Two Dice Puzzle: Part Deux
Quote: beachbumbabsQuote: AlanMendelson
Not quite. You're not using TWO DICE. This is a two dice problem.
Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.
Alan,
I've been giving you a lot of latitude based on the past, your many good conversations, and your apparent confusion, but is it possible you're NOT trolling the forum at this point?
Honest to God, BBB. Do you understand this is a question about TWO DICE with at least ONE DIE SHOWING A TWO and during this entire debate those who say the answer is 1/11 are using MORE THAN TWO DICE to prove their point?
I'm not trolling. I'm just amazed that this is still going on.
Here's the truth:
When you have only two dice and at least one die is fixed showing the 2 there are only six other combinations -- PERIOD. Now, if someone can show me that there are ELEVEN combinations with only two dice, when at least one is showing a 2, I will gladly admit I am a fool. But right now I am just amazed what is being posted here.
Quote: AlanMendelson
When you have only two dice and at least one die is fixed showing the 2 there are only six other combinations -- PERIOD. Now, if someone can show me that there are ELEVEN combinations with only two dice, when at least one is showing a 2, I will gladly admit I am a fool. But right now I am just amazed what is being posted here.
What kind of fool cannot tell the difference between:-
"At least one die"
and
"One fixed die"?
There's just no damned comparison.
metaphor time:
Two children one broken window.
At least one child broke it.
So slap one child and buy the good one some sweets. Easy.
YOU CAN'T. YOU DON'T HAVE ONE FIXED CHILD TO SLAP
The image shown above has eleven combinations of two dice with at least one die showing a 2. I really don't understand your confusion. That image shows eleven pairs of dice to illustrate all combinations at the same time (for ease of viewing), but surely you understand that you can roll each of those combinations with a single pair of dice, right? Do you think any of those eleven combinations are somehow impossible with a fair throw of a pair of standard dice?Quote: AlanMendelsonIf someone can show me that there are ELEVEN combinations with only two dice, when at least one is showing a 2, I will gladly admit I am a fool.
Quote: MathExtremistThe image shown above has eleven combinations of two dice with at least one die showing a 2. I really don't understand your confusion.
Really, ME? The original question involved two dice. I didn't read that question to mean you can pick any two dice, and change the two dice during the problem solving. I took that original question to mean that two specific dice were rolled. And of those two specific dice, at least one showed a 2.
So let's start from the beginning: you have two specific dice and at least one of them is showing a 2. How many combinations are there of that die and the other die can there be showing 2-2. I say 1/6. Do you still say 1/11?? If you say 1/11 where to the other 5 faces come from? Are they coming from the die showing a 2? And how do you manage that -- by turning the faces on the die with a 2 on it?
Quote: AlanMendelsonSo let's start from the beginning: you have two specific dice and at least one of them is showing a 2. How many combinations are there of that die and the other die can there be showing 2-2. I say 1/6.
The underlined is your problem in interpretation. There is no "of that die" -- either of the dice can be 2, or both of them can. There are 10 ways for a pair of dice to show 2 and not-2, and one way for a pair of dice to show 2 and 2. That's one in 11 total. Do you dispute any of this?
But if we're starting from the true beginning, and you're not trolling, then let's start with just one die. Please answer the following two questions:
a) I roll a single fair die and don't tell you anything about it. What is the probability of it being 2?
b) I roll a single fair die and truthfully tell you "it is an even number". What is the probability of it being 2?
Quote: OnceDearWhat kind of fool cannot tell the difference between:-
"At least one die"
and
"One fixed die"?
There's just no damned comparison.
I guess Im the fool. Because when you have a two dice problem it doesn't make a difference which of the two dice it is.
But it does make a difference when the original problem deals with two red dice (for example) but you decide to bring in a green die, or a blue die, or a yellow to show more combinations of dice. And that's what all of you have been doing.
This is why I asked all of you to illustrate your solution with a video showing the two dice in the original problem and no one has done it. You are all reaching for other answers, whether they be the odds of rolling dice or the odds of combinations using multiple dice. But not one of you has addressed the actual stated problem.
I had high hopes the Wizard would. But instead he told us that he was rolling hundreds of rolls of two dice to come up with the results of the rolls. Yes, he did not answer the original question either.
Now, in all fairness, there were a few others at the beginning who said the answer was 1/6. Ayecarumba and GWAE but I am sure they were scared away by the mob.
The mob doesn't frighten me.
it is so wonderful you boys are still having fun with two diceQuote: AlanMendelsonHonest to God, BBB. Do you understand this is a question about TWO DICE <snip>
I'm not trolling. I'm just amazed that this is still going on.
i figured, wrong, i thoughts that you would want at least 10 dice or more by now
maybe try a different game for more fun
I take one die and toss it into my 1st bedroom but B4 it lands I close the door.
i did not see the result
I take the other die and toss it into the master bedroom and watch it
I walk down stairs, i know, and say
at least one die has a 2
what is the probability that both have a 2
keep on having fun
Alan, invite us all to your next wedding!
no
it was just a thought as the Angels do not play tonight
Quote: MathExtremistQuote: AlanMendelsonSo let's start from the beginning: you have two specific dice and at least one of them is showing a 2. How many combinations are there of that die and the other die can there be showing 2-2. I say 1/6.
The underlined is your problem in interpretation. There is no "of that die" -- either of the dice can be 2, or both of them can. There are 10 ways for a pair of dice to show 2 and not-2, and one way for a pair of dice to show 2 and 2. That's one in 11 total. Do you dispute any of this?
But if we're starting from the true beginning, and you're not trolling, then let's start with just one die. Please answer the following two questions:
a) I roll a single fair die and don't tell you anything about it. What is the probability of it being 2?
b) I roll a single fair die and truthfully tell you "it is an even number". What is the probability of it being 2?
WHY DO YOU KEEP CHANGING THE QUESTION?
ALL OF YOU... WHY DO YOU KEEP CHANGING THE QUESTION?
ME, the problem involves two dice and only two dice. At least one of those two dice shows a 2. If one shows a 2 it cannot show any other face, so the question is now what the OTHER DIE shows. There are six faces on that OTHER DIE. There are only SIX FACES available for the answer. Not ten. Not eleven. Six.
Quote: AlanMendelson
WHY DO YOU KEEP CHANGING THE QUESTION?
ALL OF YOU... WHY DO YOU KEEP CHANGING THE QUESTION?
ME, the problem involves two dice and only two dice. At least one of those two dice shows a 2. If one shows a 2 it cannot show any other face, so the question is now what the OTHER DIE shows. There are six faces on that OTHER DIE. There are only SIX FACES available for the answer. Not ten. Not eleven. Six.
So when you roll two 2's, there is no "Other Die" to look at. Now what?
Quote: AlanMendelsonWHY DO YOU KEEP CHANGING THE QUESTION?
I'm asking a different, more basic question as a prelude. If you can't answer a question about a single die, you won't be able to understand questions involving multiple dice. That's what learning is about -- start at the beginning, work your way up. You did, I recall, profess to actually be interested in learning why the answer is what it is. Were you just pulling our collective legs or are you actually interested?
Answer these:
a) I roll a single fair die and don't tell you anything about it. What is the probability of it being 2?
b) I roll a single fair die and truthfully tell you "it is an even number". What is the probability of it being 2?
And stop yelling.
Quote: mipletSo when you roll two 2's, there is no "Other Die" to look at. Now what?
Good question and I addressed it about a thousand posts ago. The answer is still 1/6.
choose either of the dice showing a 2 and the answer is still 1/6 for the chance that the other die is also showing a 2.
the Wizard asked this and so did a dozen others.
Quote: AlanMendelsonQuote: beachbumbabsQuote: AlanMendelson
Not quite. You're not using TWO DICE. This is a two dice problem.
Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.
Alan,
I've been giving you a lot of latitude based on the past, your many good conversations, and your apparent confusion, but is it possible you're NOT trolling the forum at this point?
Honest to God, BBB. Do you understand this is a question about TWO DICE with at least ONE DIE SHOWING A TWO and during this entire debate those who say the answer is 1/11 are using MORE THAN TWO DICE to prove their point?
I'm not trolling. I'm just amazed that this is still going on.
Here's the truth:
When you have only two dice and at least one die is fixed showing the 2 there are only six other combinations -- PERIOD. Now, if someone can show me that there are ELEVEN combinations with only two dice, when at least one is showing a 2, I will gladly admit I am a fool. But right now I am just amazed what is being posted here.
Ok, Alan, thanks.
Given that we have AT LEAST 1 two.
Take 1 die. It's a 2, but it COULD HAVE BEEN a 1,2,3,4,5, or 6. 1 chance to be a 2. 5 chances to be a not-2.
Take a second die. It's a 2, but it COULD HAVE BEEN a 1,2,3,4,5, or 6. 1 chance to be a 2. 5 chances to be a not-2.
So, there were 10 chances to be a 2 with a not-2, and only 1 chance to be a 2-2. Totalling 11 chances.
Any possibility that helped?
Quote: MathExtremistI'm asking a different, more basic question as a prelude. If you can't answer a question about a single die, you won't be able to understand questions involving multiple dice. That's what learning is about -- start at the beginning, work your way up. You did, I recall, profess to actually be interested in learning why the answer is what it is. Were you just pulling our collective legs or are you actually interested?
Answer these:
a) I roll a single fair die and don't tell you anything about it. What is the probability of it being 2?
b) I roll a single fair die and truthfully tell you "it is an even number". What is the probability of it being 2?
And stop yelling.
I haven't yelled yet. There is no need for a prelude. The question is specific, and so is the answer. Stop changing conditions and making up scenarios.
Quote: beachbumbabs
Given that we have AT LEAST 1 two.
Take 1 die. It's a 2, but it COULD HAVE BEEN a 1,2,3,4,5, or 6. 1 chance to be a 2. 5 chances to be a not-2.
Take a second die. It's a 2, but it COULD HAVE BEEN a 1,2,3,4,5, or 6. 1 chance to be a 2. 5 chances to be a not-2.
So, there were 10 chances to be a 2 with a not-2, and only 1 chance to be a 2-2. Totalling 11 chances.
Any possibility that helped?
Absolutely not.
Try this: you rolled two dice and one of them landed on a 2. (We are told at least one landed on a 2.) You don't know what the other die is showing (remember there are only two dice in the problem). How many faces are on the unknown die? Is it ten? Is it 11? Is it 36?
What is the chance that the unknown die is also showing a 2? Is it 1/11 or 1/10 or 1/36 or 1/6 ?
That is the only question in the problem.
Quote: AlanMendelsonAbsolutely not.
Try this: you rolled two dice and one of them landed on a 2. You don't know what the other die is showing. How many faces are on the unknown die? Is it ten? Is it 11? Is it 36?
What is the chance that the unknown die is also showing a 2? Is it 1/11 or 1/10 or 1/36 or 1/6 ?
In the original question, you don't know WHICH one is a 2. Every time you've restated the problem, you've changed it from an AT LEAST question to a IF THEN question, separating out 2 dice to 1 plus 1 die. The way you rephrase it, the answer is 1/6, because you've pulled out 1 specific die from the question. But it's 1/6 for one die, AND 1/6 for the other die, depending on WHICH die is a 2. The 2-2 is a duplication (because of the AT LEAST condition), so you can only count it once, not twice.
Quote: IbeatyouracesAll I know is this. There are 36 total combinations. 11 of them show at least one 2. Of the 11, only 1 shows 2-2. The other 10 show a 2 and some other number that is NOT a 2.
And this is why there has been this lengthy exchange. You are considering 36 combinations of which 11 show at least one deuce. But that is not the problem nor the question being asked, nor the information given in the original problem.
Yet, you all continue to use that information about 36 combinations and 11 combinations when the original question clearly tells you that at least one of the two dice is showing a 2 -- which ELIMINATES that die from the problem-solving. The problem-solving rests only on the unknown die, and of that die we are told there are only six faces.
I've said it probably a hundred times. Your math regarding the answer 1/11 is good math, but it's not the answer to this particular question.
Quote: AlanMendelsonI haven't yelled yet. There is no need for a prelude. The question is specific, and so is the answer. Stop changing conditions and making up scenarios.
There's a difference between asking for an explanation and demanding one, and you just crossed the line (yes, ALL CAPS is yelling on the Internet).
Denying the fact that you even need a foundation seals the deal. You won't answer a single, specific quantitative question about your understanding of probabilities, and you haven't during the entirety of this thread. If you're not game, neither am I.
Good luck to the rest of you.
Quote: beachbumbabsIn the original question, you don't know WHICH one is a 2. Every time you've restated the problem, you've changed it from an AT LEAST question to a IF THEN question, separating out 2 dice to 1 plus 1 die. The way you rephrase it, the answer is 1/6, because you've pulled out 1 specific die from the question. But it's 1/6 for one die, AND 1/6 for the other die, depending on WHICH die is a 2. The 2-2 is a duplication (because of the AT LEAST condition), so you can only count it once, not twice.
And in doing so, you go from 2 ways in 12 combinations (which is 1 in 6) to 1 way in 11. (2-1)/(12-1) = 1/11.
Quote: beachbumbabsIn the original question, you don't know WHICH one is a 2.
I am going to ask you to make a video -- just like I asked everyone else -- and show me why it makes a difference which die is a 2 in this two-dice problem?
Because when I look at two dice, and with the left die showing a 2, I have only six choices on the right die.
And when I have the right die showing a 2, I have only six choices on the left die.
And when I roll 2-2 and look at either die, then the other die whichever die it is, only has six choices.
Quote: MathExtremistThere's a difference between asking for an explanation and demanding one, and you just crossed the line (yes, ALL CAPS is yelling on the Internet).
Denying the fact that you even need a foundation seals the deal. You won't answer a single, specific quantitative question about your understanding of probabilities, and you haven't during the entirety of this thread. If you're not game, neither am I.
Good luck to the rest of you.
Part of the problem is that you are, what I call, "overthinking" the original problem. And it's not you alone, ME. It's all of you who came up with the answer 1/10 or 1/11. To come up with 1/10 or 1/11 you have to look at graphs and various combinations of dice when very simply the problem tells you at least one die has settled on a 2.
I am going to ask you this: if one die is already settled on a 2, how can you possibly also consider the other five faces on THAT die?? Sorry, but it makes no sense.
Quote: IbeatyouracesAnd what if you don't see the dice but guess "left die had the two" when in reality it's the right die? Now what's the probability?
I guess you didn't read what I wrote a hundred times before. In a two dice problem it does not matter.
Quote: AlanMendelsonI guess you didn't read what I wrote a hundred times before. In a two dice problem it does not matter.
It really does matter. In fact, it's the crux of the misunderstanding.
Quote: AlanMendelsonI guess you didn't read what I wrote a hundred times before. In a two dice problem it does not matter.
Exactly why you're wrong.
Quote: beachbumbabsIt really does matter. In fact, it's the crux of the misunderstanding.
This is why I ask you to shoot a video, with two dice, and then show me how it matters. Fair enough?
I have nothing else to say until I see the video with your explanation.
Note: I won't be making a video, just trying to save others time.
Quote: AlanMendelsonPart of the problem is that you are, what I call, "overthinking" the original problem. And it's not you alone, ME. It's all of you who came up with the answer 1/10 or 1/11. To come up with 1/10 or 1/11 you have to look at graphs and various combinations of dice when very simply the problem tells you at least one die has settled on a 2.
(snip: yet another question)
Or maybe you're underthinking. The problem is precisely worded but you're not reading it properly.
And quid pro quo. You don't get to ask any more questions until you start answering them.
Quote: LoquaciousMoFWWould this video style be acceptable? If not, what procedural changes would need to be made?
Note: I won't be making a video, just trying to save others time.
I only watched the first minute or so of the video, but sure, this is all we would need to see the dice. Quality doesn't matter. What is important is your explanation of the following and why you think your result/answer is what it is.
THE ORIGINAL 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. 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?"
===================
1. Task #1 Presentation of the question:
In my video I rolled the two dice until one die showed a two, which was my interpretation of the original question/challenge. The original question sets up that "at least one of the dice is a 2." For your presentation you can intentionally set one of the dice as a two or even both dice showing a 2 because the original question said "at least one of the dice is a 2" and that would include both dice being a 2.
2. Task #2 Answer the question: what is the probability that both dice are showing a 2?
Simply give your answer.
3. Task #3 Explain how you came to your answer in Task #2. What methodology did you use with the two dice that we see in your video.
====================
That's all I want to see. Either roll or set the dice with at least one die showing a 2, or if you like 2s on both cubes, give your answer to the question, and then please explain how you arrived at your answer. If you want to pick up the dice and count the faces, that's good. If you want to point to the faces, that's good too. If you want to move the dice around to help explain your answer, that's good as well. Just keep it to the two dice.
Thanks. I look forward to seeing the videos.
Quote: AlanMendelsonQuote: LoquaciousMoFWWould this video style be acceptable? If not, what procedural changes would need to be made?
Note: I won't be making a video, just trying to save others time.
I only watched the first minute or so of the video, but sure, this is all we would need to see the dice. Quality doesn't matter. What is important is your explanation of the following and why you think your result/answer is what it is.
THE ORIGINAL 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. 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?"
===================
1. Task #1 Presentation of the question:
In my video I rolled the two dice until one die showed a two, which was my interpretation of the original question/challenge. The original question sets up that "at least one of the dice is a 2." For your presentation you can intentionally set one of the dice as a two or even both dice showing a 2 because the original question said "at least one of the dice is a 2" and that would include both dice being a 2.
2. Task #2 Answer the question: what is the probability that both dice are showing a 2?
Simply give your answer.
3. Task #3 Explain how you came to your answer in Task #2. What methodology did you use with the two dice that we see in your video.
====================
That's all I want to see. Either roll or set the dice with at least one die showing a 2, or if you like 2s on both cubes, give your answer to the question, and then please explain how you arrived at your answer. If you want to pick up the dice and count the faces, that's good. If you want to point to the faces, that's good too. If you want to move the dice around to help explain your answer, that's good as well. Just keep it to the two dice.
Thanks. I look forward to seeing the videos.
Alan,
Task 1 as you list it changes the question. You CAN'T set one of the dice as a 2, or you've removed 5 of your possibilities for that die before you even start. It might have been the OTHER die that came up as a 2, and the one you set, you prevented it from being anything else but a 2 by setting it.
Quote: beachbumbabs...Alan,
Task 1 as you list it changes the question. You CAN'T set one of the dice as a 2, or you've removed 5 of your possibilities for that die before you even start. It might have been the OTHER die that came up as a 2, and the one you set, you prevented it from being anything else but a 2 by setting it.
Exactly! That why I recommended changing one of the 2's to an "A" and the other to a "B". Now state the problem like this:
Shake the dice in a cup them turn it over. Your friend peeks and truthfully says "at least one of the dice is showing an "A" OR "B". What's the probability that BOTH dice are showing "A" AND "B".
Quote: IbeatyouracesExactly! That why I recommended changing one of the 2's to an "A" and the other to a "B". Now state the problem like this:
Shake the dice in a cup them turn it over. Your friend peeks and truthfully says "at least one of the dice is showing an "A" OR "B". What's the probability that the pair of dice are showing "A" AND "B".
I clarified your question a teensy bit. I hope you don't mind.
That would maybe do it IF. the A and B were stuck alongside the 2 spots and not obscuring them. so they could have little stickers on the dice (A2) and (B2)
But it might take a couple of years for Alan to acknowledge that (A2) (B2) was the same result as A-B was the same result as 2-2. Then a couple more years to agree that you haven't seriously changed the question.
Then maybe a few more years for him to accept that he is not allowed to nail one dice down, because, putting stickers on THAT ARE DIFFERENT, you gave him an impossible task. The same impossible task that he's now sidestepped for a month.
You could even start with the video with those stickers on top.
But he wouldn't understand that. If he did, you could multiply the probability of Alan Understands times Alan Acknowledges to end up with a really tiny number.
For Maths geeks, I'm not using brackets there for any mathematical intent.
It's about the meaning of the phrase 'At least..'. Two simple words.
Let's use an example from infant's school English comprehension.
If the teacher says to her assembled class.
Quote: English teacher in primary schoolWhen I look around the classroom, any day of the week, I can see that we always have at least one child off sick"
Does she mean that there is one child, one very sick child, that never shows up?
Or might it be a different sick child on different days?
If anyone really believes that it is just one sick child, the same child every day, then that person can stop pretending to read and get back to looking at the pictures.
We have a question where
Quote:'At least one die is a deuce'
It's strongly implied that it has always been and always will be, that we are interested in days where 'At least one die is a deuce'
Nowhere. ABSOLUTELY NO BLOODY WHERE, is it even mildly implied that it will be the same die every day or every throw of the dice.
Can anyone show me in any way why we should treat the 'At least one child is off sick' and the 'At least one die is a deuce' phrases any differently?
Teacher never did say
Quote: English teacher in primary school didn't sayWhen I look around the classroom, any day of the week, I can see that we have one child who is always off sick
Our dice peeker never did say
Quote: Dice peeker didn't sayWhen I look under the cup I can see that we have one die that is always a deuce
It's not said. It's not implied. To suggest that it is, is an epic FAIL at primary school English comprehension.
But that is exactly what we see in one member's assertion. That once we have picked one die, then that die can forever be considered a deuce.
It's nonsense, just as it would have been nonsense to believe in that one, forever sick, child.
Quote: OnceDearSomebody needs to realise that this is not about probability or maths or even dice.
It's about the meaning of the phrase 'At least..'. Two simple words.
That's partly true, but it's also really about math. Alan wants to hold up a pair of 2s and say "what are the chances these dice could be a pair of 2s? The answer is 1/6."
That's just wrong anyway you slice it. Either you're holding a pair of 2s already and the chances are 1/1 (100%), or you're holding two undecided dice and the chances are 1/36.
If Alan had done his "video of him holding two dice" experiment properly, he would have started with a pair of 2s and said "okay, there's only one way for this pair of 2s to happen, so how many other ways are there for the dice to show "at least one 2" but *not* be a pair of 2s?" Then hold the right die on 2 and examine five (5) other non-2 sides on the left die; and then hold the left die on 2 and examine five more (5) other non-2 sides on the right. Total of 10 other combinations that meet the "at least one die is a 2" criterion. Plus one for the pair and that's 11 total, only one of which is the pair of 2s. 1/11.
Alan's technique (and error) is to take all three of those steps individually, failing to consider that it's all part of the same scenario and that there are overlapping combinations. He is essentially double-counting the 2-2 combination and getting two ways in 12 combinations, which is 1/6. But there aren't two ways to roll 2-2. Only one. One less in the numerator and one less in the denominator goes from 2/12 to 1/11.
The question did not ask "here is a single die showing 2 that is part of a combination of two dice. What are the chances that combination is 2-2?" The answer to *that* question would be 1/6 because, though there are 11 combinations of dice with a 2 in them, the 2-2 combo has two ways to pick that 2 so double-counting is what you want to do. The question here was "here is a combination with at least one 2 in it, what are the chances that combination is 2-2?" With that phrasing, double-counting is incorrect.
You will play the game only once.
The dice are shaken in the cup and slammed on the table.
The peeker looks at one die (and only one die), and truthfully declares, "At least one of the dice is a ___".
In this instance, the die the peeker looks at happens to be a two.
What is the probability the other die is also a two?
If I understand correctly, the probability is 1/6, all because the peeker only checks one die before making the declaration.
Quote: MathExtremistThat's partly true, but it's also really about math. Alan wants to hold up a pair of 2s and say "what are the chances these dice could be a pair of 2s? The answer is 1/6." .
Hi ME,
I don't disagree with you at all. Alan's grasp of most basic probability is just staggering. However, I've been with both these threads from the outset, and come rain or shine, Alan has insisted that it is appropriate to 'Set One Die As a Deuce'. He has very very frequently told all of us that we just are not reading the question as it is posed. He's had every 'probability' type proof handed to him, spoon fed. Yet not once has he tackled his abject ignorance of the 'At Least' phrase. For that matter, he also abjectly and consistently refuses to answer any question without throwing back, WELL SHOW ME IN A DIFFERENT WAY. He should not be humoured, until, as you say, he starts doing the groundwork, rolling the dice and measuring the resulting outcomes, or even answering questions that show that he does give a damn.
I just needed to attack his stupid assertion that 'One die doesn't matter'.
But I don't hold out much hope. He must be trolling for effect, because nobody could be that..... (Insert your own expletive and adjective)
Quote: AyecarumbaFor your consideration.
You will play the game only once.
The dice are shaken in the cup and slammed on the table.
The peeker looks at one die (and only one die), and truthfully declares, "At least one of the dice is a ___".
In this instance, the die the peeker looks at happens to be a two.
What is the probability the other die is also a two?
If I understand correctly, the probability is 1/6, all because the peeker only checks one die before making the declaration.
Correct. Look at just one die. Whatever that die is, you declare it. No issue with that. That's what Alan does. He knows one die is a two, so he only rolls or considers one die.
Quote: OnceDearCorrect. Look at just one die.
And, that's where the arguments for either side of this "debacle" fall apart: The numbers which show on both dice can't be noted at once simultaneously. Simple as that.
Add on: A true understanding of the universe relies on such basic perception.
Quote: KerkebetAnd, that's where the arguments for either side of this "debacle" fall apart: The numbers showing on both dice can't be noted at once simultaneously. Simple as that.
Erm no. There's absolutely no need for them to be seen simultaneously. They only have to be seen, one after the other, but before the announcement is made. In the 'peeking interval'. Seeing one die and then the other in succession in any order would allow exactly 11 different and equally likely outcomes to be in play. It's not a play on words, it's not quantum physics or rocket science. It's not even ambiguity or open to interpretation. There is a right interpretation and there is a wrong one.
Quote: KerkebetAnd, that's where the arguments for either side of this "debacle" fall apart: The numbers which show on both dice can't be noted at once simultaneously. Simple as that.
One does not need to note or observe any numbers on any physical dice at all in order to understand and answer the question. The question can be answered analytically without resorting to examining physical objects.
I presume that you don't need to count on your fingers to get to 10, nor do you have a problem counting to 327 without having 327 physical objects in front of you. Once you move past childhood, you have (or should have) the ability to count in the abstract, untethered to any physical objects.
Probability questions involving a throw of a pair of dice are also counting problems. They're more complicated, to be sure, but ultimately discrete probability problems are about counting. You don't need fingers or toes or dice to answer them. Or at least you shouldn't.
Quote: KerkebetAnd, that's where the arguments for either side of this "debacle" fall apart: The numbers which show on both dice can't be noted at once simultaneously. Simple as that.
Add on: A true understanding of the universe relies on such basic perception.
I thought it was pretty clear that when the peeker announces that "at least one of the dice is a two" that it means that either one die is a two, the other die is a two, or both dice are a two.
The order that he looks at the dice, one at a time, isn't important. Simultaneous isn't necessary.
If the first die he looked at is a two, there are 5 ways for the 2nd die to NOT be a two.
If the first die he looked at is NOT a two, there are 5 ways for the first die to NOT be a two when the 2nd die IS a two.
there is one way for both dice to be a two.
I know the color and die A/B has already been brought up, but would this help at all?
On video, shake 2 different colored dice and slam the cup on the table. Have somebody give an announcement of whether or not there is at least one 2 showing - but keep the dice hidden from the camera. Then somebody other than the announcer (to avoid speculation of "cheating") would pull a random die out from under the cup. (This really has no effect other than to show the significance of assuming you know which die is a 2.)
I think we have established there are 11 possible combinations underneath the cup:
Blue Red
2 1
2 2
2 3
2 4
2 5
2 6
1 2
3 2
4 2
5 2
6 2
This gives 22 possibilities for the pulled die.
If the pulled die is a Blue 2 (6/22 chance) I think Alan will readily agree that there is a 1/6 chance the Red die is a 2 as well.
If the pulled die is a Red 2 (6/22 chance) again I think Alan will readily agree that there is a 1/6 chance the Blue die is a 2 as well.
Combining we get:
p(pulled Blue 2) * p(hidden Red 2) + p(pulled Red 2) * p(hidden Blue 2) =
6/22 * 1/6 + 6/22 * 1/6 = 1/22 + 1/22 =
1/11 chance that both dice are 2
Conversely, calculating the chance of only 1 die (and not both) showing a 2:
If the pulled die is a Blue non-2 (5/22 chance) there is a 1/1 chance the Red die will be a 2 since we know at least one of them must be a 2.
If the pulled die is a Red non-2 (5/22 chance) there is a 1/1 chance the Blue die will be a 2 since we know at least one of them must be a 2.
We must also consider the cases when the pulled die is a 2 but the hidden die is not:
If the pulled die is a Blue 2 (6/22 chance) I think Alan will readily agree that there is a 5/6 chance the Red die is not a 2.
If the pulled die is a Red 2 (6/22 chance) again I think Alan will readily agree that there is a 5/6 chance the Blue die is nor a 2.
Combining we get:
p(pulled Blue non-2) * p(hidden Red 2) + p(pulled Red non-2) * p(hidden Blue 2) + p(pulled Blue 2) * p(hidden Red non-2) + p(pulled Red 2) * p(hidden Blue non-2) =
5/22 * 1/1 + 5/22 * 1/1 + 6/22 * 5/6 + 6/22 * 5/6=
20/22 =
10/11
Quote: Dalex64I thought it was pretty clear that when the peeker announces that "at least one of the dice is a two" that it means that either one die is a two, the other die is a two, or both dice are a two.
The order that he looks at the dice, one at a time, isn't important. Simultaneous isn't necessary.
If the first die he looked at is a two, there are 5 ways for the 2nd die to NOT be a two.
If the first die he looked at is NOT a two, there are 5 ways for the first die to NOT be a two when the 2nd die IS a two.
there is one way for both dice to be a two.
I think the methodology is important, and is the crux of the dispute. If the peeker only called out the number on the first die he saw (which happend to be a two), the probability is 1/6 for a pair.
Quote: miplet
OMG miplet did it: he counted the same die twice. No, this video is wrong.
Let's go back to the original question: When two dice are rolled and at least one shows a 2 the question is what are the odds (chance, probability) that both dice show 2-2?
As we saw in miplet's video, originally one die showed a two and the second die showed a 5. That 5 was one of the six faces on that second die. One of the six faces on that second die was a 2, but it didn't show.
The original question said nothing about asking about the total number of combinations of two dice containing at least one two. It specifically asked for the probability that both dice would show 2-2 when at least one die is known to be a 2.
Now, consider this question: How many six-sided dice combinations contain at least one 2, and how many of those combinations show 2-2? The answer to that question would be 1/11. But that wasn't the question asked.
It is however, what you guys considered the question to be, which brings us to ME's comment here:
Quote: MathExtremistOne does not need to note or observe any numbers on any physical dice at all in order to understand and answer the question.
Of course I am really surprised ME said this. This has been the problem. All of you who say the answer is 1/11 have ignored the conditions of the question. So, I am going to ask this based on the conditions of the question:
Miplet and others: if you froze the die showing a 2 with the 2 face showing, can you still justify your answer of 1/11?
Quote: AlanMendelsonNow, consider this question: How many six-sided dice combinations contain at least one 2, and how many of those combinations show 2-2? The answer to that question would be 1/11. But that wasn't the question asked.
But that's exactly the question asked. You are told 2 dice are rolled and the rolled combination contains at least one 2. What is the probability the rolled combination shows 2-2?
Quote: wudgedBut that's exactly the question asked. You are told 2 dice are rolled and the rolled combination contains at least one 2. What is the probability the rolled combination shows 2-2?
Yes, the question is what is the probability the rolled combination shows 2-2 WHEN AT LEAST ONE DIE IS SHOWING A 2.
When one die is showing a 2 the answer is 1/6.
Otherwise, if you are not counting the die that is already showing a 2 the answer is 1/36.
The question with the answer 1/11 is NEVER asked. And I've been saying that all along.
Quote: AyecarumbaI think the methodology is important, and is the crux of the dispute. If the peeker only called out the number on the first die he saw (which happend to be a two), the probability is 1/6 for a pair.
That's true, but the premise of the problem is that the peeker does not simply call out the number on the first die and ignore the other. The premise is that the peeker says "at least one of the dice is 2," which means both dice are examined. If your evaluation is constrained to (or "conditioned upon") only those circumstances where the peeker truthfully says "at least one die is 2" then 25 of the 36 two-dice combinations disappear. That's why it's a conditional probability problem, not just one related to unconditional probabilities.
It's also why every single experimental setup (including Alan's) has involved rolling *until* you get a 2 on one die. In other words, there doesn't appear to be any real dispute about the fact that the problem is constrained to only those combinations containing at least one 2. If the intent of the problem was that the peeker would always say something, even if the roll contains no 2s, then there isn't enough information to answer the question without knowing how the peeker actually chose what to say. That was discussed way back here:
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/13/#post453494
For example, if the peeker chose the lowest number of the two dice, regardless of what they were, then the probability of a pair depends on what number the peeker called. If he said "there is at least one 6" then under those rules you know the other die is always a 6 too.
Alan is essentially saying "someone told me this die right here is 2, so I'm going to ignore it and only consider the other die." He's answering the question "what is the probability of one die being 2?" but that's the wrong question.
Quote: AyecarumbaI think the methodology is important, and is the crux of the dispute. If the peeker only called out the number on the first die he saw (which happend to be a two), the probability is 1/6 for a pair.
I don't interpret anything in the original question that would lead me to believe that the peeker would only look at one of the two dice, or if the first dice he saw wasn't a two that he wouldn't go ahead and look at the other die, or only call out "at least one of the dice is a two" if and only if the first die he saw was a two.
You can't answer "at least once of the dice is a two" if you only look at one die, and it isn't a two, but the other die that he didn't look at was a two.
