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

[kewl]

You are ignoring basic conventions of English writing to arrive at your conclusion that this is necessarily about multiple events.

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?
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.

[redietz]

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?
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?

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.

[kewl]

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?

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?

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.

[kewl]

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.

[Alan Mendelson]

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.

[regnis]

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).

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.

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.

[regnis]

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.

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).

[kewl]

Kewl if I used your question as a substitute for the original question I would say 1/6.

What do you mean, what question? The one you agree is 1/11 later in your post or a different(?) one?

Sorry, I'm smarter than you.

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...

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.

[kewl]

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).

Mmmm... what does the peeker do other than informing us we are now witnessing one of 11/36 possible event?
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?

[kewl]

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.

[Alan Mendelson]

Okay kewl, I will spell it out for you. This was your new, or second, 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?"

And this was the original (or first) question, as posted on the Wizard's forum:

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.

[kewl]

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:

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.
.

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.

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.

[Alan Mendelson]

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.

[kewl]

This is your error. Try to think it through.

Nope its your error. Please try to think it through.
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?

[kewl]

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.

[regnis]

Mmmm... what does the peeker do other than informing us we are now witnessing one of 11/36 possible event?
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 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??

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.

[Alan Mendelson]

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.

[kewl]

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.

Now, you admit the answer to the question without peeker, cup etc is 1/11. Good. We have improvement.

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.

[kewl]

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.

LOL.
Okay try and read some books, maybe it will help.

[Alan Mendelson]

That's the problem. You read books and misapplied what you learned. Now take a course in reasoning and another in reading comprehension.