How do you interpret the "dice problem"?

[redietz]

Arci, you parsed the question so as to arrive at your answer, yet a rigorous parsing of the language of the question suggests your conclusion isn't necessary. Is my parsing absurd? Maybe, but it isn't incorrect. Is your parsing absurd? Possibly, and it isn't necessarily correct.

The word "chance" gets mentioned in primary definitions of probability, but not "percentages" or "odds" or anything really mathematical. I think you should probably look up the definition of "chance" while you're at it. Not much mathematical in that primary definition, either.

Why do you think the author of the question avoided a description involving plural throws and plural peeks? Why did he not use any mathematical terms? Why was it present tense? He didn't write up this question this way by accident.

I suspect you had no idea that the primary definition of probability contains no references to odds or percentages or any math terms.

It must kill some folks to realize they're as language-blind as other folks are math-blind.

[Rob.Singer]

I'll sum it up for y'all. The original question was posed by a smart-ass because it really could be interpreted in two ways. The first one out of the gate thought he was a SMARTER-ASS (wiz) by trying to show up the smart-ass with the most technical conclusion available. The problem now transferred over to the WoV minions, who have learned to follow the wizard down that yellow brick road, no matter the consequences. As such, these guys profess only to see the problem's answer exactly as the wiz says it should be, and will not ever admit that the 1in6 answer is also valid when the question is interpreted in one of the two ways the OP meant it to be.

So Alan is right, and arci is also right, which is what I was getting at when I explained how the wizard's bet was based only on the 1in11 interpretation. Alan sees and agrees that the "math guys'" interpretation and 1in11 answer would be correct if everyone were to see & solve the problem as the wizard did, but for some reason the math guys don't want to accept the reasoning for his interpreting the question to yield a 1in6 answer.

Final conclusion: WoVers got taken to the woodshed here, first because they came in blinded by what their leader expects of them, then because they displayed a lack of civility common to a bunch of touchy theory-happy Internet prowlers who actually turned out to be nothing more than sissified versions of mickeycrimm.

[Alan Mendelson]

As soon as the Wiz and others defended the double count of the die with a two it was obvious they were not interpreting the question using basic English. As soon as someone tells me that the original question is not the same as the "dice spinner" I know it's a different mindset they're using, if not speaking a different language altogether.

Sorry -- double counting or switching values on a die is something you can't do in the real world.
I read the original question as a real world problem and not some exercise in Probability which requires multiple throws.

Probability math might work on a forum or in a classroom but just try telling a table crew in Vegas that "that two can also be a three."

[Alan Mendelson]

If the first die is not a 2, then there is still the possibility of the original question applying. That is my point.

The spinner question would not represent the "original question" unless the first die to settle was a two. Remember the peeker sees at least one two. If the peeker saw a six plus a spinner the question would not be activated.

[arcimede$]

Arci, you parsed the question so as to arrive at your answer, yet a rigorous parsing of the language of the question suggests your conclusion isn't necessary. Is my parsing absurd? Maybe, but it isn't incorrect. Is your parsing absurd? Possibly, and it isn't necessarily correct.

The word "chance" gets mentioned in primary definitions of probability, but not "percentages" or "odds" or anything really mathematical. I think you should probably look up the definition of "chance" while you're at it. Not much mathematical in that primary definition, either.

Why do you think the author of the question avoided a description involving plural throws and plural peeks? Why did he not use any mathematical terms? Why was it present tense? He didn't write up this question this way by accident.

I suspect you had no idea that the primary definition of probability contains no references to odds or percentages or any math terms.

It must kill some folks to realize they're as language-blind as other folks are math-blind.

You're grasping at straws. The entire context of the question is based on throwing dice which is pure mathematics. Good grief.

[arcimede$]

As soon as the Wiz and others defended the double count of the die with a two it was obvious they were not interpreting the question using basic English. As soon as someone tells me that the original question is not the same as the "dice spinner" I know it's a different mindset they're using, if not speaking a different language altogether.

Sorry -- double counting or switching values on a die is something you can't do in the real world.
I read the original question as a real world problem and not some exercise in Probability which requires multiple throws.

Probability math might work on a forum or in a classroom but just try telling a table crew in Vegas that "that two can also be a three."

Complete and utter nonsense. The entire context is dice roll. Why do you persist in your silly attempt to make this anything but pure a probability (aka mathematics) question? You cannot admit your were wrong. Sad.

[Alan Mendelson]

You're grasping at straws. The entire context of the question is based on throwing dice which is pure mathematics. Good grief.

But your entire answer is based on double counting and altering the value shown.

[Alan Mendelson]

Complete and utter nonsense. The entire context is dice roll. Why do you persist in your silly attempt to make this anything but pure a probability (aka mathematics) question? You cannot admit your were wrong. Sad.

As the Wizard said I can't get past a die having six sides. And I can't double count either.

[redietz]

"Good grief," and "absurd" are usually what one says when one doesn't have much of a response. I'm not even an advocate of the 1/6 camp, so "grasping at straws" to make an argument isn't my thing here.

Arci parsed the question and claimed others couldn't do math. I'm parsing his parsing and telling him that he doesn't read very well. If someone tried to prove in a US court that the question as stated referred to multiple events, I actually think they would fail.

It's a trick question. Suggestion: don't hammer at people about their lack of expertise in probability before you know what the primary definition of probability is.

[1in11]

The spinner question would not represent the "original question" unless the first die to settle was a two. Remember the peeker sees at least one two. If the peeker saw a six plus a spinner the question would not be activated.

Please explain to me how a die that immediately settles on 6 plus a spinner that eventually settles on 2 does not conform to the condition "at least one of the dice is a 2."

[Alan Mendelson]

Because you can't tell what the "spinner" is.

The "peeker" will see that the first die landed on a 2 and that "starts" the problem.

The "spinner situation" is only meant to give an example.

But okay, I will look at it your way. Let's say the first die settled on a 6, but the "spinner" landed on a 2. Even under those circumstances the die that landed as a 6 would still be only a 1/6 chance that it also didn't land on a 2.

Either way, it's 1/6. 1/6. 1/6.

It's a one roll "problem." It does not involve multiple rolls. You cannot switch values of any die to accommodate your 1/11 answer. Now go back to the WOV and let them know their methodology was wrong. Their math is good -- but their methodology and interpretation is wrong.

[1in11]

It is correct that when the first die landed on 2, there is a 1/6 chance that the second die also landed on 2 (6 possible outcomes, 1 success).
It is correct that when the second die landed on 2, there is a 1/6 chance that the first die also landed on 2 (6 possible outcomes, 1 success).

So, naturally we want to combine these probabilities to look at the problem where at least one of the dice landed on 2 - because we know either the first or the second die landed on 2. If we combine the two conditions, we get 2 successes out of 12 outcomes, or 1/6. However, there is an issue when we do so, in that the two events are not mutually exclusive. The roll 2-2 satisfies both conditions that the "first die is a 2" and the "second die is a 2," so when we combine the outcomes, we have double counted 2-2, and that is not okay. So we remove one of the 2-2s that we double counted, removed 1 success and 1 outcome. Leaving 1 success out of 11 outcomes, or 1/11.

1/6 - We know that a specific die landed on 2, what is the probability that the other die landed on 2?
1/11 - We know that at least one die landed on 2, what is the probability that they both landed on 2?

[Alan Mendelson]

There you go again -- combining and double-counting. Stop it. Only in Chicago and Cuban elections is double counting allowed and even Chicago is trying to stop the practice.

[OneHitWonder]

Their math is good -- but their methodology and interpretation is wrong.

"It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so."

- Mark Twain

[1in11]

There you go again -- combining and double-counting. Stop it. Only in Chicago and Cuban elections is double counting allowed and even Chicago is trying to stop the practice.

Please explain to me the error in my analysis. All I've ever heard from you is some form of "No, that's wrong" with no actual explanation of why it is wrong.

[Alan Mendelson]

The error is you can't change the value of a die. And that's what you do each and every time in order to justify your 1/11 answer.

This is why I asked the Wizard and the WOV forum members to shoot a video using two real, physical dice. Because when you use two real, physical dice and at least one die is showing a two the answer can only be one of the six sides on the other die.

In your theoretical world you can change die faces all you want and that's what the Wizard did in his video. But in reality you can't change the die face. If a die lands on a 2 it must stay showing a 2.

That's the error you are making.

We are told in the original question that at least one die is a 2 -- and there is no getting around that. Knowing that you must subtract one die from consideration leaving only one die with six faces.

When using two dice and given the info that at least one die is a two you must not consider changing that die.

You will argue: "but we don't know which die it is?"

And I say: "it doesn't matter when there are only two dice. Subtract from consideration either one -- it's the same thing." That might not be the "mathematical" answer but that is the "logical" answer.

And it is totally illogical to think that you can change the face of a die.

In short you should not be double-counting and adding, you should be SUBTRACTING:

2 (total number of dice) - 1 (one die showing a 2) = 1 (die with six faces to be considered)

Now, how's that for some complex math?

[1in11]

Alright, I'll do it your way. I'll take "the one showing a 2" and fix it as the first die. So there are 6 outcomes: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.

There are 2 ways to roll each of 2-1, 2-3, 2-4, 2-5, 2-6, and only 1 way to roll 2-2.

And since there's 1 way to roll a 2-2, and 11 ways total to roll 2-X: 1 in 11.

If your complaint with this is that I "double-counted" 2-1, 2-3, 2-4, 2-5 and 2-6, then I will agree with you: Yes, I "double-counted" those. I "double-counted" those because I need to cover every situation where "at least one of the dice is a 2" and there are 2 ways to roll each of those.

[Alan Mendelson]

Alright, I'll do it your way. I'll take "the one showing a 2" and fix it as the first die. So there are 6 outcomes: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.

There are 2 ways to roll each of 2-1, 2-3, 2-4, 2-5, 2-6, and only 1 way to roll 2-2.

And since there's 1 way to roll a 2-2, and 11 ways total to roll 2-X: 1 in 11.

If your complaint with this is that I "double-counted" 2-1, 2-3, 2-4, 2-5 and 2-6, then I will agree with you: Yes, I "double-counted" those. I "double-counted" those because I need to cover every situation where "at least one of the dice is a 2" and there are 2 ways to roll each of those.

But there is only ONE OTHER DIE and you are not matching the 2 with more than ONE OTHER DIE. So why would you bother with the "two ways to roll each of those"?

It's only a 2 dice problem. All of the justifications of 1/11 involve either using more than 2 dice or changing the face of the die already showing a 2.

Look, if you try that over at Vito's house when he has his Saturday nights craps party, they'll be feeding your body parts to his dogs out back.

[arcimede$]

But your entire answer is based on double counting and altering the value shown.

No, it is based on considering all possible combinations. That is how probabilities are computed. It is no different than how VP strategies are computed that you use every time you go to the casino.

[arcimede$]

"Good grief," and "absurd" are usually what one says when one doesn't have much of a response. I'm not even an advocate of the 1/6 camp, so "grasping at straws" to make an argument isn't my thing here.

Arci parsed the question and claimed others couldn't do math. I'm parsing his parsing and telling him that he doesn't read very well. If someone tried to prove in a US court that the question as stated referred to multiple events, I actually think they would fail.

It's a trick question. Suggestion: don't hammer at people about their lack of expertise in probability before you know what the primary definition of probability is.

I know what the definition of probability is when the context is clear. You are trying to ignore the obvious context and muddy the water. That is why I called your comment absurd.