It was two years ago today...

[Alan Mendelson]

This thread may go down in history as the dumbest of all time. Alan must have been in a mood to start an argument. Maybe it it could be studied by a group of sociologists/psychiatrists, it would keep them busy for years.

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

[a2a3dseddie]

And when you are told that ONE of the dice is a 2, what are the odds that the second die is also a 2?

That is not the original question. This is.

What is the probability that both dice are showing a 2?


You said it yourself:

When you roll two dice, there are 11 combinations on those two dice that include a 2, and one of those 11 combinations is 2-2. That's 1/11.

[coach belly]

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

[Alan Mendelson]

You are told TRUTHFULLY that at least one of the two dice in the cup is showing a 2. So again, what are the odds that the second die is also showing a 2? It is still 1/11 for a six sided die? Or is it 1/6?

Respond to the question and its conditions.

The condition is one die is known to be a two. It is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two. Repeat: it is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two.

This is a reading comprehension issue for those of you who think the answer is 1/11.

[Alan Mendelson]

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

Sorry, it is not a computer simulation problem. It's a simple arithmetic problem. You only need to count to six on one die.

[arcimede$]

No one is twisting words. You are simply denying the obvious fact that the peeker has to see both die making Regnis statement nonsense and showing you will say anything to protect your ego.

I see we have a real problem here with the English language. So I will answer this... slowly.

The peeker only needs to see one die with a 2 to say, truthfully, "at least one die is a two." It is not necessary to see both dice. The peeker only needs to see one die.

If the peeker sees a 2 and a 6, he can truthfully say "at least one die is a two."
If the peeker sees a 2 and never stops to look at the second die, he can truthfully say "at least one die is a two."

The term "at least" means, according to Dictionary.com: "at the lowest estimate or figure" and they give this example: "The repairs will cost at least $100."

All the peeker has to do is see one die.

Now, why are you even raising this? We know from the original question that truthfully one of the dice is showing a 2. Challenging the presence of a 2 serves what purpose?

There are two dice. There is a two on one of those two dice. That die with a two cannot change. Deal with that reality.

I can't wait to see what regnis says.

Alan, your lack of anything approaching logic is quite hilarious. You even admit .... "If the peeker sees a 2 and a 6, he can truthfully say "at least one die is a two."" In other words, he cannot ALWAYS make the determination until he sees both die when one of them is not a 2. If he sees the 6 first then he has to look at the other die. The fact there are a few times he could make the determination by seeing only one die is NOT sufficient to ALWAYS make the determination. Hence, just looking at one die will not allow him to ALWAYS make the claim "at least one die is a 2".

This entirely negates Regnis claim.

Alan, you really are making a complete fool out of yourself.

[a2a3dseddie]

You are told TRUTHFULLY that at least one of the two dice in the cup is showing a 2. So again, what are the odds that the second die is also showing a 2? It is still 1/11 for a six sided die? Or is it 1/6?

Respond to the question and its conditions.

The condition is one die is known to be a two. It is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two. Repeat: it is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two.

This is a reading comprehension issue for those of you who think the answer is 1/11.

Once again here is the original problem:

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?

That is the original question.

NOT

"what are the odds that the second die is also showing a 2?"

[arcimede$]

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

Mickey did a real life simulation and it showed 1 in 11.

[a2a3dseddie]

You guys keep making this ridiculous statement:

Since we don't know which dice has landed on the 2, we need to consider all combinations

If you have two dice, and you know one die is a 2, how many possible combinations could there be? There are only two dice. One is a 2.

It doesn't matter which of the two dice is showing the 2.

Maybe on a chart showing the combinations of dice it matters, but when you are dealing with two physical dice it doesn't matter which is showing a two.

Take your minds off your charts and graphs and put two dice in front of you. Set one of the dice as a 2. What's the answer to the problem? It's one out of six.

Alan, you shoot a lot of craps.

Are you telling me that when the dice are thrown, and at least one of them lands on a 2, you see the hard 4 get paid off 1 time in 6?

[jbjb]

You guys keep making this ridiculous statement:

Since we don't know which dice has landed on the 2, we need to consider all combinations

If you have two dice, and you know one die is a 2, how many possible combinations could there be? There are only two dice. One is a 2.

It doesn't matter which of the two dice is showing the 2.

Maybe on a chart showing the combinations of dice it matters, but when you are dealing with two physical dice it doesn't matter which is showing a two.

Take your minds off your charts and graphs and put two dice in front of you. Set one of the dice as a 2. What's the answer to the problem? It's one out of six.

Alan, you shoot a lot of craps.

Are you telling me that when the dice are thrown, and at least one of them lands on a 2, you see the hard 4 get paid off 1 time in 6?

He sees a lot of things that are impossible/improbable. Does 18 11's in a row ring a bell? :-D

[quahaug]

You guys keep making this ridiculous statement:



If you have two dice, and you know one die is a 2, how many possible combinations could there be? There are only two dice. One is a 2.Here we go.

It doesn't matter which of the two dice is showing the 2.

Maybe on a chart showing the combinations of dice it matters, but when you are dealing with two physical dice it doesn't matter which is showing a two.

Take your minds off your charts and graphs and put two dice in front of you. Set one of the dice as a 2. What's the answer to the problem? It's one out of six.

Alan, you shoot a lot of craps.

Are you telling me that when the dice are thrown, and at least one of them lands on a 2, you see the hard 4 get paid off 1 time in 6?

He sees a lot of things that are impossible/improbable. Does 18 11's in a row ring a bell? :-D

Here we go.

[a2a3dseddie]

He sees a lot of things that are impossible/improbable. Does 18 11's in a row ring a bell? :-D

Alan, you really are making a complete fool out of yourself.

arcimede$, in Alan's defence(!) the question doesn't mention multiple dice throws. If just 1 throw is considered, the peeker could have just seen 1 die and truthfully say at least 1 of the dice is a 2.

The peeker might have lucked out, had one of the 11 possible combinations of two dice where at least 1 of the 2 dice is a 2 under that cup and saw that one first when he peeked.

[a2a3dseddie]

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

Mickey did a real life simulation and it showed 1 in 11.

Here's a link to a billion trial simulation

http://wizardofvegas.com/forum/quest...ice-puzzle/41/

[RS__]

It's a conspiracy launched by me with the cooperation of regnis and a few others.

Maybe it's to show just how much logic it takes to understand two physical dice.

I'm surprised that nobody has run a computer simulation for this.

I have read here about million+ hand VP simulations, using all different strategies.

This dice problem seems easier to simulate, but nobody in the 11/1 crowd has offered one.

Pages and pages of debate on at least 2 different forums, but what does the computer simulation show?

I have run a simulation and posted the results 2 years ago or whenever this all started. More interestingly, no one in the 1/6 crowd has the capability of running a simulation. [That was mostly sarcasm, but there's some truth to it.]

The way the sim worked is this -- the dice keep getting rolled over and over again, only counting rolls where at least one die was a '2'. Since at least one die was a 2, it then recorded if both dice were a 2 or not both 2.

[jbjb]

You don't need a simulation. Roll two dice, don't set one to a two. When at least one two shows and the other one isn't, mark column A. When at least one two shows and the other one is, mark column B. Column A will out number column B by 11 to 1 after a long sequence of trials.

[Rob.Singer]

You are told TRUTHFULLY that at least one of the two dice in the cup is showing a 2. So again, what are the odds that the second die is also showing a 2? It is still 1/11 for a six sided die? Or is it 1/6?

Respond to the question and its conditions.

The condition is one die is known to be a two. It is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two. Repeat: it is not the same as throwing two dice simultaneously and knowing that there are 11 combinations showing at least one two.

This is a reading comprehension issue for those of you who think the answer is 1/11.

Once again here is the original problem:

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?

That is the original question.

NOT

"what are the odds that the second die is also showing a 2?"

Amazing to check in to see this still rages on. Eddie's post here and the one in which he stated "arci, in Alan's defense" really defines everything this dice issue is about.

There are absolutely two answers to this "problem" and either is correct depending on how the reader wants to understand the question. Redietz actually identified the complexities of how different individuals might understand the problem, albeit a bit awkwardly as usual.

The "math people" like arci as well as several of these uneducated self-proclaimed "AP's who've been imported from the WoV site, along with the mensa "geniuses" that come out of the infamous pool of libtard atheists, agnostics, queers, and weirdo transgenders over there, all see this problem as a one-way street leading to an 11 to 1 conclusion. In arci's defense, he's a hard core technical person, and he is infinitely trained to see this problem--and its final solution--in only one way: to be an 11:1 result and that is all there is to it. The rest of the 11:1 crew? RS__ & jbjb are simply responding the way they best believe they won't be chided or mocked over on WoV. In other words, they really cannot think for themselves. Mickey however is doing nothing more than taking the opposite position of whatever Alan comes up with because he's always been jealous of Alan and his money, his string of girlfriends, his job, and his basic everyday normal life compared to that of the lowlife, non-productive slug existence mickey endures.

On the other, more grounded side of this resides the people who have no interest in reading anything more or less into the problem than it simply states....while 100% ignoring anything that it may imply. These people seek out an everyday-type answer laced with common sense...and nothing more. And while their adversaries' chief goal is very likely to be making others think how deeply intelligent they can be while the rest of the "dummies" can never attain their special level of being able to dissect and analyze such a problem in ways only a class of Einsteins may begin to understand them, the dumbos know how to "keep it simple, stupid".

So who's got it right? Well, both sides have made an indisputable case for their argument (belief). But it is not really that difficult a problem to figure out--one need not have an engineering degree or be a math or psychology professor at Stanford or Boston College (I had to throw that in ) to come up with a conclusion. It all depends on your reading comprehension.

HOWEVER, if I were a court appointed arbitrator I would rule solidly in favor of the "6" crowd, while giving the "11" crowd honorable mention for their deep insight, unforeseen complex problem solving abilities, ingenuity, and overall argumentative tenacity. Why? Because NOWHERE IN THE ORIGINAL STATED PROBLEM DOES IT MENTION ANYTHING OTHER THAN A ONE TIME TRIAL. If the peeker sees only one die and it is a 2, or if he sees both dice and at least one of them is a 2--and because he doesn't identify which is the case--"6" wins as the simple, common sense conclusion.

Good luck.

[redietz]

We fucking went through this before. Of course the peeker has to see a 2 in order to make the statement "at least one die is a two."

You made this ridiculous comment the first time responding to regnis who discussed seeing a 2.

Then you tried to twist your words.

Quit it. You pulled this same shit in your battles with Singer over the years and I'm not going to fall for it.

Fuck off.

And you cheated on your taxes.

No one is twisting words. You are simply denying the obvious fact that the peeker has to see both die making Regnis statement nonsense and showing you will say anything to protect your ego.

Now you are bringing up my taxes which you obviously don't understand very well either (is anyone surprised?)? So, I guess we can add a complete lack of integrity to your growing list of faults.

So, once again .... How can the peeker answer the question "truthfully" if he only sees one die? If you contnue to avoid the question that is just as good as admitting he can't answer truthfully and therefore looking at one die is NOT an option.

Now, quit your incessant whining and just go away. Isn't that your usual mode of operation?

Why does the peeker need to see both dice? That is not explicitly stated.

If the above line was stated "awkwardly," I apologize for no mention of transgenders and libtards.

[a2a3dseddie]

HOWEVER, if I were a court appointed arbitrator I would rule solidly in favor of the "6" crowd, while giving the "11" crowd honorable mention for their deep insight, unforeseen complex problem solving abilities, ingenuity, and overall argumentative tenacity. Why? Because NOWHERE IN THE ORIGINAL STATED PROBLEM DOES IT MENTION ANYTHING OTHER THAN A ONE TIME TRIAL. If the peeker sees only one die and it is a 2, or if he sees both dice and at least one of them is a 2--and because he doesn't identify which is the case--"6" wins as the simple, common sense conclusion.

from the looks of things, the wizard is only paying 9 units for something that has 11-1 odds.

If we're at this point, I'll explain why. When two dice are peeked at and we're informed that one of them is a 2, if that die is removed from the scene then we're left with a single die that no longer has a numerical relationship to the die showing a 2. This is the way I interpret the problem with the wording from the OP, and the odds that 2nd die will be a 2 is 6-1.

However, if you take the WoV interpretation, simply telling you that one of the dice shows a 2 does not remove it from its numerical relationship with the die with the unknown number showing. Whereas rolling 2-2 is a 36-1 possibility, knowing half the outcome, ie. one of the dice is showing a 2, reduces the odds to 11-1 since it is no longer a single die event. You now have to consider all combinations of the two dice where either one of them shows a 2.

[arcimede$]

arcimede$, in Alan's defence(!) the question doesn't mention multiple dice throws. If just 1 throw is considered, the peeker could have just seen 1 die and truthfully say at least 1 of the dice is a 2.

The peeker might have lucked out, had one of the 11 possible combinations of two dice where at least 1 of the 2 dice is a 2 under that cup and saw that one first when he peeked.

You have to understand this gets back to when Alan stated:

"Yes, a "peeker" can see either one die or both dice and say truthfully and correctly "at least one die is a 2." Do you know how to speak the English language Arc?"

He did this in response to me saying he couldn't "ALWAYS" make a determination unless he sees both die. That was my point and for some stupid reason Alan thought his response refuted what I stated. Obviously, Alan doesn't understand what ALWAYS means or he is just belligerent. Since then I've just been baiting him to see what other nonsense he would spew..

[Alan Mendelson]

You're right Arc, I don't always understand what you say. So let me say it my way:

The peeker only has to see ONE die to truthfully say "at least one of the dice is a 2" if the first die he sees shows a 2.

The peeker does not have to see the second die to say "at least one of the dice is a 2" if the first die he sees shows a 2.

Is this clear now to everyone?

I'm still waiting for jbjb to clarify the conditions in his question regarding the pile of spades and the pile of hearts.
redietz perhaps you should explain one more time the concept of sequential observation.
Rob Singer you gave the 1/6 answer correctly except that you couldn't resist throwing in your usual insults.

Yes everyone in the 1/11 gang -- one out of 11 is what you have when you throw two dice: there are 11 combinations of those two dice that could show a 2 and 1 of those combinations could show 2-2. That's 1/11.

But after two dice are thrown and a peeker sees that at least one die is showing a 2 the odds that the other die is also showing a 2 is one out of the six sides on that single die. That's 1/6.

You must understand the question. You must use the math (or in this case the ability to count to 6) to answer the question that is asked.

You cannot read a question and then apply math that doesn't answer the question being asked. For the thousandth time.

Read.