The Wizard will bank this bet: 1/6 vs 1/11

[synergistic]

Both dice show a 2 1/36 times.

Answer the original question with your video. Enough arguing. The problem is specific. Shoot your video and explanation according to the problem and then try to justify 1/11.

I'm tired of this crap. I've been nice for too long.

No, both dice would show a two 1/36 times if you didn't have your peeker limiting the action to only rolls that yield at least one two. If you maintain the setup, could you add the "how often" version to your list of questions that could be answered with "one out of eleven"?

[Alan Mendelson]

Whoa synergistic. Don't start changing questions.

You wrote this:

Would you agree to an answer of "one time out of eleven" if the question was rephrased to "How often will both dice show a two?" instead of the original phrasing?

And I responded to this: "How often will both dice show a two?" And that answer is 1/36.

Now, I am telling you right now that I have been pushed to the wall. There is a specific question and I want you and everyone else who says the answer is 1/11 to justify their answer by first recreating the question with a video and then explaining their answer. You can't do it -- unless you lie through your teeth and make up a barrel of hogwash. Because when you present the question with props, and do a hands-on show and tell the ONLY answer you are going to come up with is 1/6.

Enough of this gibberish about creating scenarios. As I said from the beginning: the question is specific. Don't change anything to facilitate your 1/11 answer. Show me how you create the problem on a video and then justify your 1/11 answer if you can.

Frankly, I wonder why the Wizard didn't post the video he said he recorded? I wonder if when he did the video he discovered how 1/11 was impossible. I want to see HIS video. And I asked him on his forum.

[synergistic]

You know, initially, I thought you were trying to sucker someone you didn't particularly care for into a bad bet. I really have no idea what your motivation is at this point.

[Alan Mendelson]

My motivation has nothing to do with the bet. That's a side issue.

This is all about the original question. If you think it's 1/11 show your stuff.

Everyone on the WOV forum has been trying to justify their 1/11 answers by changing the question. I want someone to show me the answer to the original question is 1/11... if you can.

[RS__]

Alan, would you accept the results of the following: If I were to roll the dice many many times, record the times where at least one 2 was present, then show the results (ie: frequency of "other" die being a 2).

If the correct answer is 1/6, then we should see the other die show up as a deuce 1/6 of the time at least one die is present, yes?


Edit. Alan, we've showed (shown?) you many times why the answer is 1/11 but you are not willing to accept it. You can lead a horse to water but you can't make him drink.

PS: Did you ever wonder, why just about everyone on the WOV forum was willing to bank the bet I presented a while ago? Why would a bunch of APs be willing to risk their money (thousands of $$$) for somethijg where they don't expect to win?

[Alan Mendelson]

Alan, would you accept the results of the following: If I were to roll the dice many many times, record the times where at least one 2 was present, then show the results (ie: frequency of "other" die being a 2).

If the correct answer is 1/6, then we should see the other die show up as a deuce 1/6 of the time at least one die is present, yes?


Edit. Alan, we've showed (shown?) you many times why the answer is 1/11 but you are not willing to accept it. You can lead a horse to water but you can't make him drink.

PS: Did you ever wonder, why just about everyone on the WOV forum was willing to bank the bet I presented a while ago? Why would a bunch of APs be willing to risk their money (thousands of $$$) for somethijg where they don't expect to win?

No I will not accept any number of rolls. And it's because that was not the question that was asked. YOU and many others have said the answer is 1/11 to the original question. Show me how you arrived at that answer given the information in the original problem. And again that information is: there are two dice and at least one is showing a 2. No additional rolling needed. You said the chance of both dice showing 2-2 is 1/11, so go ahead and demonstrate that to me just as the original question lays out the problem.

I look forward to it.

[RS__]

Why won't you accept many rolls of the dice? If what you say is true (1/6 answer), then you should expect 1/6 of the "at least one die is showing a deuce" for the other die to also be showing a deuce....correct?

I mean, you can't honestly believe something has a 1/6 chance of occurring....but not accept a simulation?

What's next, you're going to say there's a 33% chance of flipping a coin on heads...and when a "math guy" tries to prove you wrong by flipping a coin many times, you say "No! That won't do! Only flip the coin once, not many times!!" ?

[Alan Mendelson]

Why won't you accept many rolls of the dice? If what you say is true (1/6 answer), then you should expect 1/6 of the "at least one die is showing a deuce" for the other die to also be showing a deuce....correct?

there you go again... changing the question to fit your answer. ANSWER THE ORIGINAL QUESTION. SHOW ME.

[RS__]

I do believe the original question was something like "what is the chance the other die is a 2" (or maybe it was "what is the chance both dice are a 2").

How is this any different? No math involved here [you should get this right at least..].

[Alan Mendelson]

I posted a video using two dice showing that the answer to the question was 1/6. You and others maintain the answer is 1/11. The question involves two dice, with at least one showing a 2. Let's see your video. Very simple. You have an answer -- show it, just as I showed it.

[arcimede$]

Arc do the video. Put two dice on a table and with at least one of them showing a two, explain in your video how you came up with the answer of 1/11. Do it.

By the way, no one on the WOV forum will rise to the challenge. Like Arc they are all falling back on their previous explanations. But they won't present the problem with a video demonstration that clearly says at least one of the dice is showing a 2.

Why in the world would you need a video? This is as simple as it gets. All the possible throws have been identified. The difference in the two scenarios has been explained to you. I did it again above. If you specify one particular die is a 2 then the answer is 1-6, if either die can be a 2 then it is 1-11. Why do you persist in arguing?

[arcimede$]

I posted a video using two dice showing that the answer to the question was 1/6. You and others maintain the answer is 1/11. The question involves two dice, with at least one showing a 2. Let's see your video. Very simple. You have an answer -- show it, just as I showed it.

Your video showed nothing.

[Alan Mendelson]

The video will show how the question is interpreted. In reality when you have two dice with a 2 showing -- can the 2 jump from one die to another? This reality destroys the 1/11 answer.

Edited to add: I just responded to another post on the WOV forum. It seems that no one wants to do a video but they are all continuing to argue that the answer is 1/11 because it isn't specified which of the two dice is a 2. The WOVer's question is in bold, and my response is in italics:

Quote: Dalex64
You can't just show a two on one of them - you have to show a two on either one of them.

The question states 'at least one' but doesn't specify which one.



I see that there is still some discussion about "at least one of the dice shows a 2." Let's deal in the real world, please, because in the real world the "2" cannot jump from one die to another. So you have the following options when creating your video (unless you are using Disney animation):

You can set Die A to show a 2 and Die B to be any of the six faces;
You can set Die B to show a 2 and Die A to be any of the six faces;
Or you can set both Die A and Die B to show a 2.

Any of those three will fulfill the information in the original question that "at least one of the dice shows a 2."

Now please, quiet on the set, and begin your video production.

Thanks.

[Alan Mendelson]

If you specify one particular die is a 2 then the answer is 1-6,

Arc, would you agree that in the real, physical world, that when two dice are rolled and you are told that at least one of the dice is showing a 2, that the answer is 1/6??

Again, this is the real, physical world -- not a world where the "2 face showing" can jump from one die to another.

[redietz]

There must be a glitch in the WoV video equipment. Or Schrodinger's Dice are on interlibrary loan.

[arcimede$]

Arc, would you agree that in the real, physical world, that when two dice are rolled and you are told that at least one of the dice is showing a 2, that the answer is 1/6??

Of course not, the answer is 1-11. The key phrase is "at least". That means either one could be the 2. To get 1-6 you have to specify one particular die, then the odds the other die is a 2 is pretty obvious.

Let's say you have a red die and a blue one. If the red die is a 2 then the odds the blue one is a 2 is clearly 1-6. However, no one is claiming that one particular die is a 2. That is what "at least" means. Either die could be a 2 and hence you have consider BOTH situations which leads to the 1-11 conclusion. Think about it. How can the answer be the same when either die could be a 2 vs. when one particular die is a 2.

[Alan Mendelson]

Arc let me clue you in on the English language. The key phrase "at least" means this:

There is a 2 on one die, or there is a 2 on both dice.

That is what "at least" means. It does not mean the #2 can jump from one die to another.

[arcimede$]

Arc let me clue you in on the English language. The key phrase "at least" means this:

There is a 2 on one die, or there is a 2 on both dice.

True but not the complete picture. There is a 2 on one die (the red one), or there is a 2 on the other die (the blue one) or there is a 2 on both of them.

That is what "at least" means. It does not mean the #2 can jump from one die to another.

No one said it did. Back to the red/blue dice. The 2 can be on the red die or the blue die. That is what "at least one die" means. If the 2 is on the red die then the blue die can be 1-6, that gives you 6 possibilities. If the 2 is on the blue die then the red die can have 1-6, that also gives you 6 possibilities. Since the 2-2 is a duplicate we eliminate one and you are left with 11 possible results.

This has all been explained to you multiple times.

[Alan Mendelson]

Arc try to think this through: as long as one die shows a 2 the solution rests on the second die in a two dice problem.

[Alan Mendelson]

True but not the complete picture. There is a 2 on one die (the red one), or there is a 2 on the other die (the blue one) or there is a 2 on both of them.



No one said it did. Back to the red/blue dice. The 2 can be on the red die or the blue die. That is what "at least one die" means. If the 2 is on the red die then the blue die can be 1-6, that gives you 6 possibilities. If the 2 is on the blue die then the red die can have 1-6, that also gives you 6 possibilities. Since the 2-2 is a duplicate we eliminate one and you are left with 11 possible results.

This has all been explained to you multiple times.

OMG Arc you can't use the same die twice to solve the problem.