Of course.Quote: AlanMendelsonQuote: OnceDear
If one of the dice is a two, then we cannot truthfully say that you set one of the dice as a two unless we had evidence. No evidence. No knowing.
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Are you sure this is your position?
Explain why it's a trick question. No one in the original question said anything about anyone setting dice.Quote:
Because if it is your position then what I said early on in this discussion was correct: it's a trick question.
No. I'm not saying that might not be true. It was true. At least one of the dice was a two. Just that it was not preset to be a two before the peeking.Quote:And now I'll add this: the original question is an invalid question because it contained misinformation.
Remember-- in the original question we are told TRUTHFULLY that at least one die is a 2.
But now you're saying that might not be true?
At least one of the dice was a two.Quote:Either at least one of the two dice is a two or isn't. Which is it?
What question? The assertion that at least one of the dice was a two was both truthful and valid, so I guess this discussion is not over and this question is not stricken. Whatever that means to you.Quote:
And if you tell me the question is not truthful or valid or doesn't correctly illustrate the shaken dice in the cup then the discussion is over and this question is stricken.
Quote:
Quote: AlanMendelsonQuote: DieterQuote: AlanMendelson
I just would like to know if at least one of the two dice in the cup is really a 2?
link to original post
I believe 7:1 has been offered, and it's a push if none of the dice show a 2 (or whatever point has been selected for the current round).
link to original post
I'm asking about the original question in the original post.
Is there a 2 or not?
link to original post
At least one 2 is showing.
However, endlessly discussing the matter is fruitless, and we have been cautioned to settle this experimentally.
Quote: DweenAlong the vein of the Two Coin Puzzle, I wanted to put this in another thread as to not clog up the other.
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?
Based on the discussion in the Two Coin thread, I'm interested to see how people solve this one.Pleased to meet you, name's Francis Pumphandle, but everyone calls me Pip. Cheese balls are one of my all-time favorite foods. I always seem to meet the most interesting people when I'm around them, too. In fact, cheese balls bring to mind the time I met Bob Barker. Yes, indeed -- Bob Barker, star of the most popular morning game show. He's an emcee, a host and a celebrity all rolled into one. Anyway, eight months ago, it was Tuesday the 17th, I believe, or it might have been the 18th -- no, no, it was definitely the 17th, because it was precisely one week after my Aunt Lucretia's birthday, which is the 10th. Aunt Lucretia's quite a woman -- loves to cook. She prepares a fabulous "war-shoo-off" -- that's a Chinese duck dish. I love Chinese food. I once went to a party where they served Chinese food and cheese balls. Now that was a Catch-22 situation. Catch-22 was a movie, you know. It was long -- VERY long. They say the book was better, but it was a novel, and I never finish reading those things. Of course, a lot of people don't read much nowadays; they watch television. I caught a program on PBS last night -- a very good show on chimpanzees in the media. They had a clip of Jay Fred Mug and a chimp on the Today Show, but it was Fred's chimpanzee's girlfriend that had me stumped. I couldn't remember her name, so I looked it up. Her name was Fibi B. Bibi. Anyway, as I was saying, eight months ago, Tuesday the 17th, I went downtown on a nice relaxing stroll. I love to relax. In fact, relaxing is a hobby of mine. Some people play golf, others like tennis, horseshoes, bridge, canasta, and other such fancy hobbies. Now another hobby enjoyed by many is knitting. My grandmother was a great knitter -- knitted this sweater I'm wearing. It's red, which is not my favorite color. I prefer mauve or a mustard yellow. Now, don't get me wrong, red is o.k. for ties and suspenders, but with sweaters I prefer more neutral colors. But when I'm relaxing, I don't care WHAT I wear -- long pants, bermuda shorts, t-shirts, or formal attire, you name it -- anything goes. Now, on the 17th, during my relaxing stroll, I recall wearing my herringbone jacket, my Laughlin, Nevada souvenir tie, and my charcoal grey slacks -- or was it the navy slacks? Oh, I suppose it doesn't really matter, does it? What matters is comfort. You know, I once stayed at a Comfort Inn -- warm, cozy, comfortable. I love comfort. It goes along with that pastime of mine -- relaxing. Now, for me, there's nothing more relaxing than a nice leisurely stroll like the one I took eight months ago on the 17th. It was a bright, sunny day, which of course is the optimum condition for relaxed strolling. And as I walked along, I found myself humming a haunting melody. I kept humming and humming and humming and humming. I couldn't get the tune out of my head. I racked my brain to come up with the title, but to no avail. You see, I'm not terribly musical -- and yet, I'd always wanted to play a musical instrument and be like my musical hero, Leo Sayer. But who can compete with Leo? I think I was just scared I'd fail. Well, I decided right then and there to go buy a musical instrument. So on the particular Tuesday the 17th to which I was referring, I went down to the Sixth Street Musical Emporium to buy a new tambourine, a terribly soothing instrument contrary to popular opinion. And as I was strolling along, I detected a wonderful scent in the morning air. "What could it be?" I asked myself. So I went toward that marvelous scent, distracted by its aroma from my musical mission. The odor was a mix of orchid flowers and bologna, which, of course, is one of the world's most underappreciated luncheon meats -- that and pimento loaf. I love a good pimento-loaf-and-mayo sandwich -- the more pimentos, the better. Why just the mention of pimentos makes my taste buds stand up and say "Howdy". Now there's an interesting word -- "howdy". Is it from "How are you?", or maybe "How ya doing?" "Howdy"'s one of those strange words that really HAS no origin. I like saying, "How do" more than "Howdy" -- more formal, I think -- not too flowery. But the flowery aroma of that particular Tuesday morning carried me on my fragrant quest. Now, the smell was actually less bologna and more orchid, the beautiful flower found on the island state of Hawaii. Of course, I wasn't in Hawaii, so I needed to search out the location of the nearest orchid. So, I visited every flower shop in town. Well, to make a long story short, not a SINGLE flower shop in town had ANY orchids in stock, which seemed mighty curious to me. Now, as we all know, curiosity killed the cat, but since I'm not feline, I wasn't too worried. Felines are funny creatures, don't you think? I had a cat once. It used its claws to tear my living room couch to shreds. It was a comfy couch, too -- had a sleepaway bed in it with a foam rubber mattress. Now, I bought the couch AND the mattress at Levine's department store on Third Avenue the very same afternoon of that relaxing stroll aforementioned. I also bought myself a lovely tambourine on that same shopping expedition. Anyway, I didn't want to pay extra for the delivery of the couch, so I decided to carry the couch home myself. It was quite cumbersome, and getting it through the store's revolving doors was a bit of a challenge. And just as I emerged onto the street, by accident I bumped into a well-dressed man with an orchid in his lapel. It was Bob Barker, and he was eating and bologna-and-cheese-ball sandwich. Well, it's been nice chatting with you. Bye!
link to original post
Tell me what's the difference between this question as stated and setting at least one of the two dice on 2?
The previous responses said "we dont know which of the dice in the cup is a two."
I'm saying it doesnt matter.
Quote: DieterQuote: AlanMendelsonQuote: DieterQuote: AlanMendelson
I just would like to know if at least one of the two dice in the cup is really a 2?
link to original post
I believe 7:1 has been offered, and it's a push if none of the dice show a 2 (or whatever point has been selected for the current round).
link to original post
I'm asking about the original question in the original post.
Is there a 2 or not?
link to original post
At least one 2 is showing.
However, endlessly discussing the matter is fruitless, and we have been cautioned to settle this experimentally.
link to original post
Your "experiment" does not represent the original question.
Quote: AlanMendelsonHere is the original question:
Quote: DweenAlong the vein of the Two Coin Puzzle, I wanted to put this in another thread as to not clog up the other.
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?
Based on the discussion in the Two Coin thread, I'm interested to see how people solve this one.Pleased to meet you, name's Francis Pumphandle, but everyone calls me Pip. Cheese balls are one of my all-time favorite foods. I always seem to meet the most interesting people when I'm around them, too. In fact, cheese balls bring to mind the time I met Bob Barker. Yes, indeed -- Bob Barker, star of the most popular morning game show. He's an emcee, a host and a celebrity all rolled into one. Anyway, eight months ago, it was Tuesday the 17th, I believe, or it might have been the 18th -- no, no, it was definitely the 17th, because it was precisely one week after my Aunt Lucretia's birthday, which is the 10th. Aunt Lucretia's quite a woman -- loves to cook. She prepares a fabulous "war-shoo-off" -- that's a Chinese duck dish. I love Chinese food. I once went to a party where they served Chinese food and cheese balls. Now that was a Catch-22 situation. Catch-22 was a movie, you know. It was long -- VERY long. They say the book was better, but it was a novel, and I never finish reading those things. Of course, a lot of people don't read much nowadays; they watch television. I caught a program on PBS last night -- a very good show on chimpanzees in the media. They had a clip of Jay Fred Mug and a chimp on the Today Show, but it was Fred's chimpanzee's girlfriend that had me stumped. I couldn't remember her name, so I looked it up. Her name was Fibi B. Bibi. Anyway, as I was saying, eight months ago, Tuesday the 17th, I went downtown on a nice relaxing stroll. I love to relax. In fact, relaxing is a hobby of mine. Some people play golf, others like tennis, horseshoes, bridge, canasta, and other such fancy hobbies. Now another hobby enjoyed by many is knitting. My grandmother was a great knitter -- knitted this sweater I'm wearing. It's red, which is not my favorite color. I prefer mauve or a mustard yellow. Now, don't get me wrong, red is o.k. for ties and suspenders, but with sweaters I prefer more neutral colors. But when I'm relaxing, I don't care WHAT I wear -- long pants, bermuda shorts, t-shirts, or formal attire, you name it -- anything goes. Now, on the 17th, during my relaxing stroll, I recall wearing my herringbone jacket, my Laughlin, Nevada souvenir tie, and my charcoal grey slacks -- or was it the navy slacks? Oh, I suppose it doesn't really matter, does it? What matters is comfort. You know, I once stayed at a Comfort Inn -- warm, cozy, comfortable. I love comfort. It goes along with that pastime of mine -- relaxing. Now, for me, there's nothing more relaxing than a nice leisurely stroll like the one I took eight months ago on the 17th. It was a bright, sunny day, which of course is the optimum condition for relaxed strolling. And as I walked along, I found myself humming a haunting melody. I kept humming and humming and humming and humming. I couldn't get the tune out of my head. I racked my brain to come up with the title, but to no avail. You see, I'm not terribly musical -- and yet, I'd always wanted to play a musical instrument and be like my musical hero, Leo Sayer. But who can compete with Leo? I think I was just scared I'd fail. Well, I decided right then and there to go buy a musical instrument. So on the particular Tuesday the 17th to which I was referring, I went down to the Sixth Street Musical Emporium to buy a new tambourine, a terribly soothing instrument contrary to popular opinion. And as I was strolling along, I detected a wonderful scent in the morning air. "What could it be?" I asked myself. So I went toward that marvelous scent, distracted by its aroma from my musical mission. The odor was a mix of orchid flowers and bologna, which, of course, is one of the world's most underappreciated luncheon meats -- that and pimento loaf. I love a good pimento-loaf-and-mayo sandwich -- the more pimentos, the better. Why just the mention of pimentos makes my taste buds stand up and say "Howdy". Now there's an interesting word -- "howdy". Is it from "How are you?", or maybe "How ya doing?" "Howdy"'s one of those strange words that really HAS no origin. I like saying, "How do" more than "Howdy" -- more formal, I think -- not too flowery. But the flowery aroma of that particular Tuesday morning carried me on my fragrant quest. Now, the smell was actually less bologna and more orchid, the beautiful flower found on the island state of Hawaii. Of course, I wasn't in Hawaii, so I needed to search out the location of the nearest orchid. So, I visited every flower shop in town. Well, to make a long story short, not a SINGLE flower shop in town had ANY orchids in stock, which seemed mighty curious to me. Now, as we all know, curiosity killed the cat, but since I'm not feline, I wasn't too worried. Felines are funny creatures, don't you think? I had a cat once. It used its claws to tear my living room couch to shreds. It was a comfy couch, too -- had a sleepaway bed in it with a foam rubber mattress. Now, I bought the couch AND the mattress at Levine's department store on Third Avenue the very same afternoon of that relaxing stroll aforementioned. I also bought myself a lovely tambourine on that same shopping expedition. Anyway, I didn't want to pay extra for the delivery of the couch, so I decided to carry the couch home myself. It was quite cumbersome, and getting it through the store's revolving doors was a bit of a challenge. And just as I emerged onto the street, by accident I bumped into a well-dressed man with an orchid in his lapel. It was Bob Barker, and he was eating and bologna-and-cheese-ball sandwich. Well, it's been nice chatting with you. Bye!
link to original post
Tell me what's the difference between this question as stated and setting at least one of the two dice on 2?
The previous responses said "we dont know which of the dice in the cup is a two."
I'm saying it doesnt matter.
link to original post
We don't. It does.
Quote: AlanMendelsonQuote: DieterQuote: AlanMendelsonQuote: DieterQuote: AlanMendelson
I just would like to know if at least one of the two dice in the cup is really a 2?
link to original post
I believe 7:1 has been offered, and it's a push if none of the dice show a 2 (or whatever point has been selected for the current round).
link to original post
I'm asking about the original question in the original post.
Is there a 2 or not?
link to original post
At least one 2 is showing.
However, endlessly discussing the matter is fruitless, and we have been cautioned to settle this experimentally.
link to original post
Your "experiment" does not represent the original question.
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Alan,
Does hiding the dice under a cocktail shaker, backgammon cup, or cover of darkness somehow differ from a roll not resolving a wager if there's no 2 showing?
I'm sorry, but I need to shut up before I catch trouble.
Do you assert that "if we roll a pair of dice and at least one of them is a two after they both land, that one time in six, they will both be a two?"
Remember. WE IGNORE every occasion where neither dice was a two.
Yay or Nay?
No paraphrasing, No analogy, No saying that's the same as blah blah blah. or that's not what the original... blah blah blah
Do you assert that?
Yay or Nay
Quote: DieterQuote: AlanMendelsonQuote: DieterQuote: AlanMendelsonQuote: DieterQuote: AlanMendelson
I just would like to know if at least one of the two dice in the cup is really a 2?
link to original post
I believe 7:1 has been offered, and it's a push if none of the dice show a 2 (or whatever point has been selected for the current round).
link to original post
I'm asking about the original question in the original post.
Is there a 2 or not?
link to original post
At least one 2 is showing.
However, endlessly discussing the matter is fruitless, and we have been cautioned to settle this experimentally.
link to original post
Your "experiment" does not represent the original question.
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Alan,
Does hiding the dice under a cocktail shaker, backgammon cup, or cover of darkness somehow differ from a roll not resolving a wager if there's no 2 showing?
I'm sorry, but I need to shut up before I catch trouble.
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I don't claim to be smart. I will say I have no idea what you're talking about.
Quote: AlanMendelsonI don't claim to be smart. I will say I have no idea what you're talking about.
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Preemptive warning to Dieter. dont type what you are screaming at the screen. It may be true, but we are not allowed to type obscenities nor call people names. $:o)
Quote: OnceDearQuote: AlanMendelsonI don't claim to be smart. I will say I have no idea what you're talking about.
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Preemptive warning to Dieter. dont type what you are screaming at the screen. It may be true, but we are not allowed to type obscenities nor call people names. $:o)
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In response to both you and Alan,
"I get that a lot."
Quote: OnceDearNow Alan, I'll question your position. Please answer this clearly and unambiguously. It's not a trick question....
Do you assert that "if we roll a pair of dice and at least one of them is a two after they both land, that one time in six, they will both be a two?"
No. If you roll a pair of dice, once in 36 you'll get 2-2.
If you shake two dice in a cup, and at least one is a 2, then you have a 1/6 chance that the other die will also be a 2 which would give you 2-2.
Quote: AlanMendelsonQuote: OnceDearNow Alan, I'll question your position. Please answer this clearly and unambiguously. It's not a trick question....
Do you assert that "if we roll a pair of dice and at least one of them is a two after they both land, that one time in six, they will both be a two?"
No. If you roll a pair of dice, once in 36 you'll get 2-2.
If you shake two dice in a cup, and at least one is a 2, then you have a 1/6 chance that the other die will also be a 2 which would give you 2-2.
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Does "shaking the dice in a cup, and re-rolling if there is not at least one 2" differ from "throwing the dice, and re-rolling if there is not at least one 2"?
Quote: DieterQuote: AlanMendelsonQuote: OnceDearNow Alan, I'll question your position. Please answer this clearly and unambiguously. It's not a trick question....
Do you assert that "if we roll a pair of dice and at least one of them is a two after they both land, that one time in six, they will both be a two?"
No. If you roll a pair of dice, once in 36 you'll get 2-2.
If you shake two dice in a cup, and at least one is a 2, then you have a 1/6 chance that the other die will also be a 2 which would give you 2-2.
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Does "shaking the dice in a cup, and re-rolling if there is not at least one 2" differ from "throwing the dice, and re-rolling if there is not at least one 2"?
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When you re-roll one die, you have a 1/6 chance of getting any of the 6 numbers.
If you are re-rolling both dice we are back to a 1/36 chance of getting 2-2.
I said it all along that this is a matter of English and not Math.
I suspect someone might now be starting to understand.
As I've mentioned before there is another question which would be answered with 1/11.
Dieter I'm sorry but I do not understand the relevance of your question.
I dont care if you roll or shake two dice.
To get 2-2 at once is 1/36.
To have one die showing a 2, it's a 1/6 chance that the other die will be a two on the current roll/shake or the next one.
Quote: AlanMendelsonQuote: OnceDearDo you assert that "if we roll a pair of dice and at least one of them is a two after they both land, that one time in six, they will both be a two?"
No. [edit out irrelevance for brevity]
If you shake two dice in a cup, and at least one is a 2, then you have a 1/6 chance that the other die will also be a 2 which would give you 2-2.
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You said NO. you do not assert that. Then you do exactly assert that.
So, I'll play this game.
Alan. Do you assert that "If you shake two dice in a cup, and at least one is a 2, then you have a 1/6 chance that the other die will also be a 2 which would give you 2-2."
Bonus question. How is that different to
"if we roll a pair of dice and at least one of them is a two after they both land, [that] one time in six, they will both be a two?"
apart from your using a cup. You roll your dice in a cup. At least one is a two. I roll two dice. At least one is a two.
Do you see some subtle difference in the scenarios? Please explain.
Never mind your 1 in 36 irrelevance.
Do you assert that "if we roll a pair of dice and at least one of them is a two after they both land, that you then have a 1/6 chance that the other die will also be a 2 which would give you 2-2."?"
Quote: AlanMendelson
I said it all along that this is a matter of English and not Math.
I suspect someone might now be starting to understand.
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Alan, You are not getting close to starting to understanding. Everyone else understood 9 years ago.
Bored now. Catch you all later.
Q: Alan. Do you assert that "If you shake two dice in a cup, and at least one is a 2, then you have a 1/6 chance that the other die will also be a 2 which would give you 2-2."
A: Yes.
Q: Bonus question. How is that different to
"if we roll a pair of dice and at least one of them is a two after they both land, [that] one time in six, they will both be a two?"
A: I know this is difficult but I'm afraid we have a problem here. When you roll 2 dice there's a 1/36 chance the result will be 2-2. If we roll two dice and one lands on a 2 there is a 1/6 chance that the other die will also be a 2, but together getting 2-2 is a 1/36 event.
Q: (and I'm lumping together everything else)
apart from your using a cup. You roll your dice in a cup. At least one is a two. I roll two dice. At least one is a two.
Do you see some subtle difference in the scenarios? Please explain.
Never mind your 1 in 36 irrelevance.
Do you assert that "if we roll a pair of dice and at least one of them is a two after they both land, that you then have a 1/6 chance that the other die will also be a 2 which would give you 2-2."?"
A: i think I've answered this. There are different events. There are different odds for each event.
Chance of rolling 2-2 is 1/36
Chance of rolling a 2 on one die is 1/6
Chance of rolling a 2 on one die after another die shows a 2 is 1/6
Chance of a second die showing a 2 after another die has been rolled as a 2 is 1/6
There are 11 combinations on two dice with at least one 2.
There is one combination from that 11 combinations that shows 2-2.
All of these statements are correct.
Apply them to the correct question.
Quote: AlanMendelsonOkay, I'll play along.
Q: Alan. Do you assert that "If you shake two dice in a cup, and at least one is a 2, then you have a 1/6 chance that the other die will also be a 2 which would give you 2-2."
A: Yes.
...
Chance of rolling 2-2 is 1/36
Chance of rolling a 2 on one die is 1/6
link to original post
Chance of rolling a 2 on at least one die is 11/36
Chance of rolling a pair of 2's where at least one die is a 2 is 1/11
https://duckduckgo.com/?q=monty+python+argument+sketch+youtube&iax=videos&ia=videos&iai=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DohDB5gbtaEQ
Really bored now.
Chance of rolling a pair of 2's where at least one die is a 2 is 1/11"
And the reason I wont take part in the "bet" is that I'd lose $100 to win $80.
But if you rolled a single die.... ah!
And if my mother had wheels, she'd be a bicycle.Quote: AlanMendelsonBut if you rolled a single die.... ah!
link to original post
In the original post, the roller did roll a two. They were under a cup.
At least one die really really WAS a two.
"Chance of rolling a pair of 2's where at least one die is a 2 is 1/11" You said that.
but we just said "At least one die WAS a two."
That's what logicians call a TRUE statement. I understand your American English has a similar concept
The chance that he had rolled "a pair of 2's [where our truth is true] was 1/11"
simplifying.
The chance that he had rolled "a pair of 2's was 1/11"
But, you still won't get it.
Quote: OnceDearAnd if my mother had wheels, she'd be a bicycle.Quote: AlanMendelsonBut if you rolled a single die.... ah!
link to original post
In the original post, the roller did roll a two. They were under a cup.
At least one die really really WAS a two.
"Chance of rolling a pair of 2's where at least one die is a 2 is 1/11" You said that.
but we just said "At least one die WAS a two."
That's what logicians call a TRUE statement. I understand your American English has a similar concept
The chance that he had rolled "a pair of 2's [where our truth is true] was 1/11"
simplifying.
The chance that he had rolled "a pair of 2's was 1/11"
But, you still won't get it.
link to original post
Actually we're very close. I don't disagree with what you posted. But what you posted doesn't answer the question that was asked.
Here again is the question that was asked:
"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?"
It is true that when one die shows a 2 there is a 1/11 chance that both dice are showing 2s.
But the question didnt ask you for that answer.
The question asked you that knowing that one die is a 2 what is required of the other die? And that is that 1/6 sides must be a two.
The correct answer is determined by the question.
Why bother setting up the question with shaking two dice in a cup and seeing one die IF the real question was "there are 11 combinations of two dice showing at least one 2. How many combinations show 2-2?"
Why bother? Because it is not immediately apparent there are 11 combinations of two dice showing at least one 2.Quote: AlanMendelson...Why bother setting up the question with shaking two dice in a cup and seeing one die IF the real question was "there are 11 combinations of two dice showing at least one 2. How many combinations show 2-2?"
And for those that fail to notice this, they will arrive at an incorrect answer. That's part of the fun. That's why this can be a considered a puzzle / brain teaser / logic problem / math problem.
Yes, the real question amounts to exactly that. How many combinations show 2-2, given that we know there are 11 combinations that show at least one 2. But if pose the question that way, you're already alerting the reader to the number of valid combinations. You're giving them a hint to the correct answer when no hint should be required.
Quote: EdCollinsWhy bother? Because it is not immediately apparent there are 11 combinations of two dice showing at least one 2.Quote: AlanMendelson...Why bother setting up the question with shaking two dice in a cup and seeing one die IF the real question was "there are 11 combinations of two dice showing at least one 2. How many combinations show 2-2?"
link to original post
Exactly right. And the simplest way to answer the question is to consider the remaining die... which has six sides with one side showing a 2.
Doesnt the simplest answer win? Or do we have to go through the lengthy explanation?
No. Your statement "The simplest way to answer the question is to consider the remaining die..." is not correct.Quote: AlanMendelsonExactly right. And the simplest way to answer the question is to consider the remaining die... which has six sides with one side showing a 2.
Doesnt the simplest answer win? Or do we have to go through the lengthy explanation?
That's not the method used to solve this question/problem. That's what gets you into trouble. That's where and why many people will believe that 1/6 is the correct answer.
This logic problem / brain teaser / math problem (call if what you prefer... all can be considered valid) does not need to be re-worded. It's clear and concise. It's about as simple a logic problem as you can get. (I've seen lots of logic problems... we all have.. and some of them require many sentences, even paragraphs, to set up/explain.)
The question doesn't even attempt to misdirect you, as other brain teaser-type problems do.
To arrive at the proper answer, it DOES require you to realize that we are only interested in 11 of 36 dice combinations. And yet that will not be apparent to the many people reading the problem for the first time. A typical person on the street, for example, will immediately answer 1/6.
And they would be wrong.
Again, this is what makes it an interesting / fun problem. The correct answer is not readily apparent or obvious. And yet the question should not, in any way, be re-worded to clarify or point out this fact... that to arrive at the correct answer, we need to realize we are are only really interested in 11 of the initial 36 combinations.
That's up the the reader to figure out on their own.
Quote: AlanMendelsonActually we're very close.
Here again is the question that was asked:
"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?"
It is true that when one die shows a 2 there is a 1/11 chance that both dice are showing 2s.
But the question didn't ask you for that answer.
Yes Alan. it did ask that quite explicitly. It did NOT ask anything about any 'Other' die. It told you explicitly that "one die showed a two" and then asked very explicitly What is the probability that both dice are showing a 2?"
True it did use the words "At Least "one of the dice is a two, rather than "exactly" one of the dice is a two (Which would be a pointlessly absurd other question)
Compare and contrast the Question and your answer...
Quote: Your latest answer"when one die shows a 2 there is a 1/11 chance that both dice are showing 2s.
You tell me how, exactly, your answer there is NOT the answer to the exact original question.
Quote: Original question as asked"At least one of the dice is a 2.What is the probability that both dice are showing a 2?"
Of course, if it had said "Exactly one" or "Only one" then probability would have been zero. But that's another question that was not asked. It was actually generously explicit in saying "at least" to remove one ambiguity.
It does not answer the question of what are the odds for a 2 showing on one die.
That's my statement. Thank you.
Lunch at Red Rock is on me.
Quote: AlanMendelsonWizard. The way you have this bet structured you are expected to win $100 for every $80 you lose.
It does not answer the question of what are the odds for a 2 showing on one die.
That's my statement. Thank you.
Lunch at Red Rock is on me.
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Note: bold font added to above quote.
Alan, no person on this thread, or anywhere on Earth, has been asking or wondering about "what are the odds for a 2 showing on one die?" The question has been, "if two dice are rolled, and you know that at least one of the dice is a two, what are the odds that both dice are a two."
I view your post above as an admission that you have been wrong. I credit EdCollins with the well-worded posts that convinced you.
Quote: gordonm888Quote: AlanMendelsonWizard. The way you have this bet structured you are expected to win $100 for every $80 you lose.
It does not answer the question of what are the odds for a 2 showing on one die.
That's my statement. Thank you.
Lunch at Red Rock is on me.
link to original post
Note: bold font added to above quote.
Alan, no person on this thread, or anywhere on Earth, has been asking or wondering about "what are the odds for a 2 showing on one die?" The question has been, "if two dice are rolled, and you know that at least one of the dice is a two, what are the odds that both dice are a two."
I view your post above as an admission that you have been wrong. I credit EdCollins with the well-worded posts that convinced you.
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Sorry. But the original question asks you to consider one unknown die.
I stand by my answer.
Where?Quote: AlanMendelson
Sorry. But the original question asks you to consider one unknown die.
I stand by my answer.
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Where does it use the words "the other die" or anything similar?
We don't know the value of either die. We only know that between the two of them theres at least one showing a two.
Which part of Pair of dice don't you grasp
Quote: OnceDearWhere?Quote: AlanMendelson
Sorry. But the original question asks you to consider one unknown die.
I stand by my answer.
link to original post
Where does it use the words "the other die" or anything similar?
We don't know the value of either die. We only know that between the two of them theres at least one showing a two.
Which part of Pair of dice don't you grasp
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I already said it's in how the question was worded.
Quote: AlanMendelsonQuote: gordonm888Quote: AlanMendelsonWizard. The way you have this bet structured you are expected to win $100 for every $80 you lose.
It does not answer the question of what are the odds for a 2 showing on one die.
That's my statement. Thank you.
Lunch at Red Rock is on me.
link to original post
Note: bold font added to above quote.
Alan, no person on this thread, or anywhere on Earth, has been asking or wondering about "what are the odds for a 2 showing on one die?" The question has been, "if two dice are rolled, and you know that at least one of the dice is a two, what are the odds that both dice are a two."
I view your post above as an admission that you have been wrong. I credit EdCollins with the well-worded posts that convinced you.
link to original post
Sorry. But the original question asks you to consider one unknown die.
I stand by my answer.
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Knowing one of two dies is a 2 is not the same as knowing one die is a 2. The way you are looking at it seems to be that when two are thrown and you know one is a two, you simply put that die to the side and worry about the six faces left on the remaining die. Since there are six sides, you think the answer has to be one in six.
The problem is we don't know which of the two die has rolled the two.
When you know one of the two dies lands on the number two, it means there are 11 other combinations remaining. Only one of those eleven will give you a pair of twos..
It's counter-intuitive but correct. It very well may be that people who think in math don't find it counter-intuitive at all. My limited encounters with true mathematicians has convinced me they think differently than most of us.
Quote: billryanKnowing one of two dies is a 2 is not the same as knowing one die is a 2. The way you are looking at it seems to be that when two are thrown and you know one is a two, you simply put that die to the side and worry about the six faces left on the remaining die. Since there are six sides, you think the answer has to be one in six.
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Hey Bill... when you roll two dice and you're told one of them is showing a 2, how many more dice are there to consider?
Or, do you count one die twice?
And if you count one die twice, why isn't the answer 1/12 because you don't know if its really a 2?
Quote: WizardIt seems to me Alan interprets the question as the peeker looks at one die only, as if the other die were invisible.
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No. The peeker has said truthfully that at least one die is showing a 2.
That leaves you to consider what could be on one die.
Since a die has six faces the answer is 1/6.
It doesn't matter which die it is. It's a die with six faces.
Quote: AlanMendelson
I already said it's in how the question was worded.
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Nothing wrong with the question as worded, except the issue of missing calling rules, which is an aspect Alan never even acknowledged.
No Alan. I'm not having that. You fabricated and answered a question of your own creation wherein one die was preset or observed first. For 9 years now we have been telling you that your scenario bears no resemblance to the original question as originally worded. And for 9 years you have been insisting that your question has the exact same meaning and route to an answer. An answer which you derived as 1/6.
Right answer to your question. Wrong answer to this one.
You were wrong then. You are wrong now. You will be forever wrong. You are clearly incapable of seeing that you are wrong, or maybe you see it and you are incapable of admitting you are wrong.
Which is it? "I am right" is not an option for you.
If you want to answer your own question correctly, you cannot blame the true original question's wording. That wording which you so embarrassingly cannot understand, or cannot admit to understanding.
Just go on embarrassing yourself. Plenty will be happy to help you, including myself. I had chided wizard for embarrassing you by re-flagging your previous record on this topic, as it seemed to be borderline bullying. But you seem to revel in your .... what shall I call it.... 'inability'
To Gordon and Ed,
Nice wording Ed, but both you and Gordon have overestimated some aspect of Alan by several orders of magnitude. You underestimated his tenacity to cling to his point of view.
Alan is not convinced. Never will be convinced. Cannot be convinced. He is hard wired to be wrong on this matter.
Not just a tiny bit inflexible. He has invested too much of himself into his alternative question and answer pair.
Now..... I must return to the real world, briefly.
I now retract what I said in that post. Specifically the different wording is irrelevant, and both ways have a 1/11 result for the second die being a 2 when at least one is a 2.
And now I propose a new way to look at it:
Without any prompting from the observer, a person shakes the dice under the cup, peeks, and truthfully announces, "At least one die is a 3. What's the odds that the other die is a 4?"
Note: There's no reason I picked 3 or 4 other than to avoid confusing it with the problem of the two 2s.
In all scenarios, there are 36 possible combinations of the roll outcome. Once the value of one die is known, you're reduced to 11 of those combinations. 1 of which is a pair while 10 are not - 2 for each of the other values.
Therefore, if one die is a 3, there are 2/11 chances that the other die is a 4.
You count the pair of dice once. You don't know anything about either of the dice, only that between them they can muster a two.Quote: AlanMendelson
Or, do you count one die twice?
And if you count one die twice, why isn't the answer 1/12 because you don't know if its really a 2?
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Why isn't it 1/12. Why the heck should it be 1/12? Because there are 12 faces all together? What's that got to do with the price of fish?
You disagree. That's fine with me.
We disagree.
If you want to think the correct answer to THIS QUESTION is 1/11 then so be it:
**************
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?
****************
Have a nice day.
Quote: AlanMendelsonQuote: WizardIt seems to me Alan interprets the question as the peeker looks at one die only, as if the other die were invisible.
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No. The peeker has said truthfully that at least one die is showing a 2.
That leaves you to consider what could be on one die.
Since a die has six faces the answer is 1/6.
It doesn't matter which die it is. It's a die with six faces.
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How do you choose which die to consider?. There are two in front of you. If the left die has the two, then the right die has a 1/6 chance of giving you a pair of twos. But what if left die has a six? If the left die has a six, there is zero chance of getting a pair of twos so it doesn't matter if the chances of it being a two are one in two or one in a million. You can't get a pair unless you successfully saved the die with the two to begin with.
You think that you can just put aside the die that has the two and concentrate on the other die. Since you don't know which of the dice is the correct die, you'll pick the wrong one half the time and choose to set aside the die that didn't have a two, eliminating any possibility of getting the pair.
Quote: OnceDearQuote: AlanMendelson
I already said it's in how the question was worded.
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Nothing wrong with the question as worded, except the issue of missing calling rules, which is an aspect Alan never even acknowledged.
No Alan. I'm not having that. You fabricated and answered a question of your own creation wherein one die was preset or observed first. For 9 years now we have been telling you that your scenario bears no resemblance to the original question as originally worded. And for 9 years you have been insisting that your question has the exact same meaning and route to an answer. An answer which you derived as 1/6.
Right answer to your question. Wrong answer to this one.
You were wrong then. You are wrong now. You will be forever wrong. You are clearly incapable of seeing that you are wrong, or maybe you see it and you are incapable of admitting you are wrong.
Which is it? "I am right" is not an option for you.
If you want to answer your own question correctly, you cannot blame the true original question's wording. That wording which you so embarrassingly cannot understand, or cannot admit to understanding.
Just go on embarrassing yourself. Plenty will be happy to help you, including myself. I had chided wizard for embarrassing you by re-flagging your previous record on this topic, as it seemed to be borderline bullying. But you seem to revel in your .... what shall I call it.... 'inability'
To Gordon and Ed,
Nice wording Ed, but both you and Gordon have overestimated some aspect of Alan by several orders of magnitude. You underestimated his tenacity to cling to his point of view.
Alan is not convinced. Never will be convinced. Cannot be convinced. He is hard wired to be wrong on this matter.
Not just a tiny bit inflexible. He has invested too much of himself into his alternative question and answer pair.
Now..... I must return to the real world, briefly.
link to original post
What's it like? The real world? I'm hoping to visit it someday , after I've managed to fix the internet.
Quote: DJTeddyBearBack on page 9, in this post, I stated that Alan's position on the wording forced me to give this thing a re-think. I'm not entirely sure I captured what Alan was trying to say, but what I said in that post seemed logical in my pea-brain, but it was wrong.
I now retract what I said in that post. Specifically the different wording is irrelevant, and both ways have a 1/11 result for the second die being a 2 when at least one is a 2.
And now I propose a new way to look at it:
Without any prompting from the observer, a person shakes the dice under the cup, peeks, and truthfully announces, "At least one die is a 3. What's the odds that the other die is a 4?"
Note: There's no reason I picked 3 or 4 other than to avoid confusing it with the problem of the two 2s.2/11Despite what I wrote in my prior post, it doesn't matter if the other person asks the peeker about a specific value or not, except to remove the times that the answer is neither. Nor does it matter if the person peeking is asking if the other die is the same as the announced die or not.
In all scenarios, there are 36 possible combinations of the roll outcome. Once the value of one die is known, you're reduced to 11 of those combinations. 1 of which is a pair while 10 are not - 2 for each of the other values.
Therefore, if one die is a 3, there are 2/11 chances that the other die is a 4.
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Sorry. This still depends on the calling rules. If you only see one slamming you do NOT know those rules.
Let the ruling be "If he sees a 4 he confirms it: If he doesn't he reports nothing"
1,1
1,2
1,3 He calls "At least one of the dice is a 3"
1,4
1,5
1,6
2,1
2,2
2,3 He calls "At least one of the dice is a 3"
2,4
2,5
2,6
3,1 He calls "At least one of the dice is a 3"
3,2 He calls "At least one of the dice is a 3"
3,3 He calls "At least one of the dice is a 3"
3,4 He calls "At least one of the dice is a 3" AND The other dice is a 4
3,5 He calls "At least one of the dice is a 3"
3,6 He calls "At least one of the dice is a 3"
4,1
4,2
4,3 He calls "At least one of the dice is a 3" AND the other is a 4
4,4
4,5
4,6
5,1
5,2
5,3 He calls "At least one of the dice is a 3"
5,4
5,5
5,6
6,1
6,2
6,3 He calls "At least one of the dice is a 3"
6,4
6,5
6,6
Odds of the other dice being a 4 are 2/11
Now. If the calling rules were "I call the first die I see, nearest to me. I call every time"
And NO REASON to assume those are not the rules.
1,1 He calls "At least one of the dice is a 1"
1,2 He calls "At least one of the dice is a 1"
1,3 He calls "At least one of the dice is a 1"
1,4 He calls "At least one of the dice is a 1"
1,5 He calls "At least one of the dice is a 1"
1,6 He calls "At least one of the dice is a 1"
2,1 He calls "At least one of the dice is a 2"
2,2 He calls "At least one of the dice is a 2"
2,3 He calls "At least one of the dice is a 2"
2,4 He calls "At least one of the dice is a 2"
2,5 He calls "At least one of the dice is a 2"
2,6 He calls "At least one of the dice is a 2"
3,1 He calls "At least one of the dice is a 3"
3,2 He calls "At least one of the dice is a 3"
3,3 He calls "At least one of the dice is a 3"
3,4 He calls "At least one of the dice is a 3" And the other die is a 4
3,5 He calls "At least one of the dice is a 3"
3,6 He calls "At least one of the dice is a 3"
4,1 He calls "At least one of the dice is a 4"
4,2 He calls "At least one of the dice is a 4"
4,3 He calls "At least one of the dice is a 4"
4,4 He calls "At least one of the dice is a 4"
4,5 He calls "At least one of the dice is a 4"
4,6 He calls "At least one of the dice is a 4"
5,1 He calls "At least one of the dice is a 5"
5,2 He calls "At least one of the dice is a 5"
5,3 He calls "At least one of the dice is a 5"
5,4 He calls "At least one of the dice is a 5"
5,5 He calls "At least one of the dice is a 5"
5,6 He calls "At least one of the dice is a 5"
6,1 He calls "At least one of the dice is a 6"
6,2 He calls "At least one of the dice is a 6"
6,3 He calls "At least one of the dice is a 6"
6,4 He calls "At least one of the dice is a 6"
6,5 He calls "At least one of the dice is a 6"
6,6 He calls "At least one of the dice is a 5"
You have observed only once. He had 6 opportunities to call a 3 and on one of those the other die was a 4
Probability 1 in 6
1/11:
Two dice are rolled in a cup. You ask someone to look at the dice and say whether at least one die is a 2. The person answers “yes.” Odds of 2/2 is 1/11.
1/6
Two dice are rolled in a cup. You ask someone to look at the dice and say whether the die on the left is a 2. The person answers “yes.” Odds of 2/2 is 1/6. (Assume when the cup is lifted one of the die is more to the left and one more to the right.)
See?
Why the difference?
In the second problem the answer would be “no” if the die on the right was a 2 but the die on the left was not a 2. So we are excluding 5 possible rolls from the denominator. (1/2; 3/2; 4/2; 5/2; and 6/2). However, in the first problem, the answer to the question would be “yes” for those five rolls.
Voila.
The possible outcomes of the flips were:
(1st: Heads; 2nd: Tails)
(1st: Heads; 2nd: Heads)
(1st: Tails; 2nd: Heads)
(1st: Tails; 2nd: Tails)
Then I tell you that at least one of the two flips was a Heads.
What are the odds that the first flip was a tails?
Because all you have learned is that the two flips weren't (1st: Tails; 2nd: Tails).
But the two flips could still be:
(1st: Heads; 2nd: Tails)
(1st: Heads; 2nd: Heads)
(1st: Tails; 2nd: Heads)
So the first flip has a 1/3 chance of being tails despite the fact that Tails is one of 2 sides on the first coin.
What are the odds that the 2nd flip was a Tails?
What are the odds they both flips were heads?
Telling you that "at least one of the two flips was a Heads." is not the same information as telling you that the first flip was a Heads. The difference in the information is what causes the departure from the answers being 50/50.
How many of you have been at a table where the shooter throws the dice and while one of them has stopped the other dice is on the table spinning?
Has it happened to you?
I think it's happened to everyone.
Has the die that's stopped showed a 5? Or a 6? Or a 1?
Have you said to yourself... please spinning die, don't stop on a 2... or a 1... or a 6... or whatever number would add up to 7.
When that die was spinning did you look at 1 die with 6 faces?
That's what the original question asked you to do.
And if it didnt, ask the ORIGINAL POSTER why he worded it that way?
And now for the $60,000 question:
Where is the original poster and WHAT ANSWER WAS HE LOOKING FOR?
Never happened to me. Happened to me every time I played Craps.Quote: AlanMendelsonAttention craps players.
I think it's happened to everyone.
True: True
Nope. Not for one second.Quote: AlanMendelson
That's what the original question asked you to do.
Funny, I feel confident we've heard this nonsense before.
Quote:And if it didnt, ask the ORIGINAL POSTER why he worded it that way?
And now for the $60,000 question:
Where is the original poster and WHAT ANSWER WAS HE LOOKING FOR?
link to original post
Zuga owes him a pint. Anyone searching the internet for two dice will get directed to this site. You can't buy advertising that good.
Quote: unJon[I thought Coachbelly accepted the wager also. Maybe he can come to town that day or use LMFW as proxy.
June 8 no good for me. No proxies for me either, in person only.
Terms as offered below, so no agreement to play until losing...my strategy will be to quit when ahead.
Quote: WizardI'll bet whatever stakes you wish. I'll offer you 8 to 1
I hope you guys make this happen.Quote: coachbellyQuote: unJon[I thought Coachbelly accepted the wager also. Maybe he can come to town that day or use LMFW as proxy.
June 8 no good for me. No proxies for me either, in person only.
Terms as offered below, so no agreement to play until losing...my strategy will be to quit when ahead.Quote: WizardI'll bet whatever stakes you wish. I'll offer you 8 to 1
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Quitting while ahead with a progressive wagering system could be a high probability prospect. It would prove nothing, of course.
I see Wizard offering this challenge, but I think we should have double confirmation before anyone gets suspended for backing out, just in the interests of keeping it friendly.
Out of curiosity, Coachbelly. Are you a 1/6 adherent? If not, what would be your motivation for playing this slightly -ev proposition?
Quote: WizardYes, this whole problem is the same thing as the "two boys" problem, but with six genders. Anyone who gets one will get the other. Anyone who doesn't get one won't get the other.
In other news, I'm happy to report that LoquaciousMoFW has accepted Alan's side of the challenge. Here are some tentative details:
- Challenge to take place June 8 in Las Vegas
- LoquaciousMoFW will wager $5 per shake
- I have kindly upped the win for two 2's to 8 to 1.
- LoquaciousMoFW has agreed to keep playing until losing $200. At that point, he/she has the option to back out.
- If for some reason LoquaciousMoFW can't be present, he/she has agreed to front money and have a proxy play for him/her
Gents,
I look forward to seeing this proceed. I suggest that only after double confirmation, would we even consider any penalties for backing out.
What do you have in mind for ballpark min/max bets?
And will you both have the right to call a halt?
If anyone would like to piggyback on LoquaciousMoFW's side, by all means, please express an interest. I'm happy to bet more on my side.
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Quote: OnceDear
To Gordon and Ed,
Nice wording Ed, but both you and Gordon have overestimated some aspect of Alan by several orders of magnitude. You underestimated his tenacity to cling to his point of view.
Alan is not convinced. Never will be convinced. Cannot be convinced. He is hard wired to be wrong on this matter.
Not just a tiny bit inflexible. He has invested too much of himself into his alternative question and answer pair.
Now..... I must return to the real world, briefly.
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I am issuing a warning to OnceDear and everyone else on this thread. Avoid making insulting personal comments about people whose posts you disagree with. Earlier, Wizard was suspended for this - so no one should feel exempt.