Two dice puzzle -- Part Trois
Ed, you do make one mistake. Calling rules again.Quote: EdCollins
At least one of the dice is a 2. That's the information we are given. You aren't told which die, the black die or the white die. You aren't told which die, the left die or the right die. You aren't told which die, the large die or the small die. All you are told was at least one of dice was a 2.
What is the probability that both dice are a 2? As has been demonstrated, the answer is clearly 1/11.
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Let's say they were a black die and a white die.
Let's also say that whenever he announces "at least one of the dice were a two, he reaches under and pulls that die out and presents it. You get to know its colour.
He might say "at least one of the dice is a two" and show it to you and say "and here it is look. On this occasion it was the white die". You would know which one it was.
It's back to calling rules. Would he have potentially called "At least one of the dice was a two" and pulled out and showed the black die?" Or would he have kept schtum?
We could further change what is called and on the one occasion that we observe, he might have announced
"There is a black die and a white die and the white die is a two". It's still 1/11 if he always calls "at least one of the dice is a two" but then appends "and here it is look. On this occasion it was the xxxxx die"
Ah. Simply put, because your approach fails to take into account all of the possible scenarios of what could have happened... of exactly how those two dice could have been rolled.Quote: AlanMendelsonWhy is my approach which is so simple so wrong?
Only after you do that, and then see how many of those scenarios answer the question, "What is the probability that BOTH dice are showing a 2?" can you arrive at the answer.
See unJon's dice diagrams again, for a visual of the 36 different ways two dice can be rolled. (2-1 is not the same roll as 1-2)
Is it safe to assume you haven't yet bothered to take the time to simulate the problem yourself, on your desk or kitchen table, with a couple of dice, and pencil and paper to keep track of the results?
Once you do, you will discover only 1/11 of the time will both dice show a 2, after first learning that at least one of the dies is a 2.
Quote: AlanMendelsonGosh guys... every time I look at a die... and it doesnt matter which face I'm looking at... my brain tells me that any other die has six faces and one of those faces will match it.
Then I think back to the original question. And the original question seems to ask me the same thing: when you have a face on a die what are the odds another die will match it?
Of course I didnt go to a math class that was trying to teach you some lesson in probability. All I did was use my memory to remember that a die has six faces with a different number, one thru six, on each face.
I'm also not comparing this question to how other questions are asked or answered. I'm not comparing this to census takers or TV hosts checking on kids or goats.
I'm simply remembering that if one die with six faces is showing one number, any other die with six faces will have one face matching that number.
Why is my approach which is so simple so wrong?
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You know one of the die is a two.
You remove one die, leaving one on the table.
What do you know about the remaining die?
I'm afraid I just don't understand... or you are giving unnecessary information and complicating a simple question?
Quote: AlanMendelsonGuys... again.... when I look at any one die there are six faces, and one of them will match.
I'm afraid I just don't understand... or you are giving unnecessary information and complicating a simple question?
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Alan can I ask you why you are resistant to trying the experiment of rolling two dice and seeing how often you get one 2 and how often you get two 2s?
Quote: unJonQuote: AlanMendelsonGuys... again.... when I look at any one die there are six faces, and one of them will match.
I'm afraid I just don't understand... or you are giving unnecessary information and complicating a simple question?
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Alan can I ask you why you are resistant to trying the experiment of rolling two dice and seeing how often you get one 2 and how often you get two 2s?
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I've answered this. Rolling two dice and getting 2-2 is a 1/36 event. It has nothing to do with the original question.
The Wizard's bet has nothing to do with the original question. And the Wizard will win 10 times for every one he loses.
So I will now repeat my original position. This was a trick question. It was worded in such a way as to make the mathematicians jump through unnecessary hoops. It was a set up.
Quote: AlanMendelsonQuote: unJonQuote: AlanMendelsonGuys... again.... when I look at any one die there are six faces, and one of them will match.
I'm afraid I just don't understand... or you are giving unnecessary information and complicating a simple question?
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Alan can I ask you why you are resistant to trying the experiment of rolling two dice and seeing how often you get one 2 and how often you get two 2s?
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I've answered this. Rolling two dice and getting 2-2 is a 1/36 event. It has nothing to do with the original question.
The Wizard's bet has nothing to do with the original question. And the Wizard will win 10 times for every one he loses.
So I will now repeat my original position. This was a trick question. It was worded in such a way as to make the mathematicians jump through unnecessary hoops. It was a set up.
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I don’t understand. Will you do me in favor? Go back up this thread to my posts with the three pictures. And let me know if you agree it’s 1/11 in the third roll.
That doesnt happen in the real world.
In the real world you roll two dice and then they stop rolling.
Look at one of the two dice. What number did it land on? Is it showing a 6?
If it is showing a six any other six sided has a 1/6 chance of showing a 6.
Quote: AlanMendelsonGuys... again.... when I look at any one die there are six faces, and one of them will match.
I'm afraid I just don't understand... or you are giving unnecessary information and complicating a simple question?
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One of them will match What?
There are 36 models with cases. Each case has a possible roll of the die. One has 1-1, another 1,2 so all 36 combinations are covered.
The jackpot is the case with two sixes.
Howie Mandel peeks and sees the winning number. He gives you one and only one clue.
All the models whose case doesn't have a six in it leaves.
25 of the 36 models leave the stage. 11 remain.
What are the chances you choose the right one?
Quote: billryanHow about this.
There are 36 models with cases. Each case has a possible roll of the die. One has 1-1, another 1,2 so all 36 combinations are covered.
The jackpot is the case with two sixes.
Howie Mandel peeks and sees the winning number. He gives you one and only one clue.
All the models whose case doesn't have a six in it leaves.
25 of the 36 models leave the stage. 11 remain.
What are the chances you choose the right one?
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This is becoming the theater of the absurd.
We went from having one die showing a two to counting children and their sex to models with cases.
How about going back to one die showing a two.
If another six sided die has its faces numbered 1, 2, 3, 4, 5, 6 what are the odds of having a 2 showing?
When you say "if the dice on the left is a 2 the probability that the other dice is a 2 must be 1 in 6" you must remember that the dice on the left has about a 54% chance of being a 2.
When you say "if the dice on the right is a 2 the probability that the other dice is a 2 must be 1 in 6" you must remember that the dice on the right also has about a 54% chance of being a 2.
And, so, because the probability of those two scenarios
- "the dice on the right is a 2"
- "the dice on the left is a 2"
add up to > 108%, you can't just claim that 1 in 6 for the other dice is the correct answer. It is actually approximately (1/6)/1.08
Quote: AlanMendelsonQuote: billryanHow about this.
There are 36 models with cases. Each case has a possible roll of the die. One has 1-1, another 1,2 so all 36 combinations are covered.
The jackpot is the case with two sixes.
Howie Mandel peeks and sees the winning number. He gives you one and only one clue.
All the models whose case doesn't have a six in it leaves.
25 of the 36 models leave the stage. 11 remain.
What are the chances you choose the right one?
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This is becoming the theater of the absurd.
We went from having one die showing a two to counting children and their sex to models with cases.
How about going back to one die showing a two.
If another six sided die has its faces numbered 1, 2, 3, 4, 5, 6 what are the odds of having a 2 showing?
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1/6. If you saw one die roll a two, then the chance the next one rolls a 2 is 1/6.
But if you don’t see one die roll a 2, but only know that at least one of two die (no idea which one or both) has a 2, then the chance the other die has a 2 is 1/11.
Or try this:
Chance of rolling two 2s. 1/36
Chance of rolling at least one 2. 11/36
Chance of “your partner” being able to remove a die with a 2 from the cup of at least one 2 is rolled. 100%
Chance of the die in the cup being a 2: 1/36 divided by 11/36 * 100% = 1/11
Quote: gordonm888Stop looking at one die and look at two dice. This is a problem with two dice,
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One more time... why is this a problem with two dice?
You also know the answer is 1/6 knowing one die. Why must you complicate the question? Because you're trying to justify your 1/11 answer perhaps?
Setup. Two players and a dealer. The dealer will roll two dice in a cup. The first player looks at the dice and if the can pull out a die that has a 2 on top he wins $1. Then the first player looks at the remaining die and if it is a 2, the second player will win $1.
What happens over time.
25/36 of the time no one wins a dollar.
11/36 of the time player 1 wins a dollar.
1/36 of the time player 2 wins a dollar. (Player 2 only gets a dollar if it was a hard four, as otherwise, player 1 takes the winning 2, if any.)
Over time player 1 wins $11 for every $1 player 2 wins.
Why?
Because only 1/11 of the time when there is one 2 will there be a second 2.
I'd also like to thank Gordon, Unjon, BillRyan and Ed and others for their entertaining attempts to help Alan.
The theatre in this supposedly serious thread is comedy gold. Comedy, Tragedy, Drama. It's the War and Peace length version of the Monty Python Argument sketch.
Were you gentlemen around these parts in 2013 through 2015?
Did you read the previous two threads here. Did you read the 18 page thread over on what was Alan's very own forum? You should..... It might waste a few hours of your time, but save you 9 years of therapy. $:o)
I challenge you to come up with a proof that has not already been rejected by Alan. Bonus point if it works to have him see the light.
We've had big dice, red dice, nearest dice, first dice, other dice. We've had spreadsheets, we've had charts, we've even had videos and venn diagrams. Wizard did a video, Alan did one, two. We've had two children analogies, We've had two coin flip analogies. And now we have ladies with boxes.
Somebody tried maths. Somebody tried semantic analysis. Somebody tried..... Well everything.
I understand that Alan even quoted his PHd buddies over on Facebook.
One man stood tall and firm through it all. One man refused to concede. One man proved his mettle. It must prove something after all these 9 years.
Stay Safe,
Stay Sane.
Quote: AlanMendelsonGuys... again.... when I look at any one die there are six faces, and one of them will match.
I'm afraid I just don't understand... or you are giving unnecessary information and complicating a simple question?
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But you DON'T deal with one die in the original problem, you are dealing with two dice.
When you throw two dice EITHER of them can show a 2. Sometimes the white die, sometimes the black die. In total 11 possible outcomes will have your partner answering "yes" to the question "does at least one die show a 2?" and one time out of those 11 times "the other" die will also show a 2. There is no predetermined "the other" die, they take turns so to speak.
So when you try to only deal with "the other" die to try and match "the first" die after the roll in your head you are making a mistake.
BOTH dice work together to produce eleven outcomes with at least one 2 showing. And EITHER die can show a non-two in those eleven outcomes thus evading the 2-2 combo for longer than 1/6 of the time, namely only one in eleven times will they both show a 2.
unJon's diagrams visualize the above very neatly.
There is no trickery or miswording, the puzzle is not a tricky question. You just need to realize there is a combined probability between BOTH dice for producing the "at least one 2" combos AND the exactly 2-2 combo.
I remember you from the 2013 threads. Welcome back.
Thank you for your contribution to this saga. Accept the challenge of my previous post.
$;o)
Funny that. I've decided that you are wrong exactly three times in that post. Three assertions, all wrong.Quote: unJonI’ve decided Alan understands now but is messing with people. I’ll be moving on.
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Quote: OnceDearFunny that. I've decided that you are wrong exactly three times in that post. Three assertions, all wrong.Quote: unJonI’ve decided Alan understands now but is messing with people. I’ll be moving on.
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I've asked repeatedly for someone to show me in the question why two dice must be considered. No one has said why -- but only that two dice must be considered.
I know that two dice give different answers. Those answers include 1/11 and 1/36 and as Billy pointed out even 1/10.
But what I find intriguing is that you're all married to a two dice answer and I suspect this is because along the way you were told dice questions must be answered with two dice.
Answering a question with one die is so simple it must be wrong!
This goes back to the original question. Look at it. Read the "set up" which talks about a previous problem. The question guides you to avoid the simple resolution.
And again I question why the original poster never returned? Don't you see it was a set up?
Yes. It was intended to be confusing and it was.
I can tell you when 1/11 is correct and when 1/6 is correct and so can you. It was not 1/11 this time. It was an English problem not a math problem.
The original poster had a laugh.
You ask if there is at least one 2. Your “partner” answers by pulling out a 2 and showing it to you. What is chance other die is a 2?
Well if the other die was a 2 your partner would have pulled it out 50% of the time (choosing randomly between the two 2s). If the other die was a 1, 3, 4, 5 or 6, your partner would have left it in the cup 100% of the time, because he was only pulling out a 2 if he could.
So 5 non twos plus 50% of a 2 are what could be left in the cup. That’s 5.5. That’s the denominator.
The numerator is 50% of a 2. That’s 1 * 50%. That’s 0.5.
0.5 / 5.5 = 1 / 11
Quote: AlanMendelsonI think all of this proves my initial point: the question was worded so poorly that there is division as to what it asks.
I've asked repeatedly for someone to show me in the question why two dice must be considered. No one has said why -- but only that two dice must be considered.
I know that two dice give different answers. Those answers include 1/11 and 1/36 and as Billy pointed out even 1/10.
But what I find intriguing is that you're all married to a two dice answer and I suspect this is because along the way you were told dice questions must be answered with two dice.
Answering a question with one die is so simple it must be wrong!
This goes back to the original question. Look at it. Read the "set up" which talks about a previous problem. The question guides you to avoid the simple resolution.
And again I question why the original poster never returned? Don't you see it was a set up?
Yes. It was intended to be confusing and it was.
I can tell you when 1/11 is correct and when 1/6 is correct and so can you. It was not 1/11 this time. It was an English problem not a math problem.
The original poster had a laugh.
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I’ve never been talking about the original wording. The original wording is ambiguous. Could be any answer including 1/6 or 1/11 or even 1/10 or 1 or 0.
I’m talking about the revised wording that I and the Wizard gave in this thread. You’ve continued to disagree that the revised wording in 1/11.
I believe you are also having a laugh.
Just to be clear, is this the original question, and how it was worded?Quote: AlanMendelson... This goes back to the original question. Look at it. Read the "set up" which talks about a previous problem. The question guides you to avoid the simple resolution. And again I question why the original poster never returned? Don't you see it was a set up?
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?
Quote: EdCollinsJust to be clear, is this the original question, and how it was worded?Quote: AlanMendelson... This goes back to the original question. Look at it. Read the "set up" which talks about a previous problem. The question guides you to avoid the simple resolution. And again I question why the original poster never returned? Don't you see it was a set up?
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?
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Ed,
FWIW I think that phrasing is ambiguous. I can be the “partner” in that game and guarantee you will never find a 2 still under the cup after I make that truthful statement. I will just not make the statement any time a hard four is rolled.
Quote: EdCollinsJust to be clear, is this the original question, and how it was worded?Quote: AlanMendelson... This goes back to the original question. Look at it. Read the "set up" which talks about a previous problem. The question guides you to avoid the simple resolution. And again I question why the original poster never returned? Don't you see it was a set up?
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?
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Actually it's not. Missing is the set up which preceded this.
The set up, though subtle, steers the reader from the simple answer sought in the question.
Quote: AlanMendelsonQuote: EdCollinsJust to be clear, is this the original question, and how it was worded?Quote: AlanMendelson... This goes back to the original question. Look at it. Read the "set up" which talks about a previous problem. The question guides you to avoid the simple resolution. And again I question why the original poster never returned? Don't you see it was a set up?
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?
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Actually it's not. Missing is the set up which preceded this.
The set up, though subtle, steers the reader from the simple answer sought in the question.
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In any event, your responses in this thread are not in answer to that, are they? You can't be that inflexible, can you?
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats
You are allowed to pick two doors. The host, who knows what's behind the doors tells you that atleast one of your dors have a goat behind it.
What are the odds you picked two goats?
The answer of 1/11 is a correct answer to its proper question.
I've even offered am example when 1/11 would be correct.
But the way this particular question is worded it's 1/6.
But you want it to be 1/11 and I get it. Just tell me WHY you need to answer this particular question using two dice.
It reminds me of a phrase: when you go to a surgeon for an opinion he tells you to have surgery.
So... when you go to a mathematician for an answer he gives you a long mathematical answer.
I've got more...
When you go to a lawyer he'll give you an answer by the hour.
And when you go to a TV reporter his answer will run 90 seconds. (The length of a TV news story.)
Quote: TorghattenAlan:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats
You are allowed to pick two doors. The host, who knows what's behind the doors tells you that atleast one of your dors have a goat behind it.
What are the odds you picked two goats?
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We had this earlier in the discussion.
Got some new material?
Quote: AlanMendelsonQuote: TorghattenAlan:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats
You are allowed to pick two doors. The host, who knows what's behind the doors tells you that atleast one of your dors have a goat behind it.
What are the odds you picked two goats?
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We had this earlier in the discussion.
Got some new material?
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Where? And what did you answer?
If you find something different in the wording of the question please tell me what it is.
And I'll tell you what you wont find. You won't find a query about how many 2 dice combinations include at least one 2.
Quote: AlanMendelsonJust to sum up my position... (as if I havent tried already)... as soon as you're told at least one die is a 2 TRUTHFULLY its the same as knowing one die has stopped on a 2 and the second die is spinning.
If you find something different in the wording of the question please tell me what it is.
And I'll tell you what you wont find. You won't find a query about how many 2 dice combinations include at least one 2.
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And yet you won’t grab two dice and roll them, and let your ex spouse tell you whether there is at least one die with a 2 TRUTHFULLY to see whether you are correct.
Quote: unJonI’ve decided Alan understands now but is messing with people. I’ll be moving on.
How was the move? Too difficult for you?
Quote: unJonAnd yet
Quote: coachbellyQuote: unJonI’ve decided Alan understands now but is messing with people. I’ll be moving on.
How was the move? Too difficult for you?Quote: unJonAnd yet
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Ha. The fool returning to the fire.
Quote: unJonThe fool returning to the fire.
Why did you bother to announce that you'd be moving on?
Trying to talk yourself into laying the wood, when you've got no wood?
Quote: coachbellyQuote: unJon[The fool returning to the fire.
Why did you bother to announce that you'd be moving on?
Trying to talk yourself into laying the wood, when you've got no wood?
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Questions by you is one game I do not do any longer.
Quote: unJonQuestions by you is one game I do not do any longer.
LOL...OK playa...take the 5th.
The dice on the left now has a 6/11 (54.6%) chance that it is a 2. When the dice on the left is indeed a 2, then the other die will be a 2 exactly 1/6 of the time, as you keep insisting.
The dice on the right also has a 6/11 (54.6%) chance that it is a 2. When the dice on the right is indeed a 2, then the other die will be a 2 exactly 1/6 of the time, as you keep insisting.
You are correct that the so-called 'other die' has a 1/6 chance of being a 2.
But having been told that at least one of the dice is a 2, there is no "first die" that has a 100% chance of being a 2. You keep taking that as your starting point but it is wrong. No matter which of the two dice you define as your first die, that first die will have a 6/11 chance of being a 2 - and the other dice will have a 1/6 chance of being a 2.
And 6/11 for the first die times 1/6 for the second die = 1/6 * 6/11 = 1/11.
So, we are not contradicting that that the other die has a 1/6 chance. We are telling you that you cannot identify a first die that has a 100% chance of being a 2.
That's a different scenario. If you're told "at least one die is a 2" while die B is still spinning, then that's the same as saying "die A is a 2".Quote: AlanMendelsonJust to sum up my position... (as if I havent tried already)... as soon as you're told at least one die is a 2 TRUTHFULLY its the same as knowing one die has stopped on a 2 and the second die is spinning.
But you are told "at least one die is a 2" after both have stopped spinning, so that's the same as saying: "either die A is a 2, die B is a 2, or both dice are 2's."
Quote: AlanMendelsonJust to sum up my position... (as if I havent tried already)... as soon as you're told at least one die is a 2 TRUTHFULLY its the same as knowing one die has stopped on a 2 and the second die is spinning.
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That's a mistake. It is absolutely not the same as knowing one die has stopped on a 2 and the second die is spinning. Scratch that scenario from your thought process because it is wrong assumption leading you to wrong views. Because if the die which has landed already shows a non-two your partner wouldn't have said yes, yet when the spinning die lands and eventually show a 2 what then? Your partner would have said "yes" after the full roll is over, but you are denying that option beforehand. Your partner looks AT BOTH landed dice to truthfully answer "yes" or "no".
So when you equate the information that at least one 2 is showing to "one die has stopped on a 2 and the second die is spinning" you're making a basic mistake.
Quote: gordonm888I will try one more time. Given you roll the dice and are told that at least one of the dice is a 2:
The dice on the left now has a 6/11 (54.6%) chance that it is a 2. When the dice on the left is indeed a 2, then the other die will be a 2 exactly 1/6 of the time, as you keep insisting.
The dice on the right also has a 6/11 (54.6%) chance that it is a 2. When the dice on the right is indeed a 2, then the other die will be a 2 exactly 1/6 of the time, as you keep insisting.
You are correct that the so-called 'other die' has a 1/6 chance of being a 2.
But having been told that at least one of the dice is a 2, there is no "first die" that has a 100% chance of being a 2. You keep taking that as your starting point but it is wrong. No matter which of the two dice you define as your first die, that first die will have a 6/11 chance of being a 2 - and the other dice will have a 1/6 chance of being a 2.
And 6/11 for the first die times 1/6 for the second die = 1/6 * 6/11 = 1/11.
So, we are not contradicting that that the other die has a 1/6 chance. We are telling you that you cannot identify a first die that has a 100% chance of being a 2.
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This.
Everyone says the dice cup problem/question is not the same as a die spinning on a craps table but no one tells me why?
Again I'm going to say the question was poorly worded and was a setup to create controversy.
Unless someone tells me WHY two dice must be considered and the problem is not like a spinning die, I'm sticking with my answer.
Quote: unJonQuote: AlanMendelsonJust to sum up my position... (as if I havent tried already)... as soon as you're told at least one die is a 2 TRUTHFULLY its the same as knowing one die has stopped on a 2 and the second die is spinning.
If you find something different in the wording of the question please tell me what it is.
And I'll tell you what you wont find. You won't find a query about how many 2 dice combinations include at least one 2.
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And yet you won’t grab two dice and roll them, and let your ex spouse tell you whether there is at least one die with a 2 TRUTHFULLY to see whether you are correct.
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Unjon I'd love to do that but there are days I have to spoon feed her applesauce and carry her to the toilet.
Stop.
Quote: AlanMendelson
Unjon I'd love to do that but there are days I have to spoon feed her applesauce and carry her to the toilet.
Stop.
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I think it's unfortunate that Alan raised the topic of his Ex-wife's health problems.
In deference to her, I suggest that we ALL leave her out of this conversation.
Let's show a bit of decorum.
Just a suggestion.
Quote: AlanMendelsonAgain... everyone tells me I must consider two dice but no one tells me why?
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Please explain the functional difference between:
- a roll of two dice, ignored if no 2 is showing
- a roll of two dice, under a cup, and someone tells you at least one 2 is showing
Quote: DieterQuote: AlanMendelsonAgain... everyone tells me I must consider two dice but no one tells me why?
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Please explain the functional difference between:
- a roll of two dice, ignored if no 2 is showing
- a roll of two dice, under a cup, and someone tells you at least one 2 is showing
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You got me on this one. But I'll play along.
You roll two dice and and no 2 shows? Well, you cant have a hard four... that's for certain.
And when two dice are rolled under a cup and someone tells you at least one 2 is showing? You have a 1/6 chance of getting a hard four.
Alan -
What’s the odds that the person peeking will say that neither die is a 2?
