Two dice are rolled...

[Hornbein]

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.

This sort of thing seems like writing a Lean proof in mathematics, something I will never do.

 

[PeroK]

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

This is similar to the two child problem, where instead of asking questions, the problem is changed and the information is volunteered - which changes the problem fundamentally. There is plenty of material about this online. This problem is similar and highlights a pitfall in applying probability theory too simplistically.

Ten years ago, I would have given your answer to this problem. But, then, someone pointed out the flaw and I changed my mind and realised that the data must come from direct questions. That was on this forum.

I do admit when I'm wrong. Many times on this forum. Including when I changed my mind on this problem.

Whereas, you double down and refuse to accept that you have been caught in this well-known trap.

 

[PeroK]

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.

That's definitely 1/6. Once you identify a particular die, that changes the question. The second die is independent.

 

[A.T.]

This is similar to the two child problem, where instead of asking questions, the problem is changed and the information is volunteered - which changes the problem fundamentally.

Only if you take the "you are told ..." part literally. But since then the question becomes unanswerable it's pointless to consider that interpretation further.

But if we assume the question is intended to have an answer, then we can interpret "you are told ..." as "you know", which makes the question answerable.

 

[A.T.]

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.

That's definitely 1/6. Once you identify a particular die, that changes the question. The second die is independent.

I didn't identify a particular die.

 

[PeroK]

For anyone interested, here is Wikipedia on the related two-child problem. It shows the importance of being precise about how the information has been obtained:

https://en.wikipedia.org/wiki/Boy_or_girl_paradox

 

[PeroK]

For anyone interested, here is Wikipedia on the related two-child problem. It shows the importance of being precise about how the information has been obtained:

https://en.wikipedia.org/wiki/Boy_or_girl_paradox

PS The Wikipedia article really sharpens up the point I was trying to make. We assume that the reporter has the following process: they look at the first die and, whatever it is, they say "I have at least one X".

In that case, it's a perfectly valid assumption about how the data is generated, and the probability that the second die is a six is clearly 1/6.

That's not complicating things or picking the problem apart in a negative way. It's a simple way to show that 1/11 cannot be the unambiguous answer.

 

[A.T.]

We assume that the reporter has the following process: they look at the first die and, whatever it is, they say "I have at least one X".

In that case, it's a perfectly valid assumption about how the data is generated, and the probability that the second die is a six is clearly 1/6.

It's not clear. It depends on how you interpret "first die".
- If the dice are distinguishable and the reporter always reports the same specific die, then it's 1/6
- If the reporter reports the first die they happen to find, but they are indistinguishable, or he simply doesn't tell you which one it was, then it's 1/11.

 

[DaveC426913]

You made the unjustified assumption that the reporter is honest.

I believe it is justified.

If, in a problem simple as this, you cannot trust what are told, then it is unanswerable. They might as well be rolling lemons or washing masodons for all you know.

It's not a trick question.

 

[DaveC426913]

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 memeaning.

That is a different problem.

 

[DaveC426913]

It's not clear. It depends on how you interpret "first die".
- If the dice are distinguishable and the reporter always reports the same specific die, then it's 1/6
- If the reporter reports the first die they happen to find, but they are indistinguishable,

or he simply doesn't tell you which one it was, then it's 1/11.

(I know we're essentially meta-analyzing now, as long as everybody agrees that the answer to the problem as intended is 1-in-11 and not 1-in-6).

 

[A.T.]

If, in a problem simple as this, you cannot trust what are told, then it is unanswerable.

I agree. I was just demonstrating where questioning any assumption leads to. You have to read in context.

That is a different problem.

Different than what? If the dice are indistinguishable, so you don't know which one you still see showing six, then it is the same as your interpretation of the original problem: You know that at least one is six, but not which, so 11 possibilities remain.

 

[Ibix]

If the dice are indistinguishable, so you don't know which one you still see showing six, then it is the same as you interpretation of the original problem: You know that at least one is six, but not which, so 11 possibilities remain.

The process you seem to be describing is:

1. Roll two dice.
2. Choose one at random to be visible.
3. If it shows a six, say "at least one is a six". Otherwise, skip step 4.
4. What are the chances that the other one shows a six given that you said something at step 3?

...to which the answer is 1/6. Reposting my table from earlier:$$\begin{array}{|c|c|c|c|c|c|}
\hline
1,1&1,2&1,3&1,4&1,5&1,6\\
\hline
2,1&2,2&2,3&2,4&2,5&2,6\\
\hline
3,1&3,2&3,3&3,4&3,5&3,6\\
\hline
4,1&4,2&4,3&4,4&4,5&4,6\\
\hline
5,1&5,2&5,3&5,4&5,5&5,6\\
\hline
6,1&6,2&6,3&6,4&6,5&6,6\\
\hline
\end{array}$$At step 3, you know you are in the right hand column or the bottom row (i.e. two distinct cases, not the union of the two), and hence the probability of both being a six is 1/6. Unless you are describing some other process which I haven't understood.

 

[A.T.]

At step 3, you know you are in the right hand column or the bottom row (i.e. two distinct cases, not the union of the two)

This makes no sense. The operator OR applied to two sets yields the union of the two sets.

 

[Ibix]

This makes no sense. The operator OR applied to two sets yields the union of the two sets.

You are in one set or the other with 50% probability each, not the union of the two with 100% probability (which is the assumption implicit in Dave's formulation in the OP).

Here's some python code:

import random

# Some index values
TWO_SIXES = 2
SIX_ON_VISIBLE_ONLY = 1
NO_SIX_ON_VISIBLE = 0

# One run of the experiment 
def dice():
    # Roll two dice...
    d = [random.randint(1,6), random.randint(1,6)]
    # ...pick which one we can see...
    visible = random.randint(0,1)
    # ...and classify the result
    if d[visible] == 6:
        if d[1 - visible] == 6:
            return TWO_SIXES
        return SIX_ON_VISIBLE_ONLY
    return NO_SIX_ON_VISIBLE

# Run the experiment many times and log results
counters = [0,0,0]
for i in range(100000):
     counters[dice()] += 1
atLeastOneSix = counters[SIX_ON_VISIBLE_ONLY] + counters[TWO_SIXES]

# Output results
print("No six on visible die:", counters[NO_SIX_ON_VISIBLE])
print("'At least one six':", atLeastOneSix)
print("Two sixes:", counters[TWO_SIXES])
print("Empirical probability:", counters[TWO_SIXES] / atLeastOneSix)

On my run I got an empirical probability of 0.161

 

[A.T.]

You are in one set or the other with 50% probability each, not the union of the two with 100% probability (which is the assumption implicit in Dave's formulation in the OP).

Here's some python code:

Thanks, I think now I see it: My attempt to simplify the problem statement does indeed change it, because now you miss all the reports when the invisible die is six.

 

[PeroK]

(I know we're essentially meta-analyzing now, as long as everybody agrees that the answer to the problem as intended is 1-in-11 and not 1-in-6).

Absolutely not. The problem is ambiguous (as accepted by the mathematician who first formulated the two-child problem - see the Wikipedia page). In general, in these sorts of problems, the answer depends not only on the data, but on how the data was obtained.

In order to get 1/11, you need something like the follow instructions.

Look at both dice and report whether there is at least one six. Report either "there are no sixes" or "there is at least one six".

It's crucial that whoever rolls the dice looks at them both and is only interested in sixes.

Even inferring the latter from a single trial is a fallacy. You would need to have this agreed and explicitly stated in the problem statement.

 

[PeroK]

To highlight the issue, let's consider the following sequence of trials:

At least one is a six;
At least one is a three;
At least one is a four;
At least one is a five;
At least one is a three:
At least one is a one ...

If we take each trial separately and calculate the probability of a double-six in the first case is 1/11; a double-three in the second case is 1/11 etc. Then, in general, the probability of a double (both dice showing the same) is only 1/11. Whereas, it is clearly 1/6. That, therefore, must be the correct probability in each individual case.

Then, the problem is to figure out why 1/11 is wrong!

 

[.Scott]

Some people just prefer to pick the question apart, when answering the most likely meaning is too boring for them.
...
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.

In post 2, I deduced the intent of the question based on the fact that otherwise, there was not enough information.

People who are not ready to pick apart the question will have difficulty with the well-known "Two coins add up to 30 cents and one is not a nickel" riddle.

More broadly, as a software engineer I need to be wary of the "Garbage-In-Garbage-Out" (GIGO) syndrome - and knowing the provenance of the data is part of that. Certainly, any time that any scientist or engineer is attempting to draw statistical inferences from their data, they need to know where that data came from and how it was collected.

So, thanks @DaveC426913 , you've found a really great example!!!

Continuing...
I just had a chance to catch up on my read-through of this thread.
I took the question to be: Dave believes there is only one correct answer to this problem. What is that answer?
I suppose when you've been coding against ambiguous requirements for half a century, you do that instinctively.

 

[DaveC426913]

I took the question to be: Dave believes there is only one correct answer to this problem. What is that answer?

No. My belief is thus: there is a best answer if we make some reasonable assumptions, and don't make unreasonable ones.

The debate then becomes

• which ones are reasonable (eg. the reporter does not lie),
• which ones are unreasonable (eg. the supposed ambiguous definition of fair die).

If one is attempting to answer the original question as stated, it is reasonable (IMO) to assume, for example, there is only one die roll of two dice and then the result is reported. (Attempting to address PeroK's criticism, above)



By the way, I deliberately put this in GD Fun & Games - not in math - for this very reason (to frame it as a non-rigorously mathematical puzzle).


But I do acknowledge that - on a STEM forum such as PF - "reasonable" assumptions is a fuzzy target. And we do like to plumb the depths of assumptions. The arguments for ever more esoteric interpretations of the original problem (such as does the reporter also announce "there is at least one five" ) are valid takes.

I suppose when you've been coding against ambiguous requirements for half a century, you do that instinctively.

Agreed.

In programming, such assumptions are what lead to exceptions and fatal crashes.
We can't assume all inputs are integers, or even numbers or even sanitized. We can't assume dice (rand functions) are fair, etc.

 

[PeroK]

No. My belief is thus: there is a best answer if we make some reasonable assumptions, and don't make unreasonable ones.

This is wrong. The Wikipedia page explains clearly why there in an ambiguity. It's not about unreasonable assumptions.

If one is attempting to answer the original question as stated, it is reasonable (IMO) to assume, for example, there is only one die roll of two dice and then the result is reported. (Attempting to address PeroK's criticism, above)

Insisting on only one trial does not change the probability. This is a common fallacy. It can't be the case that if you do the experiment once, the probability is 1/11; but if you repeat the experiment many times, the probability drops to 1/6.

Although, once probability is thrown into the "fun and games" domain, that is quite a common fallacy.

Another way to look at the issue is that, in all probability problems, you have to be very careful assuming different events are equally likely. Before you have any information (except that two dice have been rolled), the probability of 6-6 is the same as 3-6 and 6-3. All three are 1/36.

When you are told there is at least one six, we have to recalculate these conditional probabilities. The first possible assumption is that they are all still equally likely, each with a probability of 1/11.

The second possible assumption is that obtaining the data about the six was certain in the first case, but only 50-50 in the remaining cases (we could have received information about the 3).

This is the heart of the problem: if 3-6 has been rolled, are we certain we will be told about the 6 and not the 3? This is the assumption you need. Whether that is reasonable or not, may be a matter of taste. In my opinion, it is an unreasonable assumption. You cannot infer from a single trial that the reporter will always report on sixes. If the trial were repeated, I don't accept the impossibility that they might say "there is at least one five". I don't see that in the problem statement.

Then we have reached the "correct" answer, that the problem statement is unwittingly ambiguous. And that this needs to be fixed. A simple solution is to amend the problem statement to:

Two dice are rolled out of your sight. You ask "is at least one of them a six?", to which the answer is yes. Then, the probability of a double-six is 1/11.

 

[PeroK]

PS what's clear is that if you roll two dice repeatedly and count the number of times you get at least one six and the number of times you get a double-six, then the relative frequency is 11-1. This is basic probability theory. That it is not 6-1 is worthy of note!

But, the more you jazz this up into a flowery question with extraneous details, the more likely you are to unwittingly introduce a bias or additional, hidden conditional probability. And screw the whole thing up!

 

[Ibix]

This is the heart of the problem: if 3-6 has been rolled, are we certain we will be told about the 6 and not the 3?

I do find it interesting that the difficulty with this problem (and the two child problem, and probably other ones) is not the probability theory, which is quite trivial, but nailing down exactly what the question is.

Whether that is reasonable or not, may be a matter of taste. In my opinion, it is an unreasonable assumption. You cannot infer from a single trial that the reporter will always report on sixes.

It's an obvious assumption, but I think that is only because of the exam technique based observation that there must be a fairly straightforward answer to this question or the setter would not have posed it.

 

[DaveC426913]

This is wrong. The Wikipedia page explains clearly why there in an ambiguity. It's not about unreasonable assumptions.

OK. Fair enough.

Insisting on only one trial does not change the probability. This is a common fallacy. It can't be the case that if you do the experiment once, the probability is 1/11; but if you repeat the experiment many times, the probability drops to 1/6.

That was not what I was suggesting. People keep talking about the dice being rolled one after another, and then equivocating about when, exactly, the six is revealed. I'm trying nip that in the bud.

Although, once probability is thrown into the "fun and games" domain, that is quite a common fallacy.

Well I hope that's not the fallacy I appeared to be committing.

Another way to look at the issue is that, in all probability problems, you have to be very careful assuming different events are equally likely. Before you have any information (except that two dice have been rolled), the probability of 6-6 is the same as 3-6 and 6-3. All three are 1/36.

When you are told there is at least one six, we have to recalculate these conditional probabilities. The first possible assumption is that they are all still equally likely, each with a probability of 1/11.

The second possible assumption is that obtaining the data about the six was certain in the first case, but only 50-50 in the remaining cases (we could have received information about the 3).

See above. This seems to purport a 2-tier scenario, rather than a single scenario.

This is the heart of the problem: if 3-6 has been rolled, are we certain we will be told about the 6 and not the 3? This is the assumption you need. Whether that is reasonable or not, may be a matter of taste. In my opinion, it is an unreasonable assumption.

I can see that now. Yes, my assumption was that sixes are the only thing of interest.

A simple solution is to amend the problem statement to:

Two dice are rolled out of your sight. You ask "is at least one of them a six?", to which the answer is yes. Then, the probability of a double-six is 1/11.

That certainly eliminates the ambiguity.