Two dice puzzle -- Part Trois
Quote: gordonm888
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
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(trimmed!)
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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(also trimmed!)
Knowns:
- 2 dice, randomized, hidden (no funny business implied)
- nonzero quantity of 2's (1 or 2 inferred)
Unknowns:
- did the observer see one dice or two?
Given the people I know:
- some like the number 2, because even prime numbers are a peculiarity. They will report a 2, if possible.
- some dislike the number 2, because even prime numbers are a peculiarity. They will report anything else, if possible.
- some do not care about 2's.
OnceDear and others can continue to question "the partner's" gender identity, reasons for looking at the dice, whether 'the partner' has a health condition that affects his/her vision or cognitive ability, what the partner's motivation was for reporting on a 2 versus any other numeral, what prior agreements your partner had with you on signals and terminology and other issues. All of which I think is rubbish. My head isn't filled with hay and rags and I don't care to make a habit of conducting conversations that are uninteresting and of no importance to anyone anywhere ever.
I originally posted on this thread because I wanted to explain the mathematics. I am satisfied that I did that. I disagree with OnceDear on this matter. I don't think it's worth my time to further acknowledge his comments. I speculate he will respond with his usual hyperactivity by swamping us with another two dozen or so posts that rapturously crusade on this issue. But I won't read them. Because I have things to do and a life to live. And a partner who is asking me to come to bed.
Quote: gordonm888"Your partner peeks under the cup . . ."
OnceDear and others can continue to question "the partner's" gender identity, reasons for looking at the dice, whether 'the partner' has a health condition that affects his/her vision or cognitive ability, what the partner's motivation was for reporting on a 2 versus any other numeral, what prior agreements your partner had with you on signals and terminology and other issues. All of which I think is rubbish. My head isn't filled with hay and rags and I don't care to make a habit of conducting conversations that are uninteresting and of no importance to anyone anywhere ever.
I originally posted on this thread because I wanted to explain the mathematics. I am satisfied that I did that. I disagree with OnceDear on this matter. I don't think it's worth my time to further acknowledge his comments. I speculate he will respond with his usual hyperactivity by swamping us with another two dozen or so posts that rapturously crusade on this issue. But I won't read them. Because I have things to do and a life to live. And a partner who is asking me to come to bed.
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My challenge to you is open anytime. 20:1 is sweet odds since you are so sure 11:1 is the right answer under any circumstances.
The person who looks at the dice is forced every time to say “there is at least one [].” Where [] is any number 1-6 that the peeker sees on at least one dice. If the peeker sees two different numbers, he (or she) will randomly pick one of them.
Following that rule, the peeker calls out “I see at least one 2.”
What is the chance of two 2s?
Answer 1/6.
Do folk see how the rules the peeker follows can make the odds 1/6 or 1/11? Or frankly anything between 0% to 100%.
Be satisfied. Where ignorance is bliss, 'tis folly to be wise'Quote: gordonm888"Your partner peeks under the cup . . ."
I originally posted on this thread because I wanted to explain the mathematics. I am satisfied that I did that. I disagree with OnceDear on this matter.
Back at you.Quote:I don't think it's worth my time to further acknowledge his comments.
Wrong again.Quote:I speculate he will respond with his usual hyperactivity by swamping us with another two dozen or so posts that rapturously crusade on this issue.
Have a nice day.
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Quote: Dieter
(trimmed!)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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(also trimmed!)
Knowns:
- 2 dice, randomized, hidden (no funny business implied)
- nonzero quantity of 2's (1 or 2 inferred)
Close Dieter, but a minor area where you and I disagree.
"- nonzero quantity of 2's (1 or 2 inferred) ON THIS ONE OBSERVATION
Quote:
Unknowns:
- did the observer see one dice or two?
No. We seem to graciously accept from the physics of a cup that he/she saw both dice.
More unknowns
Whether the observer is ONLY reporting for 2's and says ?what? if 2 is not rolled? Would he/she be silent or report something? That affects the denominator in our eventual formula.
Whether the observer is ALWAYS reporting 2's or might he/she report a different number if possible.
Those two unknowns, on their own, give us many alternative answers.
Lot's of trivial stuff we don't know, such as preferences as to number called. I'm happy to ignore such trivia as out of scope.
I think that it is generally accepted that the observer will make a report of some sort every time a two is rolled.
What we cannot agree on are one simple but critical unknown.... WHAT, if anything does he/she report if no two is rolled. Wizard seems to guess that he she will report something like "Neither of the dice are twos". Others guess that he/she would report SOMETHING truthful.
There is a precedent, I believe in craps, Monopoly, snakes and ladders, ludo, etc, All rolls of the dice, once observed are declared and acted upon.
Quote:
Given the people I know:
- some like the number 2, because even prime numbers are a peculiarity. They will report a 2, if possible.
- some dislike the number 2, because even prime numbers are a peculiarity. They will report anything else, if possible.
- some do not care about 2's.The second dice has a 1/6 chance of being a 2.The second dice has a 1/11 chance of being a 2.Since she had no choice but to report a 2, both dice are 2's. 1.If they are not special, then by implication he/she is not only reporting 2's1/11.There is a 6/36 (1/6) chance of rolling doubles; 1/36 that they are both 2's. This case does not fall within the constraints of the problem.
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I think that "peeks under the cup" does not necessarily imply "observes both dice" or "randomly selects an observed value to report".
There is a difference between "The house is green" and "The house appears green on this side."
It really is a simple question which asks for a simple answer. The simple answer is best.
_____________
Quote: Dween
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?
_______________
After hearing the question, don't you put yourself into the position of the partner who sees under the cup?
The question is begging you to take the partner's point of view.
And the partner truthfully sees at least one of the dice is 2.
Everyone agrees that for the partner the answer is 1/6.
The only question is are you taking on the role of the partner?
Yes, you are.
If you weren't asked to take on the role of the partner the question would have been phrased differently. For example, it could have been phrased:
Two dice are shaken. How many combinations of the two dice contain at least one 2? How many combinations contain 2-2?
In this wording you are not led to take on the role of the partner who actually looks and sees the result of at least one die.
Quote: AlanMendelsonGuys thanks for the interesting discussion but I think you're all going way too far in picking apart the question.
It really is a simple question which asks for a simple answer. The simple answer is best.
_____________
Quote: Dween
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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Quote trimmed.
The only answer to the question as posed is: do not have enough information to determine the chance of 2 twos.
Quote: DieterI am seeing a range of answers that might not be wrong.
I think that "peeks under the cup" does not necessarily imply "observes both dice" or "randomly selects an observed value to report".
There is a difference between "The house is green" and "The house appears green on this side."
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The problem statement says that your partner is reporting truthfully, although it does not state that the partner is reporting accurately. But this line of inquiry (the unreliable reporter) does not allow one to conjure up an answer to the problem without making assumptions that are wholly unsupported by the statement of the problem.
All of this discussion is about whether the so-called truthful statement "at least one of the two dice is a 2" can be interpreted to mean something else because of the manner in which the information is obtained. You all are asking "If a partner does has seen both dice, why doesn't the partner report on both dice?" "Why is the information limited in such a bizarre way?" "Why would my partner only report on 2s and not report on the outcome of both dice? Why did he select 2's?"
My response is that this was presented as a logic problem, challenging you to understand the implication of incomplete information about the outcome of rolling two dice. If you don't buy the premise of the problem statement then you are free to ignore the problem. But all you are saying is that "I believe my partner wouldn't respond that way after looking at the dice" and complaining that Wizard didn't include statements about the peeker being a perfect logician or a Monty Hall-style game host. Meh. Get over it. We all have burned too many electrons on this.
It’s not about whether the “partner” is a partner or a logician or a Monty Hall type.
It’s literally a Bayes information question.
Also Wizard didn’t formulate the original question.
There’s a real simple restatement of the question that leads to 1/11. And I’m very sure that’s how the OP meant the question to be taken. And it’s a really interesting question.
********
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. You ask your partner if at least one of the dice show a 2. Your partner peeks under the cup at both dice, and tells you, truthfully, "Yes"
What is the probability that both dice are showing a 2?
*********
See. Not hard. And the answer now is clearly 1/11.
Quote: New wordingYou have two dice in a shaker. You ask a friend to shake the dice and peek at the outcome, which he does. You ask him, "Is at least one of the dice a two?" He says, "Yes." What is the probability both dice are a two?
I would like to say I was a bit harsh on Alan, and for that, I apologize.
Quote: WizardHi Everyone. It's nice to be out of WoV jail. I spent my time arguing with my judge and jailer about this problem. Not much else to do in there. Suffice it to say, I'm ready to quit tearing apart the original wording, which was not mine, and suggest a rewording. I submit this for the consideration of the forum:
Quote: New wordingYou have two dice in a shaker. You ask a friend to shake the dice and peek at the outcome, which he does. You ask him, "Is at least one of the dice a two?" He says, "Yes." What is the probability both dice are a two?
I would like to say I was a bit harsh on Alan, and for that, I apologize.
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Welcome back!
Yes I think that wording does it. Same idea as my wording in post above.
Quote: WizardHi Everyone. It's nice to be out of WoV jail. I spent my time arguing with my judge and jailer about this problem. Not much else to do in there. Suffice it to say, I'm ready to quit tearing apart the original wording, which was not mine, and suggest a rewording. I submit this for the consideration of the forum:
Quote: New wordingYou have two dice in a shaker. You ask a friend to shake the dice and peek at the outcome, which he does. You ask him, "Is at least one of the dice a two?" He says, "Yes." What is the probability both dice are a two?
I would like to say I was a bit harsh on Alan, and for that, I apologize.
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Wizard, Welcome back. It was fun debating you in PM. Thanks for the apology to Alan.
Your paraphrased question has the unequivocal answer of 1/11, It is geared, like the Monty Hall question, to draw out the incorrect, intuitive answer of 1/6. and I'm sure there is one here who would insist that your question should be answered 1/6. He was wrong 9 years ago and he always will be. If/when he or his fellow [Insert correct word] challenge this version, I expect fur to fly.
Yours lacks the entertainment value of the original in-explicit version. Maybe we can get some fool to accept a wager paying 8 to 1 on your version. Maybe someone will take on Unjon's 20 to 1. It's about time someone monetized this.
$:o)
Trimmed.Quote: OnceDear
Yours lacks the entertainment value of the original in-explicit version. Maybe we can get some fool to accept a wager paying 8 to 1 on your version. Maybe someone will take on Unjon's 20 to 1. It's about time someone monetized this.
$:o)
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To monetize this, I propose a craps side bet called the Harder 4. The bet pays 9:1 if a hard four is rolled before a roll that contains only one 2.
Quote: OnceDear
If/when he or his fellow [Insert correct word] challenge this version, I expect fur to fly.
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Won't be this muttonhead!
My mind is now turned to fond memories of a flugelhorn player I used to know.
Welcome back, Wizard. Thanks for sparring, Gordon. -D
You bet on all numbers. Every 6th roll (on average) you win 9 units. All other rolls you lose 2 units.
So you lose 1 unit every 6th roll on.
That sounds for me like an 1/11th chance of winning.
I'm sorry to say this but my position hasnt changed even with the new quote.
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
Unless you're allowing the two dice to continue to shake, rattle and roll? In which case 2/2 is 1/36 once the dice settle.
Sorry but I cant budge on this.
I have suggested that a different question might be answered by 1/11.
But when you have two dice and the result of one die is known, there are only six options on the second dice.
Best,
Your thick and stupid friend, Alan
Quote: AlanMendelson
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.
If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.
The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
Quote: DieterQuote: AlanMendelson
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.
If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.
The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
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Really? There are only two dice. If we know any one die is showing a 2 then the other die has six options.
Why are you making it difficult?
Quote: DieterQuote: AlanMendelson
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.
If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.
The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
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Bolded added.
That’s not right.
If the person says at least one die is a two, and then removes one of the die and shows that two to you, it’s still 1/11.
ETA: it’s a little like the Monty Hall problem in that way.
Quote: AlanMendelson…Why are you making it difficult?
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That says it all! You’ve used this “argument” before. Tell me, do you stomp your feet, as well, when you right this? I have a hard time picturing anything else.
Quote: unJonQuote: DieterQuote: AlanMendelson
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.
If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.
The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
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Bolded added.
That’s not right.
If the person says at least one die is a two, and then removes one of the die and shows that two to you, it’s still 1/11.
ETA: it’s a little like the Monty Hall problem in that way.
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I dont need to know what's on 12 faces or 11 faces. I just need to know that one die is showing a 2. So the other die has six options.
And since there are only 2 dice I dont care which die is a 2, because the answer is still 1/6 on the remaining die.
What you should find is that:
69.5% have zero 2s
27.8% have one 2
2.8% have two 2s
And when you look at the ratio of rolls that have two 2 vs “at least one” 2 you’ll see that it’s 1/11.
Quote: unJonQuote: DieterQuote: AlanMendelson
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.
If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.
The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
link to original post
Bolded added.
That’s not right.
If the person says at least one die is a two, and then removes one of the die and shows that two to you, it’s still 1/11.
ETA: it’s a little like the Monty Hall problem in that way.
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OK, I can be wrong.
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/#post452706
Part Deux problems 1 and 3 each involve a pair of dice.
In Part Deux problem 2, there is a single dice involved.
Quote: DieterIf my understanding is wrong, I invite explanation.
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/#post452706
Part Deux problems 1 and 3 each involve a pair of dice.
In Part Deux problem 2, there is a single dice involved.
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I’ll give it the college try.
There’s two ways that problem 2 might go.
Way 1:
Roll a die. If it’s not a 2, roll the same die again. If you get a 2, roll a second die under a cup.
Chance of 2/2 is 1/6.
Way 2:
Roll two dice under a cup. If no 2s then reroll. If at least one 2 then remove a die with a two to show the world that 2, and keep the second die hidden.
Chance of 2/2 is 1/11.
Follow?
Way 2 is like the Monty Hall problem. The dealer will always pick a die with a 2 to show, no matter which die it is.
Way 1 is different. The first die roll is always shown.
Quote: camaplQuote: AlanMendelson…Why are you making it difficult?
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That says it all! You’ve used this “argument” before. Tell me, do you stomp your feet, as well, when you right this? I have a hard time picturing anything else.
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I stomp my feet when I' hear about 11 faces.
I dont need to know 11 faces. I need to know if I die has a 2.
Quote: unJonQuote: DieterIf my understanding is wrong, I invite explanation.
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/#post452706
Part Deux problems 1 and 3 each involve a pair of dice.
In Part Deux problem 2, there is a single dice involved.
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I’ll give it the college try.
There’s two ways that problem 2 might go.
Way 1:
Roll a die. If it’s not a 2, roll the same die again. If you get a 2, roll a second die under a cup.
Chance of 2/2 is 1/6.
Way 2:
Roll two dice under a cup. If no 2s then reroll. If at least one 2 then remove a die with a two to show the world that 2, and keep the second die hidden.
Chance of 2/2 is 1/11.
Follow?
Way 2 is like the Monty Hall problem. The dealer will always pick a die with a 2 to show, no matter which die it is.
Way 1 is different. The first die roll is always shown.
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You're giving unnecessary steps. You've already been told that one die is showing a 2.
Why dont you pick up 2 dice.
Put one die in your pocket.
With the die not in your pocket set it to 2. This is where the problem starts. You've been told one die is a 2.
Now ask yourself... if I want to have 2-2 what does the die in my pocket have to show? Let's count the possibilities...
1, 2, 3, 4, 5, 6.
Quote: AlanMendelsonHi Wizard. No apology is needed.
I'm sorry to say this but my position hasnt changed even with the new quote.
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
Unless you're allowing the two dice to continue to shake, rattle and roll? In which case 2/2 is 1/36 once the dice settle.
Sorry but I cant budge on this.
I have suggested that a different question might be answered by 1/11.
But when you have two dice and the result of one die is known, there are only six options on the second dice.
Best,
Your thick and stupid friend, Alan
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THAT, sir, is exactly how one answers the question wrongly or answers the wrong question altogether.
Alan, You are just so totally wrong..... But more than that, you are stubborn. If you cannot understand this now, after nine years of every conceivable proof, I feel confident that you never will. Your brain's programming on this is flawed but hard wired.
Your [insert appropriate, pejorative adjective] answer to this question, and especially this clarified version of it, defines you.
There are a few proverbs that apply to this situation, but to quote them would be insulting to you.
Quote: AlanMendelsonQuote: unJonQuote: DieterIf my understanding is wrong, I invite explanation.
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/#post452706
Part Deux problems 1 and 3 each involve a pair of dice.
In Part Deux problem 2, there is a single dice involved.
link to original post
I’ll give it the college try.
There’s two ways that problem 2 might go.
Way 1:
Roll a die. If it’s not a 2, roll the same die again. If you get a 2, roll a second die under a cup.
Chance of 2/2 is 1/6.
Way 2:
Roll two dice under a cup. If no 2s then reroll. If at least one 2 then remove a die with a two to show the world that 2, and keep the second die hidden.
Chance of 2/2 is 1/11.
Follow?
Way 2 is like the Monty Hall problem. The dealer will always pick a die with a 2 to show, no matter which die it is.
Way 1 is different. The first die roll is always shown.
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You're giving unnecessary steps. You've already been told that one die is showing a 2.
Why dont you pick up 2 dice.
Put one die in your pocket.
With the die not in your pocket set it to 2. This is where the problem starts. You've been told one die is a 2.
Now ask yourself... if I want to have 2-2 what does the die in my pocket have to show? Let's count the possibilities...
1, 2, 3, 4, 5, 6.
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Do the experiment I proposed in this thread. When you get 1/11 and don’t understand why, I’ll try to explain it to you.
Do some “investigative journalism” through empirical experimentation!
Quote: OnceDearQuote: AlanMendelsonHi Wizard. No apology is needed.
I'm sorry to say this but my position hasnt changed even with the new quote.
If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
Unless you're allowing the two dice to continue to shake, rattle and roll? In which case 2/2 is 1/36 once the dice settle.
Sorry but I cant budge on this.
I have suggested that a different question might be answered by 1/11.
But when you have two dice and the result of one die is known, there are only six options on the second dice.
Best,
Your thick and stupid friend, Alan
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THAT, sir, is exactly how one answers the question wrongly or answers the wrong question altogether.
Alan, You are just so totally wrong..... But more than that, you are stubborn. If you cannot understand this now, after nine years of every conceivable proof, I feel confident that you never will. Your brain's programming on this is flawed but hard wired.
Your [insert appropriate, pejorative adjective] answer to this question, and especially this clarified version of it, defines you.
There are a few proverbs that apply to this situation, but to quote them would be insulting to you.
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Tsk tsk. You want to insult me but you wont pick up two dice and look at them?
You're not considering a roll of two dice. The roll is over.
You know one die has settled on two.
To have 2-2 what are the odds that remain?
Careful... only one die now.
Alan, Sir.....Quote: AlanMendelsonTsk tsk. You want to insult me but you wont pick up two dice and look at them?
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What I scream at the screen is acceptable under the rules here. If I typed it, then as a moderator I am held to 'A higher standard' and I would be suspended.
You are wrong. You are allowed to be wrong.
I would be a fool to try any further to set you right.
I would be a [Insert really offensive adjective] fool to believe that you would ever be set right.
No offence to unjon or anyone else here, but be assured that you will not get Alan to understand this.
I KNOW that Alan will remain forever wrong in his interpretation and answer here. If he wants to continue to make a [redacted pejorative] of himself, I'll let others in this thread entertain him.
My work here is complete.
Quote: AlanMendelson
You're giving unnecessary steps. You've already been told that one die is showing a 2.
Why dont you pick up 2 dice.
Put one die in your pocket.
With the die not in your pocket set it to 2. This is where the problem starts. You've been told one die is a 2.
Now ask yourself... if I want to have 2-2 what does the die in my pocket have to show? Let's count the possibilities...
1, 2, 3, 4, 5, 6.
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You have separated the pair of dice into two individual dice; one known, one unknown.
The problem is about a pair of dice with partial information.
Quote: AlanMendelsonIf I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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It's time to put up or shup up. I'll bet whatever stakes you wish. If you think the probability is 1/6, then fair odds would pay 5 to 1. I'll offer you 8 to 1.
It's a simple question with a simple answer.
But to get this 1/11 answer you complicate it.
So again, returning to the simple approach:
It's been announced that two dice have been rolled and at least one of those two dice is showing a 2.
That die is 2.
Now, you're asked what are the odds that the second die is also showing a 2?
You pause for a moment and realize that the second die has faces marked 1, 2, 3, 4, 5, 6. So to have both dice showing 2-2 it would be a 1/6 chance.
No one needs to concern themselves will 11 faces on two dice.
It's a one die with six faces question.
Why complicate it?
Let me ask this: why count the five other faces of the die showing a 2? That die is no longer moving. It's showing a 2.
Quote: WizardQuote: AlanMendelsonIf I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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It's time to put up or shup up. I'll bet whatever stakes you wish. If you think the probability is 1/6, then fair odds would pay 5 to 1. I'll offer you 8 to 1.
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Put up or shut up about what?
The odds of rolling 2-2 is 1/36.
The odds of rolling a 2 on one die is 1/6.
The odds of throwing two dice on a table, with one die immediately stopping on 2 and the second die spinning for ten seconds, and then stopping on 2 is also 1/6.
Where is there a result with odds of 1/11 throwing two dice?
Show me those odds on any casino bet.
Thanks.
Quote: AlanMendelsonDieter, how much more information do you need? We're talking about two six sided dice... and one of them has settled on a number that has been announced.
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In the problem, you do not know which of the two dice shows a 2.
In your rearrangement of the problem, you do know whice dice shows a 2.
That matters.
You've either been offered an amazing opportunity to win Wizard's money or an amazing opportunity to grow Wizard's bankroll. I suggest at least 40 units, and a clear statement of the rules of contest.
Only bet money you can afford to lose. Best of luck.
Quote: DieterQuote: AlanMendelsonDieter, how much more information do you need? We're talking about two six sided dice... and one of them has settled on a number that has been announced.
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In the problem, you do not know which of the two dice shows a 2.
In your rearrangement of the problem, you do know whice dice shows a 2.
That matters.
You've either been offered an amazing opportunity to win Wizard's money or an amazing opportunity to grow Wizard's bankroll. I suggest at least 40 units, and a clear statement of the rules of contest.
Only bet money you can afford to lose. Best of luck.
link to original post
Why is everyone saying you need to know which of the two dice is showing a 2?
It makes no difference in a two dice problem.
Tell me why it makes a difference. Please.
And what is the bet?
I originally thought the answer was 1/6.
I had to be shown the math to see that it's 1/11.
But when I read (Alan's?) comment that it has to do with the wording, I rattled that around for a couple days.
As a result I came up with two scenarios.
Scenario 1: The dice are shaken and rolled but hidden. One person peeks and states that at least one die is a 2.
Scenario 2: The dice are shaken and rolled but hidden. One person asks the other to verify that at least one die is a 2.
In each scenario, what are the odds that both dice are 2?
In Scenario 1 the odds are 1/6.
In Scenario 2 the odds are 1/11.
Why are there different results?
In Scenario 1, there are no ground rules that the person had to look only for 2s. I.E. No rule saying that he must say there are no 2s as opposed to saying there is at least 1 of whatever value he happens to see. In this case, he sees a die which happens to be a 2 and reports that there's at least one 2. Therefore, there's a 1/6 chance the other die matches what he happens to see.
In Scenario 2, the odds are high (25/36) that the person will be forced to report that he sees no 2s. But that's the only option. 25 times there are no 2s and 11 times there is at least one 2. And we know that of those 11 times, there is only one combination of a pair of 2s. Therefore, if he's not saying there aren't any 2s, there's a 1/11 chance of it being a pair of 2s.
Quote: DJTeddyBearLet me see if I can cut thru the clutter.
I originally thought the answer was 1/6.
I had to be shown the math to see that it's 1/11.
But when I read (Alan's?) comment that it has to do with the wording, I rattled that around for a couple days.
As a result I came up with two scenarios.
Scenario 1: The dice are shaken and rolled but hidden. One person peeks and states that at least one die is a 2.
Scenario 2: The dice are shaken and rolled but hidden. One person asks the other to verify that at least one die is a 2.
In each scenario, what are the odds that both dice are 2?
In Scenario 1 the odds are 1/6.
In Scenario 2 the odds are 1/11.
Why are there different results?
In Scenario 1, there are no ground rules that the person had to look only for 2s. I.E. No rule saying that he must say there are no 2s as opposed to saying there is at least 1 of whatever value he happens to see. In this case, he sees a die which happens to be a 2 and reports that there's at least one 2. Therefore, there's a 1/6 chance the other die matches what he happens to see.
In Scenario 2, the odds are high (25/36) that the person will be forced to report that he sees no 2s. But that's the only option. 25 times there are no 2s and 11 times there is at least one 2. And we know that of those 11 times, there is only one combination of a pair of 2s. Therefore, if he's not saying there aren't any 2s, there's a 1/11 chance of it being a pair of 2s.
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Thanks.
This is scenario #1.
Quote: AlanMendelsonAnd what is the bet?
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I'm out of town, as usual, and not taking action, as usual.
I suggest that the Deux Parte Trois structure is a reasonable starting point for your rule negotiations.
Quote: Ayecarumba
Table 3: You approach what appears to be a standard Craps table, and see that it is not Craps, but a game where the only wagers available are, "Total 4 (at least one die a two)" and "Total not 4 (at least one die a two)". The felt also states, "All rolls without at least one die a two 'en prison'". The dealer explains that this means that any roll without at least one die a two will be "locked up" until a roll with at least one die a two. She asks you to "Please place your bets". You make a wager. The stick then dumps a bowl with five dice and pushes them to you so that you can pick two and shoot em.
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Quote: AlanMendelsonPut up or shut up about what?
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The question at hand. Here is your chance to prove me wrong and make some money at the time same time. Here are the rules I propose:
Two dice are shaken.
If there are zero twos, no money changes hands.
If there is one two, you pay me whatever your wager is.
If there are two twos, then I pay you 7x your wager.
I will not discuss math with you any further. It's time to settle this with actual dice.
Quote: WizardQuote: AlanMendelsonPut up or shut up about what?
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The question at hand. Here is your chance to prove me wrong and make some money at the time same time. Here are the rules I propose:
Two dice are shaken.
If there are zero twos, no money changes hands.
If there is one two, you pay me whatever your wager is.
If there are two twos, then I pay you 7x your wager.
I will not discuss math with you any further. It's time to settle this with actual dice.
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Rolling two 2s is a 1/36 event.
This bet has nothing to do with the problem being discussed.
Quote: AlanMendelsonQuote: DieterQuote: AlanMendelsonDieter, how much more information do you need? We're talking about two six sided dice... and one of them has settled on a number that has been announced.
link to original post
In the problem, you do not know which of the two dice shows a 2.
In your rearrangement of the problem, you do know whice dice shows a 2.
That matters.
You've either been offered an amazing opportunity to win Wizard's money or an amazing opportunity to grow Wizard's bankroll. I suggest at least 40 units, and a clear statement of the rules of contest.
Only bet money you can afford to lose. Best of luck.
link to original post
Why is everyone saying you need to know which of the two dice is showing a 2?
It makes no difference in a two dice problem.
Tell me why it makes a difference. Please.
And what is the bet?
link to original post
Here is the bet. I’ll take it if the Wiz doesn’t want it.
Two dice are rolled under a cup. Someone we both trust will report the answer to the questions: is there at least one die showing a two?
If that person says “no”, then there’s no bet. They can show us the dice to verify the “no.”
If that person says “yes”, then there is a bet. Alan will win $70 if the roll is a hard four. Alan will lose $10 if the roll is not a hard four, but just a two on one die but not the other.
That’s 7:1 odds that if one die is a two the other won’t be.
That’s the bet. That’s the bet you say has 1/6 of being correct, and I and others say has 1/11 of being right.
Quote: unJonQuote: AlanMendelsonQuote: DieterQuote: AlanMendelsonDieter, how much more information do you need? We're talking about two six sided dice... and one of them has settled on a number that has been announced.
link to original post
In the problem, you do not know which of the two dice shows a 2.
In your rearrangement of the problem, you do know whice dice shows a 2.
That matters.
You've either been offered an amazing opportunity to win Wizard's money or an amazing opportunity to grow Wizard's bankroll. I suggest at least 40 units, and a clear statement of the rules of contest.
Only bet money you can afford to lose. Best of luck.
link to original post
Why is everyone saying you need to know which of the two dice is showing a 2?
It makes no difference in a two dice problem.
Tell me why it makes a difference. Please.
And what is the bet?
link to original post
Here is the bet. I’ll take it if the Wiz doesn’t want it.
Two dice are rolled under a cup. Someone we both trust will report the answer to the questions: is there at least one die showing a two?
If that person says “no”, then there’s no bet. They can show us the dice to verify the “no.”
If that person says “yes”, then there is a bet. Alan will win $70 if the roll is a hard four. Alan will lose $10 if the roll is not a hard four, but just a two on one die but not the other.
That’s 7:1 odds that if one die is a two the other won’t be.
That’s the bet. That’s the bet you say has 1/6 of being correct, and I and others say has 1/11 of being right.
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No thanks.
I do better at Red Rock.
Quote: unJonI’ll take it if the Wiz doesn’t want it.
Is this bet open to any takers, ala MDawg's challenge, or is it by invitation only?
Quote: coachbellyQuote: unJonI’ll take it if the Wiz doesn’t want it.
Is this bet open to any takers, ala MDawg's challenge, or is it by invitation only?
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Open to any challenger. Money escrowed in advance with the Wiz.
Quote: unJonOpen to any challenger. Money escrowed in advance with the Wiz.
I'm willing to make escrow arrangements with the Wizard.
Do you have any conditions not specified earlier?
I have just one...the exercise and settlement must be conducted with both participants (that's me & you) in attendance, in person.
Any witness is acceptable to me.
Quote: AlanMendelsonQuote: unJonQuote: AlanMendelsonQuote: DieterQuote: AlanMendelsonDieter, how much more information do you need? We're talking about two six sided dice... and one of them has settled on a number that has been announced.
link to original post
In the problem, you do not know which of the two dice shows a 2.
In your rearrangement of the problem, you do know whice dice shows a 2.
That matters.
You've either been offered an amazing opportunity to win Wizard's money or an amazing opportunity to grow Wizard's bankroll. I suggest at least 40 units, and a clear statement of the rules of contest.
Only bet money you can afford to lose. Best of luck.
link to original post
Why is everyone saying you need to know which of the two dice is showing a 2?
It makes no difference in a two dice problem.
Tell me why it makes a difference. Please.
And what is the bet?
link to original post
Here is the bet. I’ll take it if the Wiz doesn’t want it.
Two dice are rolled under a cup. Someone we both trust will report the answer to the questions: is there at least one die showing a two?
If that person says “no”, then there’s no bet. They can show us the dice to verify the “no.”
If that person says “yes”, then there is a bet. Alan will win $70 if the roll is a hard four. Alan will lose $10 if the roll is not a hard four, but just a two on one die but not the other.
That’s 7:1 odds that if one die is a two the other won’t be.
That’s the bet. That’s the bet you say has 1/6 of being correct, and I and others say has 1/11 of being right.
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No thanks.
I do better at Red Rock.
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Explain how you do better at Red Rock please.
I’ll make the bet even “better” for you.
Wiz rolls two dice under a cup. I’ll ask him if he has at least one two. If he says no, then we reroll.
If he says yes, then I’ll ask him to show us a two. He will show us a die with a two under cup.
Then there will be one other die under the cup. We will bet whether than one die is a two or not.
If it’s a two, you win $70. If it’s not, I win $10.
