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To all the 1/6ers hailing Onyx as president…

You do realize that he gave 1/11 as the answer for the problem in this thread?

Please quote me where I said anything about Wikipedia on the paradox thing.

I SAID to use Google and noticed that all the top searches referred to this type of a problem as a paradox.
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I did not use Wikipedia. Your reading comprehension is quite poor—which is ironic since this is what you are shouting at others for.

I have the correct calculations for the problem in post 1 of...

The problem is not the original ambiguous one. I’ve said this on page 6 in the other thread:


The problem is the one explained very clearly on page 1 of this thread.

Again, Alan (and other...

I believe you’re one of the first 1/6ers to attempt actual math. Too bad you’re wrong.

If it is given that one or both of the dice shows a deuce, then 6/11 times the left die will be a deuce. ...

Do you know if a ’math guy’ wrote this question? What if an English author wrote it? Anybody can write it. Then you’re back to “general population bad in math”.

Then the peeker will wait until the spinner has come to a rest. It was agreed that the peeker would be able to see both dice before making the “at least one deuce” announcement.

Not true.

To quote the first post…


So if the frozen die is not a deuce and the ‘spinner’ eventually rests on a duece…that DOES meet the condition of “at least one deuce”. In which case,...

For the millionth time, the answer to that question is indeed 1/6. No one is arguing against that. Jihkro even explained that in his post (by the way, thanks for jumping in Jihkro).

Alan (and...

Again, the original question does have ambiguity. Since it is opened to interpretation, your “it’s an English question” argument does have some grounds. But once the ambiguity is killed, the...

I’m quoting Wikipedia’s page on the Boy or Girl paradox…


The Boy or Girl paradox is the same problem but uses gender of children instead of dice. So ½ is analogous to the 1/6 answer for the...

Aww…I had answered your questions after you cried about people refusing to do so. Yet, you're refusing to answer my questions.

Indeed, it is the same thing we’ve been arguing about (unlike the questions that regnis proposed).

So regnis (and others)…can you answer it?

Read the very first post on this thread. It was agreed that the dice are rolled until there were a 2.


Correct


Wrong

Okay…

Yes, 1/6 and 1/1 are the correct answers to those questions. Now try this one…

I roll two dice until at least one of them is a 2.

What were the odds that it was going to be 2-2?...

@OneHitWonder

Honestly, it’s very difficult to comprehend what you are trying to say.

Regarding the ‘Alan/Wizard Lunch bet’—the number 2 was agreed upon in advance. Two dice would be rolled...

All I said was that the problem is referred as a paradox.


What were the couple of specific questions that you had asked in which I did not give an answer?


Did I say anything about...

It’s not me giving it the 'paradox' label. I have a strong background in mathematics, so the question is not counterintuitive to me—and is not a paradox in my opinion. But for most people, the...

Yea...and the peeker didn’t say that we must not wait for a two. Nor did he say that he was randomly calling out dice X. Nor did he say that he had waited until a fixed die was a two. This is why...

Again, the problem as written is ambiguous. The problem as written to produce the 1/11 answer is a veridical paradox because the 1/11 answer is very counterintuitive, yet it’s true.

Yes, the correct answer to his question is 1/48.

I proceed to make a similar mistake with my question—the answer should be 1/236.

And yea, that might seem ridiculous to you. But it is true...

Again, the original question is open to an interpretation that produces an answer of 1/6.

The problem is…Alan (and others) believes 1/6 still holds on an interpretation that ONLY produces 1/11...

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,...

This question is analogous to setting one die to a deuce and then asking what is the probability of rolling a deuce with the other die.

It is NOT analogous to the question where we roll two dice...

I’m aware of the specifics of the bet…which, as far as I know, is basically what you had described in your original post. These are the same specifics that OnceDear used in his spreadsheet.

I’m...