How do you interpret the "dice problem"?

[Harry Porter]

Harry Porter thanks for joining and for posting.

Yes that's what the question asks. But what's your answer?

First, what I'll suggest is that a trial of just 300 throws should be sufficient to lead one to the correct answer. The difference between 1/6 and 1/11 is huge; the data will quickly converge to one or the other.

FWIW, I'm confident that the data will converge to 1/11.

If the question were framed as, "Looking at a specific die (say, the first one thrown, or the left most one, or some other unambiguous designation) and it's a "2". What's the probability that the other is a "2"? The answer most certainly would be 1/6.

But the question I posed is subtly different. The die in question may be either die; this does change the math.

The best way to illustrate this is the envision (or list) the 36 possible paired outcomes (11,12,13,...,23,24,25,...64,65,66). Each is equally likely.

Of these, 11 of the combinations have at least one 2: (12,21,22,23,24,25,26,32,42,52,62). Obviously, there's only one outcome of (22).

In short, the solution is 1 in 11.

If you're inclined to challenge this, then I suggest first running that 300 throw trial.

[Alan Mendelson]

The die in question may be either die;

Unfortunately, in the real physical world we have to decide which die it is.

If you want to flip-flop and double-count be my guest... just don't try it in Vegas. It's against the rules.

[Rob.Singer]

Before this starts all over again, this is Alan's groundrule (ie, interpretation of the question/problem, and it's as valid as Harry's--which is why the WoV forum member posed it as a brain teaser):
Once the die with a 2 comes to rest, it dies there showing a 2, never allowing consideration that it has any other faces containing different numbers because they are now irrelevant. The only question remaining is what are the odds, by default, that the other die also shows a 2.

It is a real world scenario, not some theory-driven, mensa chill up the leg. WoV has been struggling with this very simple, very real interpretation for weeks, because the know-it-all mentality over there is incapable of looking at anything happening in the real world and instead, always needs to revert back to theory in its deepest concept. Naturally if you do a 300-roll sample 1in11 is the answer. BUT THIS IS JUST ONE ROLL. Period.

Good luck.

[Harry Porter]

Unfortunately, in the real physical world we have to decide which die it is.

If you want to flip-flop and double-count be my guest... just don't try it in Vegas. It's against the rules.

Fine, so kill that portion of the post.

The enumeration of candidate throws stands. There are 36 equally probable dice throws. Of these, there are 11 specific throws where someone who looked at the outcome would say, "At least one of the dice is a 2." (as posed in the original problem).

Of these 11 possible outcomes, only 1 satisfies the criteria.

If you wish to discuss this further, run the suggested trial and post the results. (it's a short trial, so "anything can happen", but odds favor a result near 1/11)

[Alan Mendelson]

There are 36 equally probable dice throws. Of these, there are 11 specific throws where someone who looked at the outcome would say, "At least one of the dice is a 2." (as posed in the original problem).

It's true. There are 11 combinations of two dice showing at least one 2.

But as soon as you throw 2 dice and one shows a 2, only six possible combinations remain. One of those will be 2-2.

[arcimede$]

But you can show a single event in present tense with a video. When are you going to understand/appreciate a piece of writing?

Quite right. You can actually show 11 single events which are the result of throwing a pair of dice with at least one of them being a 2. However, Alan would object to this as he has already.

[arcimede$]

First, what I'll suggest is that a trial of just 300 throws should be sufficient to lead one to the correct answer. The difference between 1/6 and 1/11 is huge; the data will quickly converge to one or the other.

FWIW, I'm confident that the data will converge to 1/11.

If the question were framed as, "Looking at a specific die (say, the first one thrown, or the left most one, or some other unambiguous designation) and it's a "2". What's the probability that the other is a "2"? The answer most certainly would be 1/6.

But the question I posed is subtly different. The die in question may be either die; this does change the math.

The best way to illustrate this is the envision (or list) the 36 possible paired outcomes (11,12,13,...,23,24,25,...64,65,66). Each is equally likely.

Of these, 11 of the combinations have at least one 2: (12,21,22,23,24,25,26,32,42,52,62). Obviously, there's only one outcome of (22).

In short, the solution is 1 in 11.

If you're inclined to challenge this, then I suggest first running that 300 throw trial.

Harry .... Alan has been given this information about a dozen times now. He won't accept the correct answer because he is emotionally invested in the wrong answer.

[arcimede$]

It's true. There are 11 combinations of two dice showing at least one 2.

But as soon as you throw 2 dice and one shows a 2, only six possible combinations remain. One of those will be 2-2.

Yet you continue to refuse to actually do the work. Throw the die, collect the results and report them. You are a reporter aren't you? Well, maybe not ....

[Alan Mendelson]

Arc, I've done the work. When I have a 2 showing and throw a die, one out of six times it's another 2.

That's the work because that's the question.

[Alan Mendelson]

Quite right. You can actually show 11 single events which are the result of throwing a pair of dice with at least one of them being a 2. However, Alan would object to this as he has already.

Why don't you do that? Why don't you show us eleven different videos with one die showing a two -- and keeping that die with the 2 as a 2 and not changing it -- show us how your answer is 1/11 and not 1/6 ?? Go ahead. Do it.

Wait, I'll save you and Harry and OnceDear and RS__ and the Wizard and everyone else the trouble: YOU CAN'T.

[arcimede$]

Why don't you do that? Why don't you show us eleven different videos with one die showing a two -- and keeping that die with the 2 as a 2 and not changing it -- show us how your answer is 1/11 and not 1/6 ?? Go ahead. Do it.

Why? That isn't what the question asks.

Wait, I'll save you and Harry and OnceDear and RS__ and the Wizard and everyone else the trouble: YOU CAN'T.

You can't determine probabilities in 11 throws. Why do you continue to deny the question specifically asks you for the "probability"?

[Harry Porter]

As I wrote, I have nothing more to add until someone takes the time (say 10 minutes) to run an actual trial. (A relatively small investment compared to the time spent discussing this here ...)

[OneHitWonder]

?????--can't be.

Correct.

The scientific definition of the word 'or' is the algebraic, set-definition of 'or': the union of set A and set B and... . Sets include everything distinctly possible. Double entries are removed.

Only a restrictive-or can limit the possibilities of a set, or the union of two or more sets.

The honest "peeker" is responsible for both what he saw, and what he looked for. If this is truly a probability question, then we can assume that he's looking for all of the distinctly possible rolls. Ie, all of the 1-to-6 possible dice-numbers used by the "peeker", and all of the distinctly possible rolls for each of the possible dice-numbers.

Nothing has been restricted by anyone about this 'or'. Neither the dice-number of the "at least one is an X" clause, nor the the number of distinctly possible ways in which the dice-number may be rolled. The Wizard's "dice problem" can't have it both ways, anyway. Can't limit the dice-number to 2, but turn around and not limit the dice-chart to either the row or the column. (The word 'either' in the preceding sentence changes the 'or' usage to an exclusive-or.)

Here, the dice-number is 2. This is all we were told. So, as written, the answer here is absolutely 1/6 chance of another 2. What the "peeker" says and what the "peeker" thinks, or even deceptively does, are always separate matters of no import here.

In fact, simple set theory may be easily applied to thus prove out everything I have posted about this to date. I hesitate to do so because such will as likely add to the confusion as it will help.

[OneHitWonder]

As I wrote, I have nothing more to add until someone takes the time (say 10 minutes) to run an actual trial. (A relatively small investment compared to the time spent discussing this here ...)

Simulation or an actual trial is meaningless without the correct interpretation of the problem exactly as written. The biggest problem with any simulation. A computer shall write you exactly what you want to read.

Likely the only way that the Wizard will ever arrive at any correct answer?

[OneHitWonder]

Why do you continue to deny the question specifically asks you for the "probability"?

Yes, but the general case which you want here is Mango's second case, early on in the original thread at the Wizard's.

When the dice-number isn't restricted, the answer is 1/6 chance. Even were it restricted, the "at least one 2" implies Alan's take on this.

Wake up!

[regnis]

Exactly as One Hit described in the 3 threads above.

[Rob.Singer]

It's Whack the Nerds all over again!

Someone please tell Harry that this a one-off event. And someone please tell arci AGAIN that the question is asking the probability of the OTHER die of being a 2. At least Harry knows when to quit when he gets behind.

[arcimede$]

Yes, but the general case which you want here is Mango's second case, early on in the original thread at the Wizard's.

When the dice-number isn't restricted, the answer is 1/6 chance. Even were it restricted, the "at least one 2" implies Alan's take on this.

Wake up!

Nonsense. You are making an arbitrary (and incorrect) assumption.

[arcimede$]

It's Whack the Nerds all over again!

Someone please tell Harry that this a one-off event. And someone please tell arci AGAIN that the question is asking the probability of the OTHER die of being a 2. At least Harry knows when to quit when he gets behind.

More nonsense. Nothing in the question asks anything about "the other die".

[Zedd]

Besides all of this, one item remains from the start of the original dice problem thread which hasn't been addressed at all with respect to the original problem. A number isn't agreed upon in advance of the roll. I think it was Mango who called it "forcing the number" when all of the rolls are used or none is discarded. He concluded that if a number is forced for each roll, then the chance of two numbers the same on a roll is 1/6.

I did addressed this in the other thread.

But, the original question is indeed incapable of solution without making assumptions on how the condition of 'at least one deuce' was met.

However, after we had cleared up this ambiguity and established that the number IS agreed upon in advance--Alan still believes the answer is 1/6. This is simply wrong.