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

[kewl]

....and all because they simply cannot read simple English, or more to the point, that it took someone not from their special little clicque to expose the question for what it was.

Yeah, that must be it.

They are asking the wrong question and are giving answers to it.
And " someone not from their special little clique" is not changing the question and is exposing the question for what it was.
Yes. That's it. You have an amazing abilities to read simple English. You nailed it.
The question is not what it seems. Its a trick. A deliberately vague wording with one purpose only - math geeks to try and play smart over other people.
But you exposed them. Bravo.

What was the question again?

[redietz]

Kewl,

That was halfway effective sarcasm, but you need an editor. Sarcasm loses its edge if you're a bit off on the vernacular. It reads like fashion writers reporting on sports or something. Next time, check with me and I'll clean it up for you.

[kewl]

Kewl,

That was halfway effective sarcasm, but you need an editor. Sarcasm loses its edge if you're a bit off on the vernacular. It reads like fashion writers reporting on sports or something. Next time, check with me and I'll clean it up for you.

I'm trying to learn, Sir, always truing to learn. If nothing else, at least reading and writing in this thread helps me catch a new phrase or two and accumulate impressions on how native speakers structure their thoughts in writings.

[jbjb]

KJ, don't engage these people. Remember, they pay your salary!

[redietz]

Good to know I have someone else on the payroll. Now, if "you people" give me your actual names, I'll be sure you have full health benefits and some vacation time. Rob Singer is in charge of your Christmas bonuses, so good luck with that.

P.S. See, Kewl, the sarcasm needs to flow a bit better. We can both work on it.

[OneHitWonder]

KJ, don't engage these people. Remember, they pay your salary!

I call it welfare. Welfare recipients are the ones who get out of the way in, "Lead, follow, or get out of the way."

Who's dumb enough to work - and still not accomplish anything - for welfare? Card counters.

[indignant99]

... Unfortunately, in the real world you cannot use both the vertical and the horizontal charts. That's good for theory, but in the real world with two physical dice choose either the vertical or the horizontal...

Proof please.
Show how to restrict to only horizontal (or equivalently, eliminate the vertical).
Show how to restrict to only vertical (or equivalently, eliminate the horizontal).

[indignant99]

As I said before, at a craps table when throwing two dice, then the 2-2 will appear 1/36 times...

Very nice, Senor Mendelson. Quite correct, Senor Mendelson. How about these? :

• the 2-1 will appear _______ times?
• the 2-3 will appear _______ times?
• the 2-4 will appear _______ times?
• the 2-5 will appear _______ times?
• the 2-6 will appear _______ times?

(Big, big, big, big hint: they're all the same number.)

[RS__]

Alan is basically arguing that a hard number (ie: 6-6) will show up with the same frequency as an easy number (ie: 6-5).

[regnis]

Alan is basically arguing that a hard number (ie: 6-6) will show up with the same frequency as an easy number (ie: 6-5).

The odds of any number, on the second die, are the same.

[redietz]

Again, and I know nobody listens to me, clean up your tenses and the fog begins to disperse. Regnis says, "The odds of any number, on the second die, are the same." If we say, "Were the same," we begin to define this, or at least imply this, as a one roll event that happened in the past.

The WoVers are arguing that the use of the word "probability" automatically shifts everything into a multiple event, in-the-future interpretation. However, and I am surprised nobody has mentioned this, the colloquial, common usage of the word "Probability" does not limit it to the strict mathematical definition. This is part of the problem. The WoVers fail to grasp that a common usage of the word would supersede (who knew "supercede" is a variant spelling?) the narrow mathematical definition.

That's why, in many respects, this can be considered a "math guy" versus a "non-math guy" problem. It is a trick question, in part, because the two camps are using different definitions of "probability," and neither definition is wrong.

[kewl]

The odds of any number, on the second die, are the same.

But we are talking about a throw of pair of dice here, and not only "the second die".

Try to assimilate that the whole process of determining the probability in question begins with a throw of pair of dice. The probabilities arise from here first.

Knowing that at least one die has landed on two does not eliminate any of the eleven possibilities. All those eleven outcomes where at least one die is a two are possible at that point. And will be possible the next throw and the next... Every time at least one deuce is announced we will be looking at one of those eleven possibilities. Hence 1/11.

[OneHitWonder]

The WoVers are arguing that the use of the word "probability" automatically shifts everything into a multiple event, in-the-future interpretation

It's more primitive than this. http://www.mathwords.com/i/inclusive_or.htm

To achieve the mathematical "inclusive or" for this problem, the Wizard must average out the chance that the SPECIFIC rolls of the dice will be of one 2 or two 2's. Ie, the average of the both's of such an or would be the and of the chart's SEPARATE row and column to which Alan correctly referred.

(I applied the https://sites.google.com/a/jeffcosch...-2-apostrophes system of grammar, as apparently does this editor.)

[OneHitWonder]

You guys aren't even close.

You are right about the "a throw of pair of dice" part though. But from there everything, including common sense, digressed to a lengthy ridiculous diatribe against Alan.

All in the Wizard way of course. Nobody holds more tenaciously to his own ridiculous belief that he is somehow a "Wizard of Vegas" than he and his followers. All talk and no action, over and over again.

[redietz]

Thanks, OneHit. As I mentioned earlier, my ex-girlfriend is a fine editor and makes me feel clumsy and inarticulate in any discussion like this. I'm just trying to point out some of the obvious.

Please take the baton and do a better job explaining than me, because evidently nobody understands what the heck I'm saying.

[kewl]

The WoVers are arguing that the use of the word "probability" automatically shifts everything into a multiple event, in-the-future interpretation.

No, it is just easier to understand that way. The probability is fixed and applies for those conditions - pair of dice, at least one has landed as two.

Plus, the very definition of probability dictates that a probability of event can be viewed as the odds of it happening in one trial, or the expectation(or percentage)of it happening on multiple trials, equally well and equally correctly representing the very nature of the reality we live in.

[OneHitWonder]

Please take the baton and do a better job explaining than me, because evidently nobody understands what the heck I'm saying.

There are two types of or: inclusive; and exclusive. (Don't write or exclusive, because that would be some sort of circular argument.) The Wizard's group's only chance at 1/11 here is for the inclusive or. By definition, this isn't an inclusive-or problem. That leaves the either-or, disjointed specific rolls of dice by which the or takes on a completely different meaning.

I tried to explain all of this over there by one take or another, which was followed by a few of Indignant99's strange remarks culminating with, "I don't think he understands what he's saying." I gave up. About this time, the Wiz resorted to his patented "ridiculous beliefs" argument. (He has his own patented "I'm getting madder now" scale of tip-offs; as psychological signals to the others to start to bring up the notion of banning such crazy persons.)

[RS__]

The Wizard wrote a simulation in C++. I did one in Excel as well as PHP. Others have written simulations in Excel and other languages (not that Excel is a language).

All of the simulations are giving the same answer: Approximately 1/11 or 0.090909.

I would just love to know -- why aren't any of the simulations giving us an answer around 1/6?

[kewl]

There are two types of or: inclusive; and exclusive. (Don't write or exclusive, because that would be some sort of circular argument.) The Wizard's group's only chance at 1/11 here is for the inclusive or. By definition, this isn't an inclusive-or problem. That leaves the either-or, disjointed specific rolls of dice by which the or takes on a completely different meaning.

I tried to explain all of this over there by one take or another, which was followed by a few of Indignant99's strange remarks culminating with, "I don't think he understands what he's saying." I gave up. About this time, the Wiz resorted to his patented "ridiculous beliefs" argument. (He has his own patented "I'm getting madder now" scale of tip-offs; as psychological signals to the others to start to bring up the notion of banning such crazy persons.)

Here is the simplified version of the same 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?".

Simple enough, and there is no ambiguity about any human interference.

What the 1/6 supporters fail is - they believe one 2 showing makes it equal to "we have one certain die as a 2 and are now asking whats the odds on the second die to be 2"

They fail to assimilate that "at least one 2" means twice as much possibilities for their "certain" die to be 2.

[Alan Mendelson]

But we are talking about a throw of pair of dice here, and not only "the second die".

If we are talking about a throw of a pair of dice then the answer to the question is 1/36.

If one of the two dice is known to be a two, then it is the "second die" and that is 1/6.

Your answer of 1/11 still does not apply to this question.

Knowing that at least one die has landed on two does not eliminate any of the eleven possibilities.

Yes it does, because if a 2 shows up on one die you can't change the value of that die to fit your 1/11 answer. With a 2 on one die, only 6 faces remain on the second die.

The probability is fixed and applies for those conditions -

And that's your problem -- it doesn't apply to the condition of the original question.