The Wizard will bank this bet: 1/6 vs 1/11

[RS__]

No, there is no longer the "or more is a 2" part about the specific roll in question then. Here, the roll of 6-X.

You must ask for a roll which is an element of the set of rolls which contain the number 2. The only specific roll which can satisfy "at least one die is a 2" is when only the first die is seen, and it's a 2. This allows for the roll to satisfy "at least one die is a 2". The other die could still be another or the more 2.

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.

Are you seriously arguing against this? Omg...

[Alan Mendelson]

We've gone over this about a hundred times:

What happens when the first die lands on 6? Chance of getting a 2-2 is 0. But you can still get "at least one die is a 2". Or did you not think that far ahead?

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.

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.

[arcimede$]

... Which means the other die which shows a 6 has a 1/6 chance of showing a 2.

Yeah -- that die with a 6 is showing a 6, but the chance of it showing a 2 is still 1/6.

lololololol.

[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.

[OneHitWonder]

Zedd's dead.

“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.")

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.

[OneHitWonder]

Arc, let me clue you in: a die only has six sides. And it's true, I just can't get beyond that thought.

Did you see the die I posted over at the Wizard's?



You can sum all of this up with a single four-dimensional die! Really.

[regnis]

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

[regnis]

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?

[champ724]

What happens when the first die lands on 6? Chance of getting a 2-2 is 0. But you can still get "at least one die is a 2". Or did you not think that far ahead?

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.

[Alan Mendelson]

champ724 welcome to the forum and thanks for posting.

[redietz]

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.

[arcimede$]

Arc, let me clue you in: a die only has six sides. And it's true, I just can't get beyond that thought.

How many sides do you have with a pair of dice? It appears you can't count that high.

[regnis]

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).

[kewl]

All kidding aside, if one die is a 2, why are we still considering the 5 other options on that die?

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 .
And not coming after the throw and trying to figure out what will be the possible outcomes for one particular die.

[RS__]

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).

[Alan Mendelson]

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 .
And not coming after the throw and trying to figure out what will be the possible outcomes for one particular die.

Don't you understand that this "formula" is wrong for answering this question? No, you don't.

[spojoey]

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.

[kewl]

Don't you understand that this "formula" is wrong for answering this question? No, you don't.

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.

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.

[redietz]

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.

[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.

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.
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?