unanswerable, as he could just as well decide to lie to you.

Exactly.
The premise is thatcitcis not a trick question.

 

[PeroK]

Exactly.
The premise is thatcitcis not a trick question.

You're missing the point. This is a classic misuse of probability theory. In order to use probability theory, you need to understand how the information arises.

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.

 

[PeroK]

The idea that the reporter gets to decide anything didn't even occur to me,

Precisely. You made an unjustified assumption. See previous post.

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

 

[A.T.]

The idea that the reporter gets to decide anything didn't even occur to me, because it obviously makes the whole question unanswerable, as he could just as well decide to lie to you.

Precisely. You made an unjustified assumption.

The assumption that it's a question which has an answer? Yes, I made that assumption.

 

[A.T.]

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.

You made the unjustified assumption that the reporter is honest.

 

[PeroK]

The assumption that it's a question which has an answer? Yes, I made that assumption.

I guess that's the nearest you'll ever come to admitting you're wrong!

 

[PeroK]

You made the unjustified assumption that the reporter is honest.

We can add that assumption. What we can't do is easily add an assumption about what the reporter will and will not say on each and every throw of the dice.

 

[A.T.]

You made the unjustified assumption that the reporter is honest.

We can add that assumption.

I guess that's the nearest you'll ever come to admitting you're wrong!

 

[A.T.]

Exactly. The premise is that it is not a trick question.

Some people just prefer to pick the question apart, when answering the most likely meaning is too boring for them. For example, what does "fair, six-sided dice" mean? A six-sided die with three sixes and three ones would also be fair. You can turn any simple question into a pages long problem statement, and even that will not include all assumptions.

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.