Two dice puzzle -- Part Trois
Quote: AlanMendelsonI have been accused of being a math denier only for saying what I've seen. And not because I ever challenged any math.
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You absolutely are a math denier. That is crystal clear from your hundreds and hundreds of posts in denial of the answer to the two-dice problem.
If this were an astronomy forum, you would be the one proclaiming the earth is flat.
Quote: WizardQuote: AlanMendelsonI have been accused of being a math denier only for saying what I've seen. And not because I ever challenged any math.
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You absolutely are a math denier. That is crystal clear from your hundreds and hundreds of posts in denial of the answer to the two-dice problem.
If this were an astronomy forum, you would be the one proclaiming the earth is flat.
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What I "denied" is the wording of the question. You were going to follow up by asking someone involved with English if the question was worded correctly. Did you?
Quote: AlanMendelsonWhat I "denied" is the wording of the question. You were going to follow up by asking someone involved with English if the question was worded correctly. Did you?
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I don't remember that. What was wrong with the wording?
Quote: WizardQuote: AlanMendelsonWhat I "denied" is the wording of the question. You were going to follow up by asking someone involved with English if the question was worded correctly. Did you?
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I don't remember that. What was wrong with the wording?
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You want to start this again?
Okay.
Go ahead and post the original question. I don't want to be accused of changing anything.
After you post the question I'll point out the problem with how it's worded.
Quote: AlanMendelsonYou want to start this again?
Okay.
Go ahead and post the original question. I don't want to be accused of changing anything.
After you post the question I'll point out the problem with how it's worded.
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I will decline to start a part trois to that topic. I invite you to explain how the wording was bad as a post in part deux. There, I'll be happy to entertain your complaint on the wording.
Quote: WizardQuote: AlanMendelsonYou want to start this again?
Okay.
Go ahead and post the original question. I don't want to be accused of changing anything.
After you post the question I'll point out the problem with how it's worded.
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I will decline to start a part trois to that topic. I invite you to explain how the wording was bad as a post in part deux. There, I'll be happy to entertain your complaint on the wording.
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You'll have to tell me which post contains the original question.
On the last page of that thread I mentioned the awkward wording.
Again, if the original wording was posted I'll specifically point it out.
Quote: AlanMendelsonYou'll have to tell me which post contains the original question.
On the last page of that thread I mentioned the awkward wording.
Again, if the original wording was posted I'll specifically point it out.
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The original wording is right in the first post of the original thread. Please tell me what's wrong with it in part deux thread -- not here.
Quote: WizardQuote: AlanMendelsonYou'll have to tell me which post contains the original question.
On the last page of that thread I mentioned the awkward wording.
Again, if the original wording was posted I'll specifically point it out.
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The original wording is right in the first post of the original thread. Please tell me what's wrong with it in part deux thread -- not here.
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Thanks for pointing it out.
For everyone here's the original wording:
_____________________
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
_______________________
And here's the problem with the wording. It is in this sentence.
"Your partner peeks under the cup, and tells you, truthfully, 'At least one of the dice is a 2.' "
The question is phrased in such a way that it makes you take on the role of "the partner" who looks under the cup.
Now, as "the partner" what do YOU see when you look under the cup?
As I wrote when this problem was first presented, I suggested a PROPERLY PHRASED question would be something like...
How many combinations of two dice show at least one 2, and how many of those combinations would be 2-2?
Instead, we got an awkwardly written question which elicits a different answer than what was intended.
As I pointed out before... it's not a math problem, it's an English problem. And unfortunately the bad English has been repeated so many times it has become accepted.
Ask the question MY WAY and the answer is 1 out of 11.
But if you take the position of the partner who peeks under the cup, like the partner who sees one die showing a 2 you would have to say the answer is 1 out of 6.
Of course I mentioned this before and I was told I dont understand English. So, I give up.
I dont understand English.
I dont see shooters making the ALL more often than once in 190 rolls.
I never saw a random roller throw 18 yos in a row.
I believe the world is flat.
I'm a moron.
I'm a liar.
Yeah, I'm wrong about everything.
Except, Wizard, I did send you the three photos of the stacked dice.
Here are the first two parts of this topic:
Part 1
Part 2
Quote: AlanMendelson...For everyone here's the original wording:
_____________________
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
_______________________
And here's the problem with the wording. It is in this sentence.
"Your partner peeks under the cup, and tells you, truthfully, 'At least one of the dice is a 2.' "
The question is phrased in such a way that it makes you take on the role of "the partner" who looks under the cup.
What are you talking about? The question states the partner is a separate person. You are not the one peeking under the cup.
Quote:Now, as "the partner" what do YOU see when you look under the cup?
The question doesn't state that, except that he either sees one or two two's.
Quote:As I wrote when this problem was first presented, I suggested a PROPERLY PHRASED question would be something like...
How many combinations of two dice show at least one 2, and how many of those combinations would be 2-2?
That is a very simplified version of the problem. The goal of the original wording was to get from that wording to how you stated it, or to solve it some other correct way, like Bayesian analysis.
Quote:Instead, we got an awkwardly written question which elicits a different answer than what was intended.
As I pointed out before... it's not a math problem, it's an English problem. And unfortunately the bad English has been repeated so many times it has become accepted.
There was nothing wrong with the original wording. You just didn't understand it.
Quote:But if you take the position of the partner who peeks under the cup, like the partner who sees one die showing a 2 you would have to say the answer is 1 out of 6.
No!!!!!!!!! The question never said you are the peeker and see just one die.
Quote:Of course I mentioned this before and I was told I dont understand English. So, I give up.
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That's right -- You don't understand English. Your interpretation of the problem is completely ridiculous. I admit your spelling and grammar are fine, but your reading comprehension about this and everything related to math is terrible.
If you feel this is an insult, report it, I'll accept the punishment of any other moderator. However, I had to get that off my chest.
Quote: WizardQuote: AlanMendelson...For everyone here's the original wording:
_____________________
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
_______________________
And here's the problem with the wording. It is in this sentence.
"Your partner peeks under the cup, and tells you, truthfully, 'At least one of the dice is a 2.' "
The question is phrased in such a way that it makes you take on the role of "the partner" who looks under the cup.
What are you talking about? The question states the partner is a separate person. You are not the one peeking under the cup.Quote:Now, as "the partner" what do YOU see when you look under the cup?
The question doesn't state that, except that he either sees one or two two's.Quote:As I wrote when this problem was first presented, I suggested a PROPERLY PHRASED question would be something like...
How many combinations of two dice show at least one 2, and how many of those combinations would be 2-2?
That is a very simplified version of the problem. The goal of the original wording was to get from that wording to how you stated it, or to solve it some other correct way, like Bayesian analysis.Quote:Instead, we got an awkwardly written question which elicits a different answer than what was intended.
As I pointed out before... it's not a math problem, it's an English problem. And unfortunately the bad English has been repeated so many times it has become accepted.
There was nothing wrong with the original wording. You just didn't understand it.Quote:But if you take the position of the partner who peeks under the cup, like the partner who sees one die showing a 2 you would have to say the answer is 1 out of 6.
No!!!!!!!!! The question never said you are the peeker and see just one die.Quote:Of course I mentioned this before and I was told I dont understand English. So, I give up.
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That's right -- You don't understand English. Your interpretation of the problem is completely ridiculous. I admit your spelling and grammar are fine, but your reading comprehension about this and everything related to math is terrible.
If you feel this is an insult, report it, I'll accept the punishment of any other moderator. However, I had to get that off my chest.
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I have nothing more to say. We disagree.
Quote: WizardQuote: AlanMendelsonI have been accused of being a math denier only for saying what I've seen. And not because I ever challenged any math.
link to original post
You absolutely are a math denier. That is crystal clear from your hundreds and hundreds of posts in denial of the answer to the two-dice problem.
If this were an astronomy forum, you would be the one proclaiming the earth is flat.
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Wizard,
TLDR Three days posting suspension. You may still moderate spam etc, but for any judgement calls, I ASK that you get approval after any action from one of we three other moderators.
Wizard is NOT being penalised for this...
Quote: wizardThat's right -- You don't understand English. Your interpretation of the problem is completely ridiculous. I admit your spelling and grammar are fine, but your reading comprehension about this and everything related to math is terrible.
If you feel this is an insult, report it, I'll accept the punishment of any other moderator. However, I had to get that off my chest.
He IS being penalised for this...
Quote: wizardIf this were an astronomy forum, you would be the one proclaiming the earth is flat.
Unfortunately, there is no penalty under the rules for engaging Mr Mendelson on the question which he cannot comprehend. There should be. He does not understand the question, He cannot understand the question and he will not understand the question. Maybe that makes him a math denier, or maybe he is just hard wired to not comprehend statements of logic. Attack his logic, his maths or his comprehension, but don't attack HIM.
Accusing him of being a math denier is harsh but understandable and somewhat underscored by evidence. Accusing him of being a Flat Earther, in the absence of evidence, is a personal insult. I suggest a penalty, whereby you suspend yourself for three days while you re-read the entire two dice thread, part one and two.
Incidentally, the correct answer to the original question as posed can, of course be "one in six", but not under the conditions that AM suggests. It only needs for the calling rule to be that when the peeker looks at the dice and they are of values X and Y, that he truthfully declares that 'At least one of the dice is a 'Y'' THE CALLING RULES WERE NEVER GIVEN! Only the observation of one calling event.
I.e. In the scenario observed, one dice was a two. If the dice had shown 2 and 4, he MIGHT have truthfully declared "At least one of the dice is a 4" It would have been true.
Heck, the calling rule MIGHT have been "If only one of the dice is a two, I will call "At least one of the dice is a two, but if both of the dice are two, i will call "Both of the dice are twos"
And on that note. I return to my therapist!!!!
I think, now feel sure that, I annoyed him by reawakening the nightmare that we all argued so much.
Well, Wizard, You started this thread. LOL. You also suggested I bring my argument here!
Wizard: I respect you sir. I'm your subordinate here. This post is for the amusement of the forum and for the furtherance of Maths and logic. Wizard should be able to whup my ass on logic and maths.... Well if he were not currently indisposed.
I have requested that he permit me to refer to that PM exchange in thread, because frankly I think it has entertainment value. He did suggest it early in our discussion. So, I believe that it is acceptable under the rules for me to quote a small snippet of my outgoing PM to him....
Quote: Me to Wizard by PM
Now..... You may challenge my assertion that 1 in 6 is A CORRECT answer to the question as posed, by PM even while suspended.
Red rag to a bull. LOL. He thought I was joking with my 1/6 assertion.
He might have been irritated when I suspended him, too.
Now, to those who are new to the forum, the original 'Two Dice Puzzle' Thread was massively popular and controversial. It ran through 2013 to 2015 over many hundreds of posts and the arguing went on over more than two years. It spilled out to other forums and various high flying consultants were pestered.
Moderators had nightmares over it and this one almost needed therapy $;o)
And Wizard referenced it again, thus 'yanking my chain'
To summarize the controversy
The answer to the posed question as asserted by Wizard was, and still is 1/11. He is adamant.
The answer, as insisted upon by many, was 1/6. Alan Mendelson was a particular advocate of that answer.
The answer proposed by some, including myself was 'Indeterminate because we have to make assumptions'. I stand by my answer
Oh boy. I hope he doesn't take offence at my arguing with him. As per the rules, I'm not attacking him, just countering his argument. A lot of this is about semantics. Not as much fun as Math or logic.
Note. In spite of my answer of potentially 1/6. I do NOT support Alan M's arguments as to why it is potentially 1/6
I'll leave it there until Wizard returns, because it would feel unfair to challenge his logic here while he is suspended.
Quote: OnceDearDuring Wizard's current short suspension, I've been debating with him by PM $:o)
I think, now feel sure that, I annoyed him by reawakening the nightmare that we all argued so much.
Well, Wizard, You started this thread. LOL. You also suggested I bring my argument here!
Wizard: I respect you sir. I'm your subordinate here. This post is for the amusement of the forum and for the furtherance of Maths and logic. Wizard should be able to whup my ass on logic and maths.... Well if he were not currently indisposed.
I have requested that he permit me to refer to that PM exchange in thread, because frankly I think it has entertainment value. He did suggest it early in our discussion. So, I believe that it is acceptable under the rules for me to quote a small snippet of my outgoing PM to him....Quote: Me to Wizard by PM
Now..... You may challenge my assertion that 1 in 6 is A CORRECT answer to the question as posed, by PM even while suspended.
Red rag to a bull. LOL. He thought I was joking with my 1/6 assertion.
He might have been irritated when I suspended him, too.
Now, to those who are new to the forum, the original 'Two Dice Puzzle' Thread was massively popular and controversial. It ran through 2013 to 2015 over many hundreds of posts and the arguing went on over more than two years. It spilled out to other forums and various high flying consultants were pestered.
Moderators had nightmares over it and this one almost needed therapy $;o)
And Wizard referenced it again, thus 'yanking my chain'
To summarize the controversy
The answer to the posed question as asserted by Wizard was, and still is 1/11. He is adamant.
The answer, as insisted upon by many, was 1/6. Alan Mendelson was a particular advocate of that answer.
The answer proposed by some, including myself was 'Indeterminate because we have to make assumptions'. I stand by my answer
Oh boy. I hope he doesn't take offence at my arguing with him. As per the rules, I'm not attacking him, just countering his argument. A lot of this is about semantics. Not as much fun as Math or logic.
Note. In spite of my answer of potentially 1/6. I do NOT support Alan M's arguments as to why it is potentially 1/6
I'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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I do agree with Wizard. In my opinion, the answer to the posed question is 1/11.
I also agree that the odds are 1/11. I’m sure my rationale is already posted in that thread, or in the part deux thread, so I won’t duplicate my reasons.
Nah. He’d incorrectly proclaim that the earth is round. 😵💫Quote: WizardIf this were an astronomy forum, you would be the one proclaiming the earth is flat.
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Quote: billryanCan someone walk me thru this?
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Please tell me you're asking about dice and not Ball Eartherism.
Quote: 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."
Now if the dice had been colored red and blue, and your friend had reported to you that the red die was a 2, then the probability of the blue die being a two would be 1 in 6. That's common sense. And the people arguing for 1 in 6 seem to be applying this common-sense conclusion to the problem.
But the friend has reported that "at least one of the two dice is a two." So, it's a different problem statement and you need to return to the original probabilities of the dice rolls.
1 and 2 2/36
2 and 2 1/36
3 and 2 2/36
4 and 2 2/36
5 and 2 2/36
6 and 2 2/36.
Those are the only dice roll outcomes that count because you have been informed that at least one of the dice is a 2 which corresponds to 11 of the 36 possible original dice roll outcomes.
And rolling a pair of 2's has a probability weight of 1 out of 11 for those dice roll outcomes.
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?
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.
First, I restate the EXACT wording of the first post that raised this ugly mess
Quote: DweenYou have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
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And now I stake my case completely for the potential validity of 1/6
Our calling rules are indeterminate!
I'm arguing that nowhere in the OP did it say that he was CONFIRMING twos. Wizard assumed that without good reason. He denies the possibility that he doesn't call a number at every roll, also without good reason.
Let's say he saw both dice but always called one of the dies values. Choosing randomly between the two dice seen.
That is not ruled out in the OP and is a perfectly reasonable assumption. He's not obligated to declare twos as that is not in the original premise. YES he sees both dice.
Here are all the 72 possibilities, equally weighted.
First 36 possibilities considering that he calls the first value to catch his eye, maybe being the one nearest to him
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a one!
1,3 "At least one of the dice is a one!
1,4 "At least one of the dice is a one!
1,5 "At least one of the dice is a one!
1,6 "At least one of the dice is a one!
2,1 "At least one of the dice is a two!!
2,2 "At least one of the dice is a two!!
2,3 "At least one of the dice is a two!!
2,4 "At least one of the dice is a two!!
2,5 "At least one of the dice is a two!!
2,6 "At least one of the dice is a two!!
3,1 "At least one of the dice is a three!!
3,2 "At least one of the dice is a three!!
3,3 "At least one of the dice is a three!!
3,4 "At least one of the dice is a three!!
3,5 "At least one of the dice is a three!!
3,6 "At least one of the dice is a three!!
4,1 "At least one of the dice is a four!!
4,2 "At least one of the dice is a four!!
4,3 "At least one of the dice is a four!!
4,4 "At least one of the dice is a four!!
4,5 "At least one of the dice is a four!!
4,6 "At least one of the dice is a four!!
5,1 "At least one of the dice is a five!!
5,2 "At least one of the dice is a five!!
5,3 "At least one of the dice is a five!!
5,4 "At least one of the dice is a five!!
5,5 "At least one of the dice is a five!!
5,6 "At least one of the dice is a five!!
6,1 "At least one of the dice is a six!!
6,2 "At least one of the dice is a six!!
6,3 "At least one of the dice is a six!!
6,4 "At least one of the dice is a six!!
6,5 "At least one of the dice is a six!!
6,6 "At least one of the dice is a six!!
Now another 36 possibilities where he calls the second die to catch his eye, maybe the one furthest from him.
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a two!
1,3 "At least one of the dice is a three!
1,4 "At least one of the dice is a four!
1,5 "At least one of the dice is a five!
1,6 "At least one of the dice is a six!
2,1 "At least one of the dice is a one!
2,2 "At least one of the dice is a two!
2,3 "At least one of the dice is a three!
2,4 "At least one of the dice is a four!
2,5 "At least one of the dice is a five!
2,6 "At least one of the dice is a six!
3,1 "At least one of the dice is a one!
3,2 "At least one of the dice is a two!
3,3 "At least one of the dice is a three!
3,4 "At least one of the dice is a four!
3,5 "At least one of the dice is a five!
3,6 "At least one of the dice is a six!
4,1 "At least one of the dice is a one!
4,2 "At least one of the dice is a two!
4,3 "At least one of the dice is a three!
4,4 "At least one of the dice is a four!
4,5 "At least one of the dice is a five!
4,6 "At least one of the dice is a six!
5,1 "At least one of the dice is a one!
5,2 "At least one of the dice is a two!
5,3 "At least one of the dice is a three!
5,4 "At least one of the dice is a four!
5,5 "At least one of the dice is a five!
5,6 "At least one of the dice is a six!
6,1 "At least one of the dice is a one!
6,2 "At least one of the dice is a two!
6,3 "At least one of the dice is a three!
6,4 "At least one of the dice is a four!
6,5 "At least one of the dice is a five!
6,6 "At least one of the dice is a six!
Of the 12 times that he calls "At least one of the dice is a two", they are both twos on two occasions.
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in SIX
The answer 2/12 or 1/6
QED
That he peeks at both dice is irrelevant.
The calling rules MATTER and they are unstated.
Now..... If your friend had a declared hatred of using the word Two, after a nasty incident where 2 anvils fell on his 2 feet, but was honor bound to truthfully declare in the form "At least one of the dice is a X" and at one roll, he declared "At least one of the dice is a Two", what is the probability that the other dice is a two?
Quote: OnceDearProof that 1 in 6 is a perfectly valid answer in spite of Wizard insisting otherwise
First, I restate the EXACT wording of the first post that raised this ugly messQuote: DweenYou have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
link to original post
And now I stake my case completely for the potential validity of 1/6
Our calling rules are indeterminate!
I'm arguing that nowhere in the OP did it say that he was CONFIRMING twos. Wizard assumed that without good reason. He denies the possibility that he doesn't call a number at every roll, also without good reason.
Let's say he saw both dice but always called one of the dies values. Choosing randomly between the two dice seen.
That is not ruled out in the OP and is a perfectly reasonable assumption. He's not obligated to declare twos as that is not in the original premise. YES he sees both dice.
Here are all the 72 possibilities, equally weighted.
First 36 possibilities considering that he calls the first value to catch his eye, maybe being the one nearest to him
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a one!
1,3 "At least one of the dice is a one!
1,4 "At least one of the dice is a one!
1,5 "At least one of the dice is a one!
1,6 "At least one of the dice is a one!
2,1 "At least one of the dice is a two!!
2,2 "At least one of the dice is a two!!
2,3 "At least one of the dice is a two!!
2,4 "At least one of the dice is a two!!
2,5 "At least one of the dice is a two!!
2,6 "At least one of the dice is a two!!
3,1 "At least one of the dice is a three!!
3,2 "At least one of the dice is a three!!
3,3 "At least one of the dice is a three!!
3,4 "At least one of the dice is a three!!
3,5 "At least one of the dice is a three!!
3,6 "At least one of the dice is a three!!
4,1 "At least one of the dice is a four!!
4,2 "At least one of the dice is a four!!
4,3 "At least one of the dice is a four!!
4,4 "At least one of the dice is a four!!
4,5 "At least one of the dice is a four!!
4,6 "At least one of the dice is a four!!
5,1 "At least one of the dice is a five!!
5,2 "At least one of the dice is a five!!
5,3 "At least one of the dice is a five!!
5,4 "At least one of the dice is a five!!
5,5 "At least one of the dice is a five!!
5,6 "At least one of the dice is a five!!
6,1 "At least one of the dice is a six!!
6,2 "At least one of the dice is a six!!
6,3 "At least one of the dice is a six!!
6,4 "At least one of the dice is a six!!
6,5 "At least one of the dice is a six!!
6,6 "At least one of the dice is a six!!
Now another 36 possibilities where he calls the second die to catch his eye, maybe the one furthest from him.
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a two!
1,3 "At least one of the dice is a three!
1,4 "At least one of the dice is a four!
1,5 "At least one of the dice is a five!
1,6 "At least one of the dice is a six!
2,1 "At least one of the dice is a one!
2,2 "At least one of the dice is a two!
2,3 "At least one of the dice is a three!
2,4 "At least one of the dice is a four!
2,5 "At least one of the dice is a five!
2,6 "At least one of the dice is a six!
3,1 "At least one of the dice is a one!
3,2 "At least one of the dice is a two!
3,3 "At least one of the dice is a three!
3,4 "At least one of the dice is a four!
3,5 "At least one of the dice is a five!
3,6 "At least one of the dice is a six!
4,1 "At least one of the dice is a one!
4,2 "At least one of the dice is a two!
4,3 "At least one of the dice is a three!
4,4 "At least one of the dice is a four!
4,5 "At least one of the dice is a five!
4,6 "At least one of the dice is a six!
5,1 "At least one of the dice is a one!
5,2 "At least one of the dice is a two!
5,3 "At least one of the dice is a three!
5,4 "At least one of the dice is a four!
5,5 "At least one of the dice is a five!
5,6 "At least one of the dice is a six!
6,1 "At least one of the dice is a one!
6,2 "At least one of the dice is a two!
6,3 "At least one of the dice is a three!
6,4 "At least one of the dice is a four!
6,5 "At least one of the dice is a five!
6,6 "At least one of the dice is a six!
Of the 12 times that he calls "At least one of the dice is a two", they are both twos on two occasions.
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in SIX
The answer 2/12 or 1/6
QED
That he peeks at both dice is irrelevant.
The calling rules MATTER and they are unstated.
Now..... If your friend had a declared hatred of using the word Two, after a nasty incident where 2 anvils fell on his 2 feet, but was honor bound to truthfully declare in the form "At least one of the dice is a X" and at one roll, he declared "At least one of the dice is a Two", what is the probability that the other dice is a two?100% certain. Unless both dice were two, he would have called the other number: He only calls a two when he absolutely can't avoid it
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You’re wasting time. 1 in anything is a valid answer to the original wording.
How very true.Quote: unJonQuote: OnceDearProof that 1 in 6 is a perfectly valid answer in spite of Wizard insisting otherwise
First, I restate the EXACT wording of the first post that raised this ugly messQuote: DweenYou have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
link to original post
And now I stake my case completely for the potential validity of 1/6
Our calling rules are indeterminate!
I'm arguing that nowhere in the OP did it say that he was CONFIRMING twos. Wizard assumed that without good reason. He denies the possibility that he doesn't call a number at every roll, also without good reason.
Let's say he saw both dice but always called one of the dies values. Choosing randomly between the two dice seen.
That is not ruled out in the OP and is a perfectly reasonable assumption. He's not obligated to declare twos as that is not in the original premise. YES he sees both dice.
Here are all the 72 possibilities, equally weighted.
First 36 possibilities considering that he calls the first value to catch his eye, maybe being the one nearest to him
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a one!
1,3 "At least one of the dice is a one!
1,4 "At least one of the dice is a one!
1,5 "At least one of the dice is a one!
1,6 "At least one of the dice is a one!
2,1 "At least one of the dice is a two!!
2,2 "At least one of the dice is a two!!
2,3 "At least one of the dice is a two!!
2,4 "At least one of the dice is a two!!
2,5 "At least one of the dice is a two!!
2,6 "At least one of the dice is a two!!
3,1 "At least one of the dice is a three!!
3,2 "At least one of the dice is a three!!
3,3 "At least one of the dice is a three!!
3,4 "At least one of the dice is a three!!
3,5 "At least one of the dice is a three!!
3,6 "At least one of the dice is a three!!
4,1 "At least one of the dice is a four!!
4,2 "At least one of the dice is a four!!
4,3 "At least one of the dice is a four!!
4,4 "At least one of the dice is a four!!
4,5 "At least one of the dice is a four!!
4,6 "At least one of the dice is a four!!
5,1 "At least one of the dice is a five!!
5,2 "At least one of the dice is a five!!
5,3 "At least one of the dice is a five!!
5,4 "At least one of the dice is a five!!
5,5 "At least one of the dice is a five!!
5,6 "At least one of the dice is a five!!
6,1 "At least one of the dice is a six!!
6,2 "At least one of the dice is a six!!
6,3 "At least one of the dice is a six!!
6,4 "At least one of the dice is a six!!
6,5 "At least one of the dice is a six!!
6,6 "At least one of the dice is a six!!
Now another 36 possibilities where he calls the second die to catch his eye, maybe the one furthest from him.
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a two!
1,3 "At least one of the dice is a three!
1,4 "At least one of the dice is a four!
1,5 "At least one of the dice is a five!
1,6 "At least one of the dice is a six!
2,1 "At least one of the dice is a one!
2,2 "At least one of the dice is a two!
2,3 "At least one of the dice is a three!
2,4 "At least one of the dice is a four!
2,5 "At least one of the dice is a five!
2,6 "At least one of the dice is a six!
3,1 "At least one of the dice is a one!
3,2 "At least one of the dice is a two!
3,3 "At least one of the dice is a three!
3,4 "At least one of the dice is a four!
3,5 "At least one of the dice is a five!
3,6 "At least one of the dice is a six!
4,1 "At least one of the dice is a one!
4,2 "At least one of the dice is a two!
4,3 "At least one of the dice is a three!
4,4 "At least one of the dice is a four!
4,5 "At least one of the dice is a five!
4,6 "At least one of the dice is a six!
5,1 "At least one of the dice is a one!
5,2 "At least one of the dice is a two!
5,3 "At least one of the dice is a three!
5,4 "At least one of the dice is a four!
5,5 "At least one of the dice is a five!
5,6 "At least one of the dice is a six!
6,1 "At least one of the dice is a one!
6,2 "At least one of the dice is a two!
6,3 "At least one of the dice is a three!
6,4 "At least one of the dice is a four!
6,5 "At least one of the dice is a five!
6,6 "At least one of the dice is a six!
Of the 12 times that he calls "At least one of the dice is a two", they are both twos on two occasions.
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in SIX
The answer 2/12 or 1/6
QED
That he peeks at both dice is irrelevant.
The calling rules MATTER and they are unstated.
Now..... If your friend had a declared hatred of using the word Two, after a nasty incident where 2 anvils fell on his 2 feet, but was honor bound to truthfully declare in the form "At least one of the dice is a X" and at one roll, he declared "At least one of the dice is a Two", what is the probability that the other dice is a two?100% certain. Unless both dice were two, he would have called the other number: He only calls a two when he absolutely can't avoid it
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You’re wasting time. 1 in anything is a valid answer to the original wording.
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Funny that the OP stated...9 years ago...
Quote: DweenI'm interested to see how people solve this one.
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But he never once re entered that thread!
Quote: OnceDear
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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This is exactly what I came to as I mulled it over for the last 3 or 4 hours.
Is there something special about twos in this context?
Until I know that, I am making the assumption that a dice value is randomly revealed.
Have fun Dieter. You weren't here in 2013. This threat was the source of a lot of argument and a lot of expended mental effort. It brought out analogies, simplifications, alternative interpretations and even real life wagers.Quote: DieterQuote: OnceDear
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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This is exactly what I came to as I mulled it over for the last 3 or 4 hours.
Is there something special about twos in this context?
Until I know that, I am making the assumption that a dice value is randomly revealed.I am greatly enjoying the opportunity to analyze my assumptions.
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And there are still two people that just don't get it.
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.
link to original post
The peeker might have said "one of the dice is a 5." There is nothing special about a two. Then the odds would have been 1/11 that the other die is a 5.
This 2-dice puzzle challenges you to understand the additional information that has been conveyed to you and its effect on your ability to assign probabilities to the unknown outcomes. I am teaching senior-level high-school math at home and the textbooks constantly reinforce the need to understand
1. what the known and unknown variables are in a problem statement
2. What parameter you are asked to calculate or define
3. Writing an equation that expresses the requested parameter in terms of the known and unknown variables
Approximately half the people posting on this thread seem to have never been taught the above approach in their education and don't seem to understand how to interpret a word problem. The problem statement isn't War and Peace, you don't need to analyze motivations of the "friend," you are simply asked to straightforwardly interpret new information from your friend (at face value.)
For some reason, some otherwise fairly intelligent people seem determined on this thread to provoke others by pretending to be obtuse. I understand that BillRyan has strengths other than math -that's okay, I'm cool with that. But for OnceDear to constantly denounce people on other threads who don't understand "Expected Value" and then to play at being obtuse about dice probabilities on this thread is very discouraging. No wonder Wizard lost his cool with Alan and the other flat-earthers.
Quote: gordonm888Quote: 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.
link to original post
The peeker might have said "one of the dice is a 5." There is nothing special about a two. Then the odds would have been 1/11 that the other die is a 5.
This 2-dice puzzle challenges you to understand the additional information that has been conveyed to you and its effect on your ability to assign probabilities to the unknown outcomes. I am teaching senior-level high-school math at home and the textbooks constantly reinforce the need to understand
1. what the known and unknown variables are in a problem statement
2. What parameter you are asked to calculate or define
3. Writing an equation that expresses the requested parameter in terms of the known and unknown variables
Approximately half the people posting on this thread seem to have never been taught the above approach in their education and don't seem to understand how to interpret a word problem. The problem statement isn't War and Peace, you don't need to analyze motivations of the "friend," you are simply asked to straightforwardly interpret new information from your friend (at face value.)
For some reason, some otherwise fairly intelligent people seem determined on this thread to provoke others by pretending to be obtuse. I understand that BillRyan has strengths other than math -that's okay, I'm cool with that. But for OnceDear to constantly denounce people on other threads who don't understand "Expected Value" and then to play at being obtuse about dice probabilities on this thread is very discouraging. No wonder Wizard lost his cool with Alan and the other flat-earthers.
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Gordon,
I disagree with much of what you say. The rules that the peeker follows are critically important to the problem. And those rules are not disclosed in the original formulation of the problem.
I’ll give you two examples.
1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?
2) The rule the peeker follows is:
a) if there is at least one 1, say “there is at least one 1”.
b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”
The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?
Quote: gordonm888
The peeker might have said "one of the dice is a 5." There is nothing special about a two. Then the odds would have been 1/11 that the other die is a 5.
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SO WRONG on so many levels
Gordon.....
If the peeker had seen a 5 and a 2 or a 2 and a 5, what would he have declared and why?
I assert that we do not know and we do not know.
Very similar to what I just PM'd wizard
Quote: oncedear to wizard a few minutes ago
Stop speed reading. ! Stop seeing what you want to see. You're doing an [redacted]
If you only read one sentence here, read the red one.
If the dice had been a two and a four, or a four and a two, what would the peeker have said? And why?
Quote: unJon
1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?
2) The rule the peeker follows is:
a) if there is at least one 1, say “there is at least one 1”.
b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”
The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?
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Quote: unJonI offer my challenge posted above to the Wiz, Gordon or anyone else. Willing to escrow funds. 20 to 1 odds on the bet that there are 2 twos (I take side there are not 2 twos).
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Analysis... I think one of us dropped the ball, Unjon.
Here are all the 36 possibilities for dice landing, equally weighted.
1,1 “there is at least one 1” No Bet
1,2 “there is at least one 1” No Bet
1,3 “there is at least one 1” No Bet
1,4 “there is at least one 1” No Bet
1,5 “there is at least one 1” No Bet
1,6 “there is at least one 1” No Bet
2,1 “there is at least one 1” No Bet
2,2 “there is at least one 2” Game Open Pays Out
2,3 “there is at least one 2” Game Open
2,4 “there is at least one 2” Game Open
2,5 “there is at least one 2” Game Open
2,6 “there is at least one 2” Game Open
3,1 “there is at least one 1” No Bet
3,2 “there is at least one 2.” Game Open
3,3 Silence=No bet
3,4 Silence=No bet
3,5 Silence=No Bet
3,6 Silence=No Bet
4,1 “there is at least one 1.” No Bet
4,2 “there is at least one 2.” Game Open
4,3 Silence=No Bet
4,4 Silence=No Bet
4,5 Silence=No Bet
4,6 Silence=No Bet
5,1 “there is at least one 1” No Bet
5,2 “there is at least one 2” Game open
5,3 Silence=No Bet
5,4 Silence=No Bet
5,5 Silence=No Bet
5,6 Silence=No Bet
6,1 “there is at least one 1” No Bet
6,2 “there is at least one 2” Game open
6,3 Silence=No Bet
6,4 Silence=No Bet
6,5 Silence=No Bet
6,6 Silence=No Bet
Of the 9 times that he calls "There's at least one 2", they are both twos
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in NINE
The answer 1.9
QED
Quote: OnceDearQuote: unJon
1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?
2) The rule the peeker follows is:
a) if there is at least one 1, say “there is at least one 1”.
b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”
The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?
link to original postQuote: unJonI offer my challenge posted above to the Wiz, Gordon or anyone else. Willing to escrow funds. 20 to 1 odds on the bet that there are 2 twos (I take side there are not 2 twos).
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Analysis... I think one of us dropped the ball, Unjon.
Here are all the 36 possibilities for dice landing, equally weighted.
1,1 “there is at least one 1” No Bet
1,2 “there is at least one 1” No Bet
1,3 “there is at least one 1” No Bet
1,4 “there is at least one 1” No Bet
1,5 “there is at least one 1” No Bet
1,6 “there is at least one 1” No Bet
2,1 “there is at least one 1” No Bet
2,2 “there is at least one 2” Game Open Pays Out
2,3 “there is at least one 2” Game Open
2,4 “there is at least one 2” Game Open
2,5 “there is at least one 2” Game Open
2,6 “there is at least one 2” Game Open
3,1 “there is at least one 1” No Bet
3,2 “there is at least one 2.” Game Open
3,3 Silence=No bet
3,4 Silence=No bet
3,5 Silence=No Bet
3,6 Silence=No Bet
4,1 “there is at least one 1.” No Bet
4,2 “there is at least one 2.” Game Open
4,3 Silence=No Bet
4,4 Silence=No Bet
4,5 Silence=No Bet
4,6 Silence=No Bet
5,1 “there is at least one 1” No Bet
5,2 “there is at least one 2” Game open
5,3 Silence=No Bet
5,4 Silence=No Bet
5,5 Silence=No Bet
5,6 Silence=No Bet
6,1 “there is at least one 1” No Bet
6,2 “there is at least one 2” Game open
6,3 Silence=No Bet
6,4 Silence=No Bet
6,5 Silence=No Bet
6,6 Silence=No Bet
Of the 9 times that he calls "There's at least one 2", they are both twos
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in NINE
The answer 1.9
QED
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That’s correct for my scenario 2). Good work. That has nothing to do with my scenario 1) challenge.
Quote: gordonm888But for OnceDear to constantly denounce people on other threads who don't understand "Expected Value" and then to play at being obtuse about dice probabilities on this thread is very discouraging. No wonder Wizard lost his cool with Alan and the other flat-earthers.
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1. I'm not playing at being obtuse. I'm pointing out one of those remarkable situations where I abjectly disagree with Wizard. A worthy cause against the math and stats and logic expert. Be discouraged if you wish.
2. Did you just group Alan in with other flat earthers? I just suspended Wizard for implicitly doing that.
Quote: unJon
1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?
2) The rule the peeker follows is:
a) if there is at least one 1, say “there is at least one 1”.
b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”
The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?
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Quote: unJonI offer my challenge posted above to the Wiz, Gordon or anyone else. Willing to escrow funds. 20 to 1 odds on the bet that there are 2 twos (I take side there are not 2 twos).
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Sorry Unjon. I misread para 2) as the peek rules applied to challenge at 1)Quote: unjon
That’s correct for my scenario 2). Good work. That has nothing to do with my scenario 1) challenge.
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If challenge 1 stands on its own, you would never need to pay out a bean. 1000 to one would be robbery. Only ever declare when there is exactly one 2 showing. It is within the challenge 1 rules
Quote: OnceDear
If challenge 1 stands on its own, you would never need to pay out a bean. 1000 to one would be robbery. Only ever declare when there is exactly one 2 showing. It is within the challenge 1 rules
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Bingo. I’m hoping it will help show Wiz and Gordon why ambiguity in the rules affects analysis.
Quote: billryanThe difference here might be semantics more than math.
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Much of it is. But once you get past the semantics there is a real conceptual issue in the math where people have trouble with seeing 1/11 vs 1/6.
The problem is people go back to being bogged down in the semantics.
Quote: billryanThe difference here might be semantics more than math.
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Until the semantics are ironed out, there can be no codified question and no derived explicit answer.
It's like playing a card game, not knowing the rules and losing your arse.
Quote: unJonQuote: billryanThe difference here might be semantics more than math.
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Much of it is. But once you get past the semantics there is a real conceptual issue in the math where people have trouble with seeing 1/11 vs 1/6.
The problem is people go back to being bogged down in the semantics.
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Wizard is working on a reworking of the puzzle where the semantics were ironed out. I believe he's doing that, cos like the Monty Hall problem, this is one where the answer is not as intuitive as one would expect. In his reworked version, the 1 in 11 answer is absolutely proven, but still expect AlanM and a few others to plumb for and cling on to 1/6
Quote: OnceDearQuote: unJonQuote: billryanThe difference here might be semantics more than math.
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Much of it is. But once you get past the semantics there is a real conceptual issue in the math where people have trouble with seeing 1/11 vs 1/6.
The problem is people go back to being bogged down in the semantics.
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Wizard is working on a reworking of the puzzle where the semantics were ironed out. I believe he's doing that, cos like the Monty Hall problem, this is one where the answer is not as intuitive as one would expect. In his reworked version, the 1 in 11 answer is absolutely proven, but still expect AlanM and a few others to plumb for and cling on to 1/6
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Easy to work out semantics. Just have someone else that can’t see the dice to ask the peeker “is there at least one 2” and the peeker truthfully says “yes.”
