If I see a 6-2 then "at least one of the dice are/is a 2" and "one or more dice are a 2" are valid.Quote:
Originally Posted by OneHitWonder
Are you seriously arguing against this? Omg...
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If I see a 6-2 then "at least one of the dice are/is a 2" and "one or more dice are a 2" are valid.Quote:
Originally Posted by OneHitWonder
Are you seriously arguing against this? Omg...
We've gone over this about a hundred times:
If the first die is a 6, and the second die is a 2, the solution to the problem is still 1/6. You count the die showing the 2 as the "2" which means the other die which shows a 6 has a 1/6 chance of showing a 2.Quote:
Originally Posted by RS__
Yeah -- that die with a 6 is showing a 6, but the chance of it showing a 2 is still 1/6.
The bottom line is it doesn't matter which of the two dice is showing a 2 for the question to be active. The "peeker" only had to see one die showing a 2. It didn't matter which die. And it wouldn't matter if both dice showed a 2, because the answer for the "second die" would still be 1/6.
Does this sound familiar? It was written here and on the WOV forum about a hundred times.
lololololol.Quote:
Originally Posted by alan mendelson
Arc, let me clue you in: a die only has six sides. And it's true, I just can't get beyond that thought.
“In the long run we are all dead.” - John Maynard Keynes ("Keynes wrote this in one of his earlier works, The Tract on Monetary Reform, in 1923. It should be clear that he is not arguing that we should recklessly enjoy the present and let the future go hang. He is exasperated with the view of mainstream economists that the economy is an equilibrium system which will eventually return to a point of balance, so long as the government doesn’t interfere and if we are only willing to wait. He later challenged that view in his most important work The General Theory of Employment, Interest and Money (1935). arguing that the economy can slip into a long term underemployment equilibrium from which only government policy can rescue it.")Quote:
Originally Posted by Rob.Singer
That everything must even out - and electric charge certainly appears to, though it's not at all understood yet how this might apply to anti-charge - and hence is still quite ambiguous in terms of a theory of everything to date, doesn't mean that what we do know is ambiguous per se. As much as the newer models of the universe take hold, what has been properly established shall remain somewhere in "the tapestry".
We know a lot about the "two-envelope paradox". No definitive explanation yet. There are two legs to that paradox as well, which together give rise to the other intuitive way to calculate the expected value. Obviously, the other calculation is wrong. And, the more you learn about formal logic, the more perplexing the rationale behind this seems.
In which universe, or part of this one, might the other calculation for the "two-envelope paradox" be the correct answer? As far as the "dice problem" goes, its two legs can as easily come together in the 1/11 chance answer in the same way as "two-envelope paradox's". You'll have to think about the way the "two-envelope paradox" comes together to form the other, different calculation of expected value. This is the real paradoxical part of it, where the other calculation begins to make sense. Note that the "two-envelope paradox" involves multiplication; and this one, addition.
Did you see the die I posted over at the Wizard's?Quote:
Originally Posted by Alan Mendelson
http://upload.wikimedia.org/wikipedi...0px-8-cell.gif
You can sum all of this up with a single four-dimensional die! Really.
Let's try an easy one. There is a girl standing on the corner. What are the odds she is a girl?----1-1
Now--let's place another child next to her with a bag over its body. What are the odds this one is a girl?--1 of 2
Similarly--there is one die showing a 2. If I place another concealed die next to it, what are the odds the concealed die is also a 2?---1 of 6
Now a harder one. If arci is in his dodge dart heading towards the Indian casino near Shakopee at 35 MPH, and Singer has the RV on cruise control at 65 MPH coming from parts unknown, what are the odds that any face on a die are 1 of 6?
no that is where the 1/11 comes in if you throw the first die and its anything but 2 say a 4 so now you have 0 chance of winning the bet and 1/6 losing. which the first die was 1/6 of throwing a 2 but the 4 came out so that accounts for the 5/6 odds now the next die thrown is a free roll and the wizard would have a 1/6 chance of winning the $ and 0 chance of losing $ so throwing the die 1 at a time is 1/11. but thats only if the die are thrown one at a time or the peeker doesn't look at both die just looks at one at a time. if the die are thrown and the 1 of the die is 2 then its 1/6 the other is 2. by saying one of the die have to be 2 or its a no bet takes the odds to 1/6 if they are thrown the same time.Quote:
Originally Posted by RS__
champ724 welcome to the forum and thanks for posting.
Regarding what regnis said, odds are "Pretty Damn Good."
And I figure regnis used a Dodge Dart because they're famous for their Slant 6 engines. Very dependable. No matter how you rotate the engine, it always has six cylinders.
How many sides do you have with a pair of dice? It appears you can't count that high.Quote:
Originally Posted by Alan Mendelson
All kidding aside, if one die is a 2, why are we still considering the 5 other options on that die? It is a 2---done, over, kaput.
I am holding up one or more fingers on each hand. Alan peaks and tells singer that on one hand I am holding up 2 fingers. What are the odds that I am holding up 2 fingers on the other hand?
Hint--I only have five fingers on each hand unlike a former cub pitcher that had 6 (Alfonseca).
Because the consideration of possible outcomes begins with the throw of pair of dice. That's how the probability of the event in question is computed in both classical and frequintist's definition of probability .Quote:
Originally Posted by regnis
And not coming after the throw and trying to figure out what will be the possible outcomes for one particular die.
regnis, I tried giving an example to Alan so he could understand what conditional probability is. He ignored the examples. I'm hoping you won't ignore my following example:
Let's say I have 10 marbles on the table. 5 are red and 5 are blue. I tell you to pick a random marble (you can't see them). I ask you, what is the probability you picked a blue marble?
Hopefully, you would say 50%. Do you agree?
Now, let's say we do the exact same exercise. But this time, I remove a few of the blue marbles. You still can't see the marbles. I tell you to pick a random marble and I ask what's the probability/chance that you picked a blue marble? Would you still say you have a 50% chance to have picked a blue marble?
Or rather, let's make it seem dead obvious: I remove all the blue marbles except for 1. Then you pick one. Is the probability you picked a blue marble still 50%?
This is the same thing with the dice problem.
I roll two dice. Each die has a 1/6 chance of being a 2 (like a marble has a 1/2 chance of being blue). But then we essentially remove the die that is showing a 2 (in the same way that I removed a blue marble). For the remaining die, the chance of it showing a 2 has decreased from 1/6 (just like of the remaining marbles, you have less than a 50% chance of picking a blue marble).
Don't you understand that this "formula" is wrong for answering this question? No, you don't.Quote:
Originally Posted by kewl
Lets try this... I find one die on the floor with a two showing and put it in my pocket and look for the other one.What are the odds a two is showing when I find it? By golly I found it. It is a two also. I give the problem to my 6 year old he says 1 in 6.
Unfortunately, it is you who is unable to comprehend simple stuff as that as to see it is the number of all possible outcomes for two dice versus the one possible outcome we are looking for when the question is involving two dice.Quote:
Originally Posted by Alan Mendelson
Instead, you are misinterpreting/over/under/thinking that if you were told one of the dice is a two, you get to forget about the initial state of the question and are now dealing with a single die.
The way I see it, you think you are smarter than the majority and are seeing the question for "what it is" - one die is a two, how many faces could be there for the other?
What you fail to take in account in your arrogant state of mind is that the people able to perceive the answer is 1/11 just might be one level up the ladder.
Been there where you are, saw it - no, its not it. Both semantics and mathematics do not agree with 1/6. One level up - there you go, everything lines up. Its 1/11.
Why do you think the bet simulation (the same bet you agreed represents the question in full) confirms it is 1/11? Got an answer? No, you don't. Or maybe you do, but you just need to continue the argument for whatever personal reasons you have.
I'm not sure it deserves the label "paradox." If you read it front to back, with the emphasis on the opening presentation of a single event in present tense, then one arrives at one conclusion. If you read it back to front (why you would emphasize the end is a question in and of itself), then you can draw a second conclusion that it's about multiple events, and it's not referring to a single event that already happened.
The latter is not how one usually reads English.
It's a piece of writing, not an equation. Pieces of writing conventionally, in English (not Hebrew), are read as a left top to end right, front to end narrative. So when you read about a single event in present tense, the convention is to accept the opening of the piece of writing as is and not reformulate its meaning after reading further along in the piece of writing. That's why Alan is telling you that you can't read.
You are ignoring basic conventions of English writing to arrive at your conclusion that this is necessarily about multiple events.
Let me add another angle to this. When you open a fictional narrative in present tense, and the reader reads it, and then you later continue the narrative, the convention is that what was presented earlier is now in the past, that is, it happened. It's not ongoing, and we didn't miss a bevy of similar events that may or may not have happened. Why didn't we miss them? Because the author didn't present them.
If the answer is meant to be 1/11, this is a really bad piece of writing, and any decent editor would have rewritten it. If the answer is meant to be 1/6 (which I doubt), it's not much better.
Indeed, let's try it. But this time let's just ask the actual question for a change and not the twisted variant above. I'm sure your 6 year old will get it right on first attempt.Quote:
Originally Posted by spojoey
Here it goes:
You find a pair of dice on the floor. Neither is a two. You keep walking. In fact someone has tossed out pairs of dice everywhere on the floor with all kind of random outcomes.
Now, what are the odds you'll find a 2-x pair?
And how many times out of all occasions when you find that 2-x pair will you find a 2-2 pair?
Alright, I'm not at all convinced it is a matter of semantics, but for the sake of it,let's for a moment forget about the original question, shall we?Quote:
Originally Posted by redietz
Here is another question:
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"
How do you read this question and what answer does it invoke for the 1/6 supporters?
Plain and simple evading all ambiguities there may be with the OQ.
Go.
LOL -- "matter of semantics." Lord almighty, now we have a dose of wizardry. You do realize "a matter of semantics" is like screwing up the order of signs in an equation and expecting to argue for the same values?Quote:
Originally Posted by kewl
This is what I mean by language-blind. A bit obnoxious, a bit arrogant, and somehow correct even though you have no idea about the overall context of a piece of writing. Unbelievable.
Why all the ranting? A simple question, without semantics problems, was asked - you have an answer or not?Quote:
Originally Posted by redietz
To be even more clearer - "matter of semantics." - I was referring to an eventuality that semantics might be the cause of misunderstanding even though I strongly doubt that and believe the question as written is JUST FINE especially after the ambiguity whether the outcome is announced ONLY after at least one two is present was cleared like years ago. I believe it is a mix of inability to read/think and ignorance in probability. Do I have to spoon feed every detail on top on my poor English?
Don't you people have at least a grain of shrewdness in you and are totally incapable/unwilling/trolling lovers to perceive a meaning unless it is spelled out letter by letter? Where are you in your life with such altitude? Are you even capable of any social life? Cause I don't believe you are if you need a Socrates like form in order to be able to perceive a bar chat like content.
A question was asked, here it is again:
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"
State an answer or keep trolling about the "meaning of a letter", your choice.
Kewl if I used your question as a substitute for the original question I would say 1/6. But your new question is a textbook question that leads to the 1/11 answer.
The textbook question lacks the information given in the original question which is that the peeker sees at least one 2. That defines the original question as being a one time event.
You cannot apply your textbook methodology to this one time event in the original question.
Sorry, I'm smarter than you.
I see this as completely different. Each marble has 1 face. If you remove 1 blue marble, there are still 9 separate marbles to choose from, and the odds of picking a blue one are now 4 of nine.Quote:
Originally Posted by RS__
With the dice, although you have the same 10 options (as 5 blue and 5 red on your marbles) originally. But when you remove one die, you have physically removed 5 additional numbers that were on that die. You have one die left, and the odds of any number on that one die are 1 of 6.
I am actually glad that you used the marble example because it amplifies the clear distinction. The removal of a marble, removes only one option. The removal of a die removes all 6 options on that die.
I disagree that the consideration of possible outcomes begins with the throw of a pair of dice, because we have the intervening act of the peeker and then the additional data that one is a 2. Since we now know that one is a 2, we are only concerned about the other which has a 1 of 6 chance of being a 2 (at least in an honest game).Quote:
Originally Posted by kewl
What do you mean, what question? The one you agree is 1/11 later in your post or a different(?) one?Quote:
Originally Posted by Alan Mendelson
Even though I'm one of the least vanity susceptible persons, I somehow have a hard time believing that. Especially after your show of ignorance at a 3-rd grade level here, combined with almost mentally challenged-like stubbornness...Quote:
Originally Posted by Alan Mendelson
But I will be the first person to admit someone is smarter than me as long as he/she really demonstrates abilities and level of thinking which strikes me as a first-time-see on my side.
So which other question?
And then I'll ask you a second question.
Mmmm... what does the peeker do other than informing us we are now witnessing one of 11/36 possible event?Quote:
Originally Posted by regnis
How is it different from judging the odds for an event which happens 1 time of every 11 times we are witnessing at least one 2?
And how come the consideration does not begin with the throw of pair of dice? What does the act of peeking after the roll is made do, so that the odds change midstream?
When you draw 2 random cards from a deck doesn't the odds depend on the full number of cards in the deck? And then you have an outcome - the cards are drawn.
Giving you the incomplete info about at least one of the cards somehow changes the odds for the initial drawing?
The way to get to 2-2 is dependable to the ways to get to 2-x. They are not undependable events.
See here how and why computing probability of dependable events needs to account of all ways to arrive to either one of the events:
http://www.shodor.org/interactivate/...nditionalProb/
The experiment which is described in the original question begins with throw of two dice. It does not shift to a single die experiment out of the blue.
And it does not matter whether the experiment is conducted one time or one quadrillion times not one bit. The odds are the same. One time, second time, all times.
Okay kewl, I will spell it out for you. This was your new, or second, question:
And this was the original (or first) question, as posted on the Wizard's forum:Quote:
Originally Posted by kewl
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?
The answer to your second question is the textbook question to which the answer is 1/11.
The first or original question involves a one time event: dice in a cup, slammed, peeker sees at least one 2. And that answer is 1/6.
And to be honest, even the "textbook question" is not written well. Unfortunately, the "textbook question" has been used so many times and changed (using different values for the dice) using its same sentence structure, that it is probably viewed in biblical proportions as being correct. Frankly, it needs to be rewritten.
It is not a new question, I posted it at least 5 or 6 times till now, and only now you decide to address it.Quote:
Originally Posted by Alan Mendelson
And the first question is not mine(neither is the second one but nevermind) or maybe it is a way of expressing in English I got confused by.
IT DOES NOT MATTER whether the dice are thrown once or a million times. The odds are one and the same. Always 1/11. Try to think it through.
This is your error. Try to think it through.Quote:
Originally Posted by kewl
Nope its your error. Please try to think it through.Quote:
Originally Posted by Alan Mendelson
Or are you saying that when you roll a pair of dice or toss a coin or spin a roulette wheel the odds are different depending on whether you do it one time only vrsus many times?
How do you think the simulation got to 1/11? Because one time is 1/6 but many times is 1/11 lol?
Here is another question on conditional probability involving dice:
http://www.algebra.com/algebra/homew...on.434512.html
Hopefully, if anyone even could be bothered to read and actually think, may find the analogy in the method which is used and open their eyes.
The additional facts changed the odds. If the question was there is one die slammed down and the peeker said it was a 2, would you now say the odds that it is a 2 is 1 of 6??Quote:
Originally Posted by kewl
No--it is a 2. 100%. No ifs ands or buts. The second die does not change anything. We have more facts than you are willing to acknowledge. The question is not what were the odds that that die would be a 2. The question relates to the 2 dice that were thrown this one time. We know what one of them is. We have only the other to consider and we all can agree on the odds for that one die.
The simulation involves 11 different combinations of dice in which at least 1 die is showing a two. That means you have 11 pair of two dice.
When you roll any single pair of dice, and one is known as a 2 the odds of the second die also being a 2 is 1/6.
The original question with the cup involved a single pair of dice. It did not involve 11 pair of dice.
Now, you admit the answer to the question without peeker, cup etc is 1/11. Good. We have improvement.Quote:
Originally Posted by Alan Mendelson
What in the writings of the OQ makes you think it is different than this new question? What?
It happens one time? So? What? Do you believe in black magic or something? Somehow one time is different? How? Why when you toss a coin many times you get closer and closer to 50%? You know it before hand it is 1/2 right?
Now, you have 11/36 ways to get to hear the peeker "It is at least one 2!". And one of those eleven to hear "It is a 2-2!". So? How its different for you when you read the OQ?
Where do you get that 1/6 from? Didn't you already outgrow this childish "But it is one die left to choose"? Can't you see that at the get go you are dealing with probabilities of two dice?
Read the links on conditional probability and share your thoughts. The second one is very close to our question and the methodology might strike you as very familiar.
LOL.Quote:
Originally Posted by Alan Mendelson
Okay try and read some books, maybe it will help.
That's the problem. You read books and misapplied what you learned. Now take a course in reasoning and another in reading comprehension.
Two dice make eleven combos representing 2-x, are they not?Quote:
Originally Posted by regnis
So the odds for us to see 2-x are 11/36 are they not?
Now this above IS NOT IRRELEVANT to the odds of us seeing 2-2.
You 1/6ers are all claiming it is up to one die only, once we know there is a deuce among the two dice. It is not.
You have 36 cards (that's not for Alan - he disregards anything not involving dice for whatever reason) with every single combo of pair of dice on them.
Draw a card. Would you say now that the ways to draw one of those eleven 2-x are not influencing the odds of drawing the 2-2?
Kewl,
If you present a single event in present tense and then continue the narrative beyond it, usually the event is over, finished, not one of many events. Now the 1/11ers are arguing that using the word "probability" at the end in essence says that the author is expressly telling you that it was one of many events. The word "probability," however, does not have as its primary definition the mathematical definition they are trying to assign. So for their argument to make sense, they need both a non-conventional narrative structure and a secondary definition of "probability." It's possible, but convoluted, for a reader to interpret the piece of writing this way. It's probably more natural, common, call it what you will, for a reader to interpret it the way the "non-geniuses" interpret it. An editor would, after asking the author what the hell he really meant, completely rewrite the original question because of this.
No, I'm not misapplying anything. The bet you agreed represents the question straight as it was written and clarified, proves me right.Quote:
Originally Posted by Alan Mendelson
And indeed there is nothing much to say here. The odds are indeed 1/11. In the real, physical world they are. On a single throw. They are 1/11. Many times - still 1/11. In fact that's the ultimate proof. Many a times experimenting.
It does not matter if we are talking single, one time throw or not. The odds are still the same.Quote:
Originally Posted by redietz
Why? Why do you persist in considering the various faces on both dice? When "at least one die" is identified as showing a 2, that die's other faces cannot be considered.Quote:
Originally Posted by kewl
And by the way, this is the same problem with how the "textbook version" of the question is written. It is also poorly written, and should be changed. But, I understand how the "textbook version" will have the answer of 1/11 because it was written in the context of a probability lesson. Unfortunately, how it was written as part of a probability question is not good enough for a normal conversation. I realize you don't understand what I am saying or what redietz is saying or what regnis is saying about how sentences are written but we deal with "words" and you parrot math lessons. This is why I also wrote on the Wizard's forum that I wouldn't hire anyone for my company who says the answer is 1/11 -- I don't need that kind of overthinking of a basic problem.
Because they are related to everything, why can't you get it?Quote:
Originally Posted by Alan Mendelson
What's worse parrot or bigot not capable to comprehend neither math nor words?Quote:
Originally Posted by Alan Mendelson
Here is your one time throw example:
I, Bob, John and hundreds of 1/6ers would like to empirically proof the odds on question.
We decide we will only do the experiment once. But all of us will do it.
We throw dice. I got 6-4. Bummer. No question to ask, no answer, case closed.
Bob and 300 1/6ers got 2-x. guess how many of them got 2-2 and what the ratio tends to?
What does this tells us? Yeah, it tells us the real, physical world actually says the odds on getting 2-2 given at least one 2 are indeed 1/11. Damn...
No. Technically and simply, getting 2-2 given rolls of the set of at least one 2 is indeed a 1/11 chance. And then, if and only if a 2 is the explicit objective from the start. However, 1/6 is the answer which makes up the 1/11, no matter what. The 1/11 is a compound answer, ie, illusory in some way.Quote:
Originally Posted by kewl
A specific roll which has or involves "at least one 2" doesn't include a non-2 with 2, or 2-2.
Damn! Better tell the Wizard. Lol.
A specific roll isn't a set of rolls, or the union of types of rolls. "At least one 2" refers to a set of rolls.
Well, that's one of the conditions as clarified like million times after the initial few posts on the old discussion. Have you missed it?Quote:
Originally Posted by OneHitWonder
What's this nonsense?Quote:
Originally Posted by OneHitWonder
Wow, you're loosing it. Outright nonsense like this doesn't even deserve a reply.Quote:
Originally Posted by OneHitWonder
"at least one 2" doesn't include a non-2 with 2, or 2-2.???????
I'll ignore all future nonsense like this on the spot.
Doesn't matter if the throw in question is a "one time event". The question asks for "probability". Probability is calculated by considering all possible throws. What you are showing is you are not very smart..Quote:
Originally Posted by Alan Mendelson
Good grief .... even when you answer the question with 1-6 you are giving a mathematical answer. That means you interpreted the question using the mathematical definition of probability. There is no definition of probability where 1-6 is the correct answer.Quote:
Originally Posted by redietz
Look, everyone knows how you are failing to understand the question. Never is it stated that one particular die is a 2 and you can only look at the other die. It is stated that at least one die is a 2. It doesn't specify which one. Therefore, all combinations where either die could be the 2 must be considered to get the correct answer. Your continued denial of the obvious meaning of the question only makes it obvious you have become emotionally attached to the wrong answer.Quote:
Originally Posted by Alan Mendelson
There's nothing emotional in what I am about to say: no one is presenting any new information here. On both sides.
Good grief yourself, Charlie Brown. If you find a quote where I said it was either 1/11 or 1/6, please let me know. I repeat my earlier example, "Arci winks at the cocktail waitress, slams down his drink, then peers down her blouse. What is the probability a sexfest ensues?" In your mind, is that a "math" question? And how is that clearly different from the original question in narration and what it is asking?Quote:
Originally Posted by arcimede$
You say people need to learn to do math. Yeah, and people need to understand the difference between a sentence and an equation. What's most disturbing is that some people fail to understand that if a large chunk or majority of readers read something, and come to Answer A when the answer is B, then the problem lies with the author, not the readers. Arci, I hope you've tracked down the author and posted 50 times regarding what a lousy job he did. Just because you "got it" doesn't make it correct. You are not the arbiter of how people interpret pieces of writing and how they should interpret pieces of writing.
But it doesn't matter how is this question interpreted regarding one time action or multiple experiments. The odds hold true every single time and it is not 1/6, so why are people blaming the writing?Quote:
Originally Posted by redietz
One ambiguity there is and it is regarding the peeker intention to announce whatever value is there or specifically a deuce and it was cleared for the purpose of this discussion long time ago.
Not quite sure about that.Quote:
Originally Posted by Alan Mendelson
It seems it is new information for you that whether you do a one time experiment or multiple times doesn't matter and the odds are still the same.
It is quite new info for you again, that the number of ways to get 2-x actually influences the odds on getting 2-2. only if you could educate yourself enough to see why. I'm afraid I'm running out of examples and ways to describe the why in simple logical terms and you seem to be unwilling to learn anyhow.
A specific roll which is (an element) of a defined set of rolls is different from a roll which is a defined set of rolls. A major difference, but one which may seem an unobvious nuance to someone with virtually no algebraic experience.Quote:
Originally Posted by kewl
The problem as written here doesn't address a specific roll which "is (an element) of" a set of rolls. It addresses the specific roll as, "At least one of the dice is a 2." It does not mention the words "is (an element) of" the "at least one of the dice is a 2" set of rolls of the 1/11 chance answer.
The roll in the problem by which "at least one of the dice is a 2" means a roll by which one or two of the dice is a 2. This definitely doesn't include a non-2 with a 2, or 2-2. Eg, an end roll of 6-2 isn't included in one or two of the dice is a 2, because 6-2 has no possibility of two of the dice is a 2. However, a specific roll of 2-X, where X has yet to be determined, has the possibility of one or two of the dice is a 2.
The set of rolls defined by "at least one 2" definitely does include {a non-2 with 2} or {2-2} but is different from a specific roll defined by "at least one 2". The latter doesn't include a non-2 with 2, or 2-2. It's the set of rolls of 2-X, where the X-die has yet to be looked at: {2-(1), 2-(2), 2-(3), 2-(4), 2-(5), and 2-(6)}. An end roll of this set is either 2-1, 2-2, 2-3, 2-4, 2-5, or 2-6.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?
You apply the nonsense word as "a crutch" to not make the effort to understand what I and others have written, and then to not write an intelligent reply. Even a nonsense idea deserves an intelligent reply, as I continue to reply to you.Quote:
Originally Posted by kewl
In my opinion, this is, foremost, a math question. Not a gambling forum, or court of law, exercise. To this end, I will post it up at one of the math forums, along with a summary of my own arguments of course, when I have some more free time.
It doesn't surprise me that neither the Wizard nor Alan has chosen such a form of "binding arbitration". They aren't the math and formal logic experts. Aside from redietz, and perhaps Singer, I get the feeling that a lot of this went over everyone's head. Same at the Wizard's.
Not by the original problem. That's the problem the Wizard replied to immediately, and the one at hand here. Calculations and simulations are a distant second.Quote:
Originally Posted by kewl
Let me ask the question differently: if instead of using the word probability and we used the phrase "what are the chances that the other die is also a 2" would the 1/11 answer still hold?
Probability is the math word for chance. Like faith is the religious word for trust.
It seems that probability allows for using the other five faces on the die showing a 2 but chance might isolate the second die.
I'll answer, but please Alan, try to comprehend the answer FULLY.Quote:
Originally Posted by Alan Mendelson
We are NOT EVER asked what the chances or PROBABILITY is of the other die being a two. We ARE asked IN THE ABSENCE OF KNOWING WHICH DIE IS A TWO, what the probability is that we have a pair of twos. That is different. WAYYYYY different.
You need to get your head around this rather odd set of facts.
For each die, the odds of that die being a deuce would be 1/6. You can use the words odds, chance and probability without too much disagreement.
You, not unreasonably, assert that the odds of 'the other die being a two' is 1/6
Are you happy to agree with me on that? It's not really what's being argued. Really. it's not.
BUT. GREAT BIG BUT....
That is not the question.
We are NOT asked what the probability of one other die being a deuce. You adamantly believe that's what is asked. But the question is absolutely explicit. We are asked what the probability is of A PAIR of Deuces! YOU can NOT extract one die, the die that is a deuce, because only the peeker knows which die that is, and he is not telling.
Your assertion would only follow from the original question if we could identify which die is already a deuce.
We know only one thing and that 'AT LEAST one of the dice is a deuce'.
Until the cup is lifted, we cannot ELIMINATE any die. We cannot eliminate any face on any die. All we can eliminate is outcomes of dice where there is no deuce. All we can eliminate is...
1-1
1-3
1-4
1-5
1-6
3-1
3-3
3-4
3-5
3-6
4-1
4-3
4-4
4-5
4-6
5-1
5-3
5-4
5-5
5-6
6-1
6-3
6-4
6-5
6-6
UNTIL we lift the cup, none of the above outcomes could sit under the cup, but there are still those pesky 11 other ways that the dice under the cup COULD have come to rest:-
1-2
2-1
3-2
2-3
4-2
2-4
5-2
2-5
6-2
2-6
2-2
UNTIL the cup is lifted. That is the sum total of all that we know. Having a red die and a blue die, or a left die and a right die or a big die and a small die can all help us to visualise that.
If we have a big die and a little die, you absolutely cannot rule out which one is a deuce. The peeker could put it in his pocket, but until you see it, you are none the wiser.
Once you see the cup lifted, probabilities are no longer at issue: bets are off. But until then..... 1 of 11 outcomes is hiding under that cup and you and I don't know which it is. THAT is the very nature of 'chance' or 'probability' or 'odds'. the reason no-one shows you this in a video is because once the cup is lifted, we can only see one fixed outcome of both dice. You, yourself rotated the 'other' die in some effort to show the six ways it could have landed. But you would have needed to show how both dice could have landed and for that you would have to acknowledge that until the cup is lifted, you don't know which one to rotate. Your video would have had to show you rotating both dice.
By 'setting' one specific die as a deuce, you would be imposing a condition that is not imposed.
I don't know much about craps, but would the casino let you set a die and roll just the other? Well this is why, because it is breaking the rules and the payout table would be shot to hell.
Now, the more serious question. Why has no-one taken up wagers offered by Wizard, myself and others.
Alan seems to assert that the bet offered DOES model the question appropriately.
Alan does agree that our peeker saw both dice and would only declare deuces or no deuces. It's agreed that that was an original omission from the question, but this thread is very specific.
With agreement on the peeking and declaring rules, we have a pretty darned explicit question. It has only one correct answer, unlike the very original question in a distant memory far far away.
Why is not one person taking the offered wager. It's either going to favour the player or the offerer and after a few hundred rolls, variance would become insignificant. If you want more confidence, stake $5 per resolved bet over 100,000 bets. The money is there for the taking. Maybe some 'maths guy' could tell you what you could expect to win or lose and even within what reasonable margin of error.
You could even use my spreadsheet, It's free and it even tells you what I estimate that you would lose.
Even Rob Singer saidQuote:
But, if as I suspect, we don't get to see which die has the original 2 until after the bet, then 9-1 is a punk bet because two 2's, after knowing half the outcome at that point, is an 11-1 event in this case.
yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!yes!Quote:
Originally Posted by alan mendelson
NOTHING isolates the second Die and NOTHING isolates the first die.Quote:
Originally Posted by Alan Mendelson
They are a PAIR of Dice and the one guy who can tell them apart, and who can tell you which one is a deuce is NOT telling. THAT is the nature of probability, betting on, or taking chances on what you don't know.
As long as either die can be the 2 then the answer is yes. The critical part of the question is whether you lock one die into being the 2. As in the case of the red and blue die ... If you ask what are the chances the blue die is a 2 when the red die is a 2 then the answer is 1/6.Quote:
Originally Posted by Alan Mendelson
OnceDear wrote above: "We are asked what the probability is of A PAIR of Deuces!" That answer is 1/36.
Unfortunately I have to deal with real life and not some imaginary fantasy about probability. In real life if you tell me that at least one of two dice is a 2, in my mind I define one of two dice being a 2. Yes, it is possible that both dice are a 2, but one die is definitely a 2.
Also, my real life experience tells me that given two dice and with at least one die being a 2, that the chance for the other die in the problem can only be 1/6.
Now, if this were a textbook exercise and you set up the question as a textbook exercise explaining that there are 36 dice combinations using two dice and that there are 11 different combinations with at least one die showing, then I could go for your theoretical probability gymnastics.
But I deal with reality and physical dice which is why I created my video explanation that shows the answer is 1/6.
The Wizard also did a video explanation and in it broke all the rules of reality -- because he did not accept a die showing a 2 as a 2 and flipped it around to fit the 1/11 answer.
So here is how I would like to sum it up:
Those of you who say the answer is 1/11 get a gold star for answering a theoretical probability question that was not asked but had it been asked you would have the right answer.
For the rest of us who say the answer is 1/6 we also get a gold star for keeping our minds in this real, physical world and reasoning what is real and not imaginary.
In the real physical world where odds of 9 to 1 are available on unlimited wagers where probability is 1/6 there would be a stampede of takers.
Roll up. Roll up You can wear your gold star on your dunces cap if you stay away or if you show up.
Not me.Quote:
Originally Posted by Alan Mendelson
I would rather know the single correct answer as summed up by the experts in this field. The one that at least doesn't involve the words nonsense and dunce cap.
No one ever got anywhere by giving up.
Unfortunately I'm afraid that the experts in this field will fail to read English and will not correctly reason the question. All of the 1/11ers have failed to read and understand the question and have simply applied their "probability textbook answer" to the question when in reality they shouldn't.Quote:
Originally Posted by OneHitWonder
I'm sorry but the "math guys" get a fail for understanding the actual question, but I am sure they learned their textbook probability lesson well because that's all they parrot.
I'm using "nonsense" as a short and to the point way to express my believe you are barking the wrong tree here with your desire to dismantle a simple logical math question into philosophical argument about the meaning of a phrase, detail twisting andQuote:
Originally Posted by OneHitWonder
challenging the very structure of math-semantic-logic up to the point where every effort
to make sense of anything becomes futile.
I can do that as well. You tell me something simple, such as "Good day." and I can start questioning, arguing and wooden philosophizing the "real" meaning of the phrase to the end of all days. But I choose not to, because to me, such exercise is an utter waste of time and demonstrates perfectly how wooden philosophizing on simple things can lead to total paralysis of the mind and breaks all possible ways of communication. That's all it accomplishes.
Why don't you bring the question to a math experts( or semantics experts or both) dedicated forum or blog and see how they interpret it and answer it? I don't think discussion as the one you seek belongs in a gambling forum.
I bet the math experts there will first clear one thing - the intention of the peeker, and then will give you the correct answer in identical manner as this answer:
"There are 36 combinations of two 6-sided dices, only 11 of them show at least a 2, from which one is 2-2. Hence the probability would be 1/11.
However, this would only hold if the player would say nothing when no 2 were spotted.
If (before peeking) your friend is determined to say "at least one of the dice is a X" where X is one of the numbers he may spots, the probability for a pair would be different.
If the friend sees any pair (1/6), probability of "X-X" where X is the named number, is one.
If the friend sees a non-pair (5/6), probability of "X-X" is zero whatever he choses for X.
Hence, like the Monty-Hall-Problem, if your friend must always say "At least one of the dice is a X" (while not lying), probability of a pair is 1/6."
I'm going to try to make this real easy AGAIN!!!
I roll one die. It is a 2.
What were the odds that it was going to be a 2----1 of 6.
What are the odds that it is a 2----100%.
It is all about understanding plain simple english.
This has been argued for months now and here it is in plain simple english. We (1-6ers)see the 2 and therefore only consider the odds on the remaining die. You (1-11ers)are viewing the probabilities of something that the plain english does not ask for.
Your post perfectly describes the 1-in-6ers' folly.Quote:
Originally Posted by regnis
Let me try this:
I have 4 cards: Ace and king of clubs (black). Ace and king of hearts (red). I deal you one black card and then one red card. At least one of the cards is an ace. What is the probability/chance both cards are an ace?
1/2
1/3
1/4
1/8
1/16
Other
PS: I am using the same "at least one..." requirement as in the dice problem, ie: The peeker doesn't say something about every roll or every hand, but only when a deuce (or in this case an Ace) is present.
Try to rationalize on the example with the 36 cards. It demonstrates in a very simple terms why the odds of 2-2 are to be considered only in conjunction with what the odds of 2-x were prior to the throw.Quote:
Originally Posted by regnis
We draw a random card and disregard all non 2 cards. In fact we are drawing one card out of eleven cards (that's the number of the cards that matter to us).
We know the outcome. Its 2-x. What are our odds it is 2-2?
Also, why are all 1/6-ers forgetting/not commenting (excluding Alan's ridiculous views) the fact that the bet from page 1, based word by word on the question is yielding 1/11?
Is it because they believe the question is supposed to be a one time event? What if so? This doesn't change the odds one bit.
Any thoughts on this?Quote:
Originally Posted by OnceDear
What does the real, physical world say about the odds?
What does it say when we experiment one time across many experimenters?
How come all of them agree it is 1/11?
Where does the 1/6 come from???
Oh, yeah, it is because Alan says so. What about any proof? Oh, yeah a die has six faces... Nevermind he says that in a totally irrelevant context - when the throw is made and he comes in to pick the other die and "conclude the real odds". A total brain mess...
No you don't. Only the peeker see's the dice. Only the peeker can eliminate one die, and that peeker could eliminate probability completely because he knows the full outcome.Quote:
Originally Posted by regnis
Unless YOU can see through the cup, you cannot eliminate the red die, or blue die or left die or right die or big die or small die. All dice faces are in play and unknown to you.
First re-read the wager offered in post 1 of this thread !!!
Take on the damned wager. Big stakes or little stakes or roll the dice yourself. This is not about English. Just damned well do it. Try two dice of different size or colour. After you've rolled the dice a dozen times you will realise that you were abjectly wrong to eliminate one die. Unless you are the kind of idiot who would just roll one die. That's why I said re-read post 1 !!!
If you want to use identical dice the outcome will still tend towards 1/11 it will just be a tiny bit harder to visualise why.
I don't eliminate a red die, a blue die, a left die, or a right die. I eliminate the one with a 2 on it. That leaves one die for which the odds are 1 of 6.
I clearly explained the difference between a die and cards or marbles. Once one die is eliminated, you have eliminated all the other options on that die.
The peeker said one is a 2. The other one has 6 faces. Can we please drop this nonsense.Quote:
Originally Posted by OnceDear
Please address my example of one die.Quote:
Originally Posted by kewl
Again--cards are different because removal of a card leaves the other options. Removal of a die removes the other options.
Regnis they cannot drop their nonsense because they were not taught to think in real terms and only theoretical terms. Wizard proved this in his video when he rotated the die with a two. They cannot deal with the question in a real, physical world.