Two Dice Puzzle: Part Deux
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.
Quote: AlanMendelsonYes, 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.
So now there's a difference between a two-dice combination "containing at least one 2" and a two-dice combination "showing at least one 2"?
This just keeps getting better.
Quote: MathExtremist
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.
I agree that the "peeker" saw both dice when he announced "at least one of the dice is a 2." But he didn't have to see both dice.
I treat the problem/question as if he saw both dice -- but even if he saw only one of the cubes with a 2 on it the answer would be the same.
As soon as you know that one of the cubes shows a 2 the question becomes: what does it take for the second cube to also show a 2 to have 2-2?
As I said a hundred times, even if both dice were rolled and both already showed 2-2, the second cube had a 1/6 chance of showing that 2.
Again it's the wording of the question. There was never a question asked that would give the answer of 1/11. Never was a question asked about the combinations of dice containing a 2 either.
As I said early-on in the discussion, the question was written to make you consider the 11 combinations because of this phrase that preceded the original question:
"Along the vein of the Two Coin Puzzle,..." It was that phrase that made everyone make the question more complicated than it really is.
It was that phrase "Along the vein of the Two Coin Puzzle,..." that had you all thinking you had to solve the question as you would with the two coin puzzle and looking at all of the combinations of flipping coins. But again, this wasn't a coin flipping question. There are two dice and at least one of those two dice landed on a 2.
And by the way, for those of you who question the meaning of the phrase "at least one of the dice" it means this: one of the dice could be a two or both of the dice could show 2. "At least" means one or more. It does not mean that 2 could jump from one die to another.
I want to thank miplet for doing his video, but unfortunately he did what the rest of you did: ignored the condition of the problem that at least one die is showing a two. What he did was roll two dice and after one landed on a two proceeded to show that there are 11 combinations of dice with a 2. For heaven's sake, that was not the question.
Someone please do a video and when at least one of the two dice lands on a 2 look at the dice and ask -- "with at least one die showing a two what is the probability that both dice show 2-2?" And the only answer you can come up with is simply this: when one die shows a 2 then there is a 1/6 chance that the other die is showing a two.
And please stop this nonsense about not knowing which die is showing a 2. It doesn't matter. There are only two dice. If either die shows a 2 or if both dice show a 2, when at least one is a 2 the answer is 1/6.
And if you don't accept 1/6 and you ignore that one die is showing a 2 then the answer is 1/36.
Quote: Dalex64I 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.
What I am proposing is that the peeker would call out whatever was on the observed die. In this case, it happened to be a two. The original conditions do not specifically call for the probability to be calculated only if a deuce is seen. It only happens that the peeker happened to see a deuce in that particular instance. It could have been any of the other five numbers.
And you can say truthfully, that "at least one die is a ___ " by looking at only one die in the pair. It happend to be a two this time, maybe next time it will be a five.
Just roll two dice. Whenever one of the dice is a two, write down how often the other die is a two.
This isn't about how many sides there are on the other die, it is about how often a two on the other die ends up on top.
The longer you do this, the more accurate the answer should become.
Quote: MathExtremistSo now there's a difference between a two-dice combination "containing at least one 2" and a two-dice combination "showing at least one 2"?
This just keeps getting better.
He thinks having 2 dice already set and knowledge of at least one of them showing a 2 is not in the same realm of probabilities as physically rolling 2 dice.
Quote: Dalex64Why do the people who still insist that the answer is 1/6 refuse to do the experiment?
Just roll two dice. Whenever one of the dice is a two, write down how often the other die is a two.
This isn't about how many sides there are on the other die, it is about how often a two on the other die ends up on top.
The longer you do this, the more accurate the answer should become.
Are you sure this is supposed to prove 1/11 ??
If I use loaded dice, I will not get 1/6. But every time I roll one die, I should get a 2 those 1/6 times.
Quote: AlanMendelsonAs I said a hundred times, even if both dice were rolled and both already showed 2-2, the second cube had a 1/6 chance of showing that 2.
Which second cube? The first second cube or the second second cube?
Quote:And please stop this nonsense about not knowing which die is showing a 2. It doesn't matter. There are only two dice. If either die shows a 2 or if both dice show a 2, when at least one is a 2 the answer is 1/6.
Also, if neither die shows a 2 the answer is still 1/6, because you're only looking at one die anyway.
Quote: AyecarumbaWhat I am proposing is that the peeker would call out whatever was on the observed die. In this case, it happened to be a two. The original conditions do not specifically call for the probability to be calculated only if a deuce is seen. It only happens that the peeker happened to see a deuce in that particular instance. It could have been any of the other five numbers.
And you can say truthfully, that "at least one die is a ___ " by looking at only one die in the pair. It happend to be a two this time, maybe next time it will be a five.
Again, what part of the original question implies that the peeker would only look at one of the two dice?
Sometimes does he not look at either of the dice, so he isn't able to announce a two then either?
I do think the original question only calls for the probably to be calculated if a deuce is seen, and calls for the probability to be calculated every time a deuce is seen, and that both dice are looked at to determine whether or not a deuce is seen.
Quote: AlanMendelsonAre you sure this is supposed to prove 1/11 ??
If I use loaded dice, I will not get 1/6. But every time I roll one die, I should get a 2 those 1/6 times.
Which part of the original question implies that the dice are not fair?
Which part of the original question implies that only one of the dice is rolled?
Quote: AlanMendelsonBut every time I roll one die, I should get a 2 those 1/6 times.
And there it is. Alan is only looking at a single die. Round and round this goes, at some point we'll all stop feeding it...
Just to recap yet again, Here's the problem statement:
Quote: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?
By all means, let's roll one die to figure it out.
Quote: MathExtremistSo now there's a difference between a two-dice combination "containing at least one 2" and a two-dice combination "showing at least one 2"?
This just keeps getting better.
No there is no difference. What are you talking about? The phrases "containing at least one 2" and "showing at least one 2" mean the same thing to me. Why do you think there's a difference?
I think you are starting to reach.
Quote: MathExtremist
Also, if neither die shows a 2 the answer is still 1/6, because you're only looking at one die anyway.
If neither die shows a 2, there is no question to be answered. Did you ever read the original question? If you ever did you would not be looking at your graphs showing there are 11 combinations with dice showing a 2.
This is the problem: the original question.
Quote: AlanMendelsonNo there is no difference. What are you talking about? The phrases "containing at least one 2" and "showing at least one 2" mean the same thing to me. Why do you think there's a difference?
I quoted you... Do you not even read what you write?
Quote: AlanMendelsonHow 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.
Quote: AlanMendelsonthe 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.
Quote: wudgedHe thinks having 2 dice already set and knowledge of at least one of them showing a 2 is not in the same realm of probabilities as physically rolling 2 dice.
What are you talking about?
Excuse me, "math guys" but you are all starting to reach now to look for different answers, different reasons, and none of you has even tackled the original question. Not one of you. You are literally pulling "math" out of the air to come up with answers to a question that wasn't even asked.
The original question and the question to be answered is simply this: if you have at least one 2 showing, what are the odds for the second die to be a 2 as well. And that is the same as rolling one die with a 2 showing 1/6 times.
Quote: AlanMendelsonWhat are you talking about?
Excuse me, "math guys" but you are all starting to reach now to look for different answers, different reasons, and none of you has even tackled the original question. Not one of you. You are literally pulling "math" out of the air to come up with answers to a question that wasn't even asked.
Do you have any thoughts on my video experiment? https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/55/#post456702 Yes, it contains math, but it breaks down every possible scenario where there could be at least one 2 and then either both are 2 or there is only one 2.
Quote: AlanMendelsonWhat are you talking about?
Excuse me, "math guys" but you are all starting to reach now to look for different answers, different reasons, and none of you has even tackled the original question. Not one of you. You are literally pulling "math" out of the air to come up with answers to a question that wasn't even asked.
The original question and the question to be answered is simply this: if you have at least one 2 showing, what are the odds for the second die to be a 2 as well. And that is the same as rolling one die with a 2 showing 1/6 times.
The questions specifically states that both dice are rolled then asks what is the probability they both show 2. Not "what is the probability that the second die shows a 2"
Quote: MathExtremistI quoted you... Do you not even read what you write?
Quote: AlanMendelsonHow 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.
Quote: AlanMendelsonthe 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.
I guess you really don't understand. So let me take them one at a time:
I wrote: 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. And yes, aren't there 11 dice combinations with at least one 2? And isn't there only one of those 11 combinations that is 2-2. That is 1/11. Do you not understand that?
Secondly I wrote: 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. Do you not understand that? It's what I've been saying all along. When you have rolled two dice, and one settles on a 2, or when at least one settles on a 2, the chance of the second or other die also showing on a 2 is 1/6.
My statements do not conflict with each other. Do you think they do?
By the way, as I wrote in another post:
The original question and the question to be answered is simply this: if you have at least one 2 showing, what are the odds for the second die to be a 2 as well. And that is the same as rolling one die with a 2 showing 1/6 times.
Quote: wudgedThe questions specifically states that both dice are rolled then asks what is the probability they both show 2. Not "what is the probability that the second die shows a 2"
Yes, and there is the problem with the interpretation.
If you tell me that in a two-dice problem that one die is showing a 2, I immediately think what does it take for the second die in the two-dice problem to also show a 2? And that answer is 1/6. That is how I interpret the question and that is all this entire discussion has been about.
I want you to tell me why you think that is not the way to interpret the question? And that is why I asked everyone to shoot a video. And when a video was shot, immediately I could see that the question was not interpreted correctly. miplet ignored the condition of the question which was that at least one die was showing a 2.
miplet simply rolled the dice until a 2 was showing and then continued to show there were 11 combinations of two dice showing at least one two. And all the while he ignored the question.
And all of you who say the answer is 1/11 have ignored the original question, just as miplet did.
Quote: AlanMendelsonQuote: MathExtremistI quoted you... Do you not even read what you write?
Quote: AlanMendelsonHow 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.
Quote: AlanMendelsonthe 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.
I guess you really don't understand. So let me take them one at a time:
I wrote: 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. And yes, aren't there 11 dice combinations with at least one 2? And isn't there only one of those 11 combinations that is 2-2. That is 1/11. Do you not understand that?
Secondly I wrote: 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. Do you not understand that? It's what I've been saying all along. When you have rolled two dice, and one settles on a 2, or when at least one settles on a 2, the chance of the second or other die also showing on a 2 is 1/6.
My statements do not conflict with each other. Do you think they do?
By the way, as I wrote in another post:
The original question and the question to be answered is simply this: if you have at least one 2 showing, what are the odds for the second die to be a 2 as well. And that is the same as rolling one die with a 2 showing 1/6 times.
a) Yes, your statements conflict with each other. The two quotes from you are restatements of precisely the same question, which is why the answer should be the same. You have two different answers to the same question.
b) The phrase "at least one die ... the second die" makes no sense, and that is why your analogy to rolling a single die is wrong.
I can't explain further if you don't understand what "at least one" means.
Quote: AlanMendelsonI want you to tell me why you think that is not the way to interpret the question?
Because the dice are rolled together, not independently of one another. Both dice are rolled at the same time, giving 11 possible outcomes. 1 out of those 11 is 2-2.
Your analogy of assuming one die is a 2 and then rolling the second die (which you correctly state as 1/6) is not valid as the dice are being rolled independently of one another.
MODERATORS: PLEASE SUSPEND ME. I'm done here.
Quote: Dalex64Again, what part of the original question implies that the peeker would only look at one of the two dice?
There is no need to look at both. The peeker looks at one, calls it out and asks the probability of the other, unseen die, making a pair. In the case of the original question, the revealed die happened to be a two.
Quote: Dalex64
I do think the original question only calls for the probably to be calculated if a deuce is seen, and calls for the probability to be calculated every time a deuce is seen, and that both dice are looked at to determine whether or not a deuce is seen.
I suspect Alan may be doing his computation based on the "any pair" methodology.
Quote: AyecarumbaThere is no need to look at both. The peeker looks at one, calls it out and asks the probability of the other, unseen die, making a pair. In the case of the original question, the revealed die happened to be a two.
Are you saying that is what you think the original question is asking you to do?
This here cup holds one Red Die and one Blue Die. Shake. Slam. Announce "Deuce."Quote: AlanMendelson
Could any of these results have been the actual result? :
- The Red Die showed a deuce.
- The Blue Die showed a deuce.
- Both dice showed deuces.
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.
Mathematicians inherently assume that the mind just exists, but let it to the physicists to try to figure it all out afterward. How schizophrenic is that? Perhaps, the sort of "argument" which has unfolded. (Maybe the answer to that show about the math existing in the chemicals, etc, depends on the person's scholastic background?)
Part of this revolves around "at least one die of two", which amounts to "one or more dies". After the initial truthful "peek", for lack of a better word given the continued not-well defined arguments here, "one or more dies" can't show a 2 when only one of the dies shows a 2. No "or more dies" possibility then. Strictly speaking.
Quote: AlanMendelson...MODERATORS: PLEASE SUSPEND ME. I'm done here.
When the going gets tough...
Quote: Dalex64Are you saying that is what you think the original question is asking you to do?
I'm putting it forward as a possible explanation for the dispute. The methodology fits the description of the original (and by original, I mean the original Two Dice Puzzle thread) sequence of events. Assuming that the question is only interested in one die showing a two is a leap, because there is no indication the dice would continue to be shaken and checked until at least one die was a two. The more likely scenario is that the dice are shaken, peeked at, and the declaration made, no matter what the called out die was. It just happened in this paticular instance, that the called out die was a two.
Quote: AlanMendelsonI can't explain it to you either, ME.
MODERATORS: PLEASE SUSPEND ME. I'm done here.
Before I suspend you, I'd be interested in your thoughts on my video response to your video.
If you still wish to be suspended, you may indicate a term if you wish, which must be at least 30 days. If you don't specify a period of time, then after 30 days you may request to have the voluntary suspension lifted.
Quote: WizardBefore I suspend you, I'd be interested in your thoughts on my video response to your video.
Oh Mike, forever the optimist.
Not a cat in hells chance that will persuade. Not that person. No way. Care to take a bet?
Quote: KerkebetMathematicians inherently assume that the mind just exists, but let it to the physicists to try to figure it all out afterward. How schizophrenic is that? Perhaps, the sort of "argument" which has unfolded. (Maybe the answer to that show about the math existing in the chemicals, etc, depends on the person's scholastic background?)
Part of this revolves around "at least one die of two", which amounts to "one or more dies". After the initial truthful "peek", for lack of a better word given the continued not-well defined arguments here, "one or more dies" can't show a 2 when only one of the dies shows a 2. No "or more dies" possibility then. Strictly speaking.
All communication is predicated on a shared understanding. We can't be said to be communicating about a die if I'm thinking about a six-sided acrylic cube with spots on the faces and you're thinking about a device used to cut threads on metal rods. But both could show a 2 on the face!
The assumptions behind a probability question about dice, if not stated explicitly, is that the dice are both six-sided standard dice and that they are fair. No biases, imperfections, loads, etc., and six sides numbered one through six. The probability of throwing 2-2 is obviously not 1/11 if one of the dice is a D20 and the other is a D10(00).
Also, the condition "one or more dice show 2" is met by both the cases "exactly one die shows 2" and "exactly two dice show 2." That's the point - both are part of the problem space. You don't know which any given roll may be because you only are told about one die, but what happens on any given roll doesn't actually matter.
Quote: OnceDearOh Mike, forever the optimist.
Not a cat in hells chance that will persuade. Not that person. No way. Care to take a bet?
Can I have 10 to 1 odds that it will make Alan see the light? If so, I'll bet $1.
Quote: WizardCan I have 10 to 1 odds that it will make Alan see the light? If so, I'll bet $1.
Answer by pm
Quote: AyecarumbaI'm putting it forward as a possible explanation for the dispute. The methodology fits the description of the original (and by original, I mean the original Two Dice Puzzle thread) sequence of events. Assuming that the question is only interested in one die showing a two is a leap, because there is no indication the dice would continue to be shaken and checked until at least one die was a two. The more likely scenario is that the dice are shaken, peeked at, and the declaration made, no matter what the called out die was. It just happened in this paticular instance, that the called out die was a two.
That could have been the reasoning for the original misunderstanding, but later Alan agreed with the clarified problem that the peeker is specifically looking for a two. If in real life, I came upon someone, they rolled two dice in a cup, looked under it, and said "I see a two" I would assume they were randomly telling me the value of a die.
I still have no clue what this guy is talking about 90% of the time. Also PLEASE END THE MADNESS !!!!!! I AM YELLING!!!!!!!!!!!!!!!!!!!!!!!!!!!!Quote: KerkebetMathematicians inherently assume that the mind just exists, but let it to the physicists to try to figure it all out afterward. How schizophrenic is that? Perhaps, the sort of "argument" which has unfolded. (Maybe the answer to that show about the math existing in the chemicals, etc, depends on the person's scholastic background?)
Part of this revolves around "at least one die of two", which amounts to "one or more dies". After the initial truthful "peek", for lack of a better word given the continued not-well defined arguments here, "one or more dies" can't show a 2 when only one of the dies shows a 2. No "or more dies" possibility then. Strictly speaking.
https://xkcd.com/435/
As you can see, these dice show the 36 possible ordered pairs of two dice. Rather than give a math lesson, let me just ask?
1. How many of them have at least one two?
2. How many have two twos?
3. What is the ratio of the answer from #2 to #1?
http://en.m.wikipedia.org/wiki/Monty_Hall_problem
Quote: mipletI think Alan's missing the fact that once you roll two dice at the same time until there is at least one 2, there are only 11 possible outcomes. Just one of those outcomes is a pair of 2s. If he does understand this, I don't know how he is getting 1 in 6.
I'm sure deep down Alan knows he is wrong. Otherwise, he would be betting thousands of dollars as opposed to lunch money. However, I truly believe he doesn't understand the flaw in HIS argument. I tried to explain it in my video. He is simply can't get past a die having six sides.
I'm not a "math guy" and I can clearly see this.
Quote: AyecarumbaWhat I am proposing is that the peeker would call out whatever was on the observed die.
You're having a problem, son. The scenario does not have "the observed die." It has "an observed pair of dice." Never is the peeker restricted to observing only one die. He's obligated to see both, and if any deuce manifests, he's further obligated to say "at least one of the dice is a two."
Quote: WizardI was hoping I'd get more comments on my video but here is the thrust of my "argument using only dice."
As you can see, these dice show the 36 possible ordered pairs of two dice. Rather than give a math lesson, let me just ask?
1. How many of them have at least one two?
2. How many have two twos?
3. What is the ratio of the answer from #2 to #1?
It could help, visually, if you had little dixie cups (kid kool-aid cups) concealing all eleven deuce-combinations. And a big, bold 2 drawn on the upside-down cup bottoms. And then rhetorically ask, which of these under-the-cup results could have happened; and which ones could not have happened. All the while, your meandering hand is randomly uncovering a cup, wandering to another, to reveal another Devastating Deuce.
Quote: WizardI was hoping I'd get more comments on my video
You have, from Alan, on his own forum. I'm not linking to it, because I begrudge him the free traffic. It's not looking good :o)
Well. For you, that is. It is looking good for me $:o)
Quote: Alan Mendelson;27963I said when you see a 2 playing craps, the chance of seeing 2/2 is a 1/36 event.
Alan's getting support from a world renowned linguist/mathematician. What could possibly be wrong with that?Quote: IbeatyouracesWhat's funnier is Rob Singer getting out of jail over there for the 6th, or is it the 11th :-), time over there! That's desperation folks.
