Quote:billryanI appreciate the explanation and accept it, but don't fully grasp it.

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Potentially an easier way to think about it:

It’s similar to a hard 8 bet. Chance of a hard 8 before a seven or easy 8.

Math: one way to make a hard eight. Six ways to make a 7. 4 ways to make an easy 8.

So hard 8 fair odds are 1 / (1+6+4) = 1 / 11

If above makes sense to you, you can apply it to the question in this thread letter.

Chance of making a hard four (2/2) before any other roll that contains a 2.

Math: one way to make hard four. 10 ways to make a roll that contains a 2 that isn’t a hard four.

So fair odds to the two dice problem are 1 / (1+10) = 1/11.

[ flat-ur-ther ]

📓 High School Level

noun

1. a person who adheres to the idea that the earth is flat.

2. a person who clings to an idea or theory that has long been proved wrong.

https://www.dictionary.com/browse/flat-earther

innumerate

[ ih-noo-mer-it, ih-nyoo- ]

🎓 College Level

adjective

1. unfamiliar with mathematical concepts and methods; unable to use mathematics; not numerate.

noun

2. an innumerate person.

https://www.dictionary.com/browse/innumerate

Verdict: Release The Wizard, but send him back to Middle Earth.

Quote:gordonm888Quote:billryanCan someone walk me thru this?

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Your friend looks at the two dice and truthfully reports to you that "at least one of the two dice is a 2."

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(trimmed)

I am not seeing how that changes the probability from rolling two dice and revealing only one.

I am not seeing how that it would matter if my counterpart revealed the dice on the north or on the south.

There are other things I don't see.

I can believe there are 11/36 chances to roll at least one two.

Quote:DieterQuote:gordonm888Quote:billryanCan someone walk me thru this?

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Your friend looks at the two dice and truthfully reports to you that "at least one of the two dice is a 2."

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(trimmed)I'm a muttonhead. Thanks for not going too rough on me.

I am not seeing how that changes the probability from rolling two dice and revealing only one.

I am not seeing how that it would matter if my counterpart revealed the dice on the north or on the south.

There are other things I don't see.

I can believe there are 11/36 chances to roll at least one two.

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Bold added.

If you can also believe there’s only 1/36 ways to roll two twos then you have all the pieces.

1/36 divided by 11/36 = 1/36 * 36/11 = 1/11

Quote:oncedearI'll leave it there until Wizard returns, because it would feel unfair to challenge his logic here while he is suspended.

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Quote:gordonm888

I do agree with Wizard. In my opinion, the answer to the posed question is 1/11.

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Gordon, Heres a clue as to why I assert that 1 in 6 is ONE perfectly valid answer to the EXACT WORDING OF THE ORIGINAL QUESTION.

How, from the original question do you know the answer to that?

What does the peeker say when a four and a five appear?

How, from the original question do you know the answer to that?

What does the peeker say when a four and a two appear?

Which is it? Do we know?

How, from the original question do you know the answer to that?

Quote:unJonQuote:billryanI appreciate the explanation and accept it, but don't fully grasp it.

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Potentially an easier way to think about it:...

So fair odds to the two dice problem are 1 / (1+10) = 1/11.

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BUT You assume that the peeker is confirming twos. Exactly the mistake Wizard makes.

The peeker is calling the value of one of the dice. We don't know which one. We don't know how he decides which one. We don''t know what he declares if no two is rolled. See my reply to Gordon.

Quote:OnceDearQuote:unJonQuote:billryanI appreciate the explanation and accept it, but don't fully grasp it.

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Potentially an easier way to think about it:...

So fair odds to the two dice problem are 1 / (1+10) = 1/11.

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BUT You assume that the peeker is confirming twos. Exactly the mistake Wizard makes.

The peeker is calling the value of one of the dice. We don't know which one. We don't know how he decides which one. We don''t know what he declares if no two is rolled. See my reply to Gordon.

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You are speaking orthogonal to me. I am explaining the problem to people that don’t get it, once you fix the fair semantic ambiguity that you pointed out in the original formulation on this forum.

In other words, I take as given that the actual problem is that the speaker is asked if at least one die shows a 2, and truthfully answers “yes.”

Quote:unJon

You are speaking orthogonal to me. I am explaining the problem to people that don’t get it, once you fix the fair semantic ambiguity that you pointed out in the original formulation on this forum.

In other words, I take as given that the actual problem is that the speaker is asked if at least one die shows a 2, and truthfully answers “yes.”

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I appreciate that. 1/11 is spot on if the peekers job is to report the presence or absence of a two. Wizard insists that his his role and activity. I insist that there is no reason to believe he is so doing.

Quote:OnceDearQuote:unJon

You are speaking orthogonal to me. I am explaining the problem to people that don’t get it, once you fix the fair semantic ambiguity that you pointed out in the original formulation on this forum.

In other words, I take as given that the actual problem is that the speaker is asked if at least one die shows a 2, and truthfully answers “yes.”

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I appreciate that. 1/11 is spot on if the peekers job is to report the presence or absence of a two. Wizard insists that his his role and activity. I insist that there is no reason to believe he is so doing.

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The Wiz is wrong based on the original formulation of the problem.

Quote:AlanMendelsonBilly...

Suppose I told you I had two dice in a cup. After shaking I told you at least one of the dice is a 2.

Then I asked you, what are the odds that the other die is also a 2?

What would you say?

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I would say one in six but realize I'd be wrong. I may not understand math, but I believe in it.