It was two years ago today...

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Originally Posted by Alan Mendelson

In this case the dealers have a better idea of the actual problem. And they are not using one die to have multiple values. In other words, they have a better grip on the reality of the question.

You "1/11 guys" still don't understand how you are misinterpreting simple English.

I guess when you take probability classes they forget to teach how to understand the English language.

I gave you the question that would have an answer of 1/11.

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One die is a fixed parameter, the other is variable.

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Originally Posted by Alan Mendelson

In this case the dealers have a better idea of the actual problem. And they are not using one die to have multiple values. In other words, they have a better grip on the reality of the question.

You "1/11 guys" still don't understand how you are misinterpreting simple English.

I guess when you take probability classes they forget to teach how to understand the English language.

I gave you the question that would have an answer of 1/11.

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You're back peddling. The only reason you brought up the dice crew is because you thought they have some kind of authority or expertise on the subject. But, that's clearly not true. So now you're saying they just understand or "have a better grip" on the question than......people who actually use probability to make a living.

You're a reporter. Stick with reporting instead of arguing yet another case you're absolutely wrong in.

I ran another sample space of 100 decisions this morning. The other die was a two 13 times in 100 decisions. So now we have a sample space of 300 decisions. The other die was a two 32 times. That's an average of once very 9.375 decisions. Are the numbers now regressing towards 1 in 6, or did the solid four just catch some positive variance in the third sample space.

I have some driving to do today and some slot parlors in a couple of small towns to sweep through. I'll be in a different hotel tonight and run some more sample spaces.

Alan has already admitted that if the question is worded one way the answer is 1:11. Hence, the argument here is what is the wording of the question. Alan needs to show everyone the precise difference in wording that would yield 1:11 vs. 1:6 and explain why.

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Originally Posted by RS__

Alan.....dealers, floor people, and pit bosses are generally not experts on probability. To say they are experts on dice is like saying the cooks at McDonalds are expert chefs.

Hell, one time my boss backed off our biggest bettor because he was winning.....he always played the same way: Hop hard ways, regular hard ways, horn bets, etc. for the max he could, and play until he was stuck $20k.....he could be up $40k but he'd still lose it all back, if it took 1 hour or 10 hours.

I asked the other dealers I was dealing with the same question, word for word. Three dealers total. One almost instantly said 1/6 because the other die had 6 sides. Another waited a few moments, his brain working, and said 1/36 because that's the chance of a 2-2 being rolled. The other was still thinking, said 1/12. I asked why 1/12 and he said because you could roll a 1-2, 2-2, 3-2, 4-2, 5-2, 6-2, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6. Then thought about it for a second more and said no, it's 1/11 because he double counted 2-2.

I asked a boss and he just said "it's random, dice are random". I said yes they're random but what do you think the probability is or chance? Like flipping a coin and it landing on heads is 50% or 1/2. Again, he said its random.

I asked them if you are dealt a blackjack, what's the probability the first card is a ten or face card. The guy who said the dice prob was 1/6 said it was 4/13 for the BJ problem, because 4 of 13 cards are tens. The other dealer had some convoluted math and came up with an answer worse than Rob could come up with. The one who said the dice are 1/11, that there's a 1/2 chance your first card is a ten if you have a blackjack, because there are 2 combinations to get a BJ, either Ace+Ten or Ten+Ace. He was 2 for 2.

I asked a few other probability questions, riddles, etc. as did they. It became apparent the guy who said the dice probability is 1/11 actually knew how to do math, while the other two were a little slow, to say the least.

The fact you think dice dealers and floor people are considered "experts" is a bit telling.



I'd be careful about asking a group of math professors any question and accepting (sort of) the answer from just one. I had a computer science professor who gave us a homework assignment that was literally impossible in the language we were using. I don't remember the specifics now, but he wanted us to write a function which would use pass-by-reference which, done the way he wanted, was impossible (per the book and any expert or beginner). I asked for an example of code which would be sufficient, and the code he gave did not meet the criteria in the assignment (pass by reference). I ended up getting a 100% on the assignment, but only after many emails back and forth as well as getting the head of the department involved. The head of the department said something like, "This assignment was impossible. The language simply doesn't work that way."

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During a dispute at the craps table I once made them call the gaming board agent on to the boat (in Illinois we originally had to have our casinos on big boats). After it became clear that he didn't have a clue and was just a typical government employee, I made a point to embarrass him by asking him to tell me what the proper payoff on a nine was. He was unable to do so.

I did used to take great pleasure in delaying the game for about 45 minutes over a minor amount of money. These casinos and their employees don't seem to understand that every roll, every spin, every deal is $$$ and that they should never let the game slow down. So they could argue with me for 45 minutes over a few dollars and it costs them hundreds.

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Originally Posted by arcimede$

Alan has already admitted that if the question is worded one way the answer is 1:11. Hence, the argument here is what is the wording of the question. Alan needs to show everyone the precise difference in wording that would yield 1:11 vs. 1:6 and explain why.

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Arci, good to see you! I appear to have inherited your mantle as Rob's most beloved poster, only he wants me to put my driver's license online. That'll be the day. At least he believed you were you!

Anyway, arci and I disagree on this, as it is the responsibility of the writer to make clear the math of the answer. If the way a question is written leads a large number of reasonably educated readers to the wrong answer, then the writer is more at fault than the reader. Math may be about finding the correct answers, but writing is about making the correct answers understandable.

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Originally Posted by redietz

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Originally Posted by arcimede$

Alan has already admitted that if the question is worded one way the answer is 1:11. Hence, the argument here is what is the wording of the question. Alan needs to show everyone the precise difference in wording that would yield 1:11 vs. 1:6 and explain why.

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Arci, good to see you! I appear to have inherited your mantle as Rob's most beloved poster, only he wants me to put my driver's license online. That'll be the day. At least he believed you were you!

Anyway, arci and I disagree on this, as it is the responsibility of the writer to make clear the math of the answer. If the way a question is written leads a large number of reasonably educated readers to the wrong answer, then the writer is more at fault than the reader. Math may be about finding the correct answers, but writing is about making the correct answers understandable.

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The question couldn't be clearer. You say it's poorly written. Please explain why that is. And please don't count Alan in as a reasonable person.

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Originally Posted by quahaug

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Originally Posted by redietz

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Originally Posted by arcimede$

Alan has already admitted that if the question is worded one way the answer is 1:11. Hence, the argument here is what is the wording of the question. Alan needs to show everyone the precise difference in wording that would yield 1:11 vs. 1:6 and explain why.

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Arci, good to see you! I appear to have inherited your mantle as Rob's most beloved poster, only he wants me to put my driver's license online. That'll be the day. At least he believed you were you!

Anyway, arci and I disagree on this, as it is the responsibility of the writer to make clear the math of the answer. If the way a question is written leads a large number of reasonably educated readers to the wrong answer, then the writer is more at fault than the reader. Math may be about finding the correct answers, but writing is about making the correct answers understandable.

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The question couldn't be clearer. You say it's poorly written. Please explain why that is. And please don't count Alan in as a reasonable person.

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First of all, your last sentence reminds me of everybody with a doctorate who thinks they can write. Just because something is clear TO YOU doesn't mean it's clear to a general population. Write a different variation of it that makes it clear and doesn't allow a good chunk of the readership to draw an erroneous conclusion. That's the writer's job. YOU are not the general population. As a writer, your job isn't (1) to be tricky, (2) to show how smart you are, or (3) to dichotomize the readership into those who get it and those who don't.

When this was first posted two years ago, one of the posters described the rules of writing concerning conditionals and tense uses. I am not going to go back and review it because (1) I'm not an expert on it and (2) I remember my ex-girlfriend (who is an editor at Penn State) describing the same rules. From what he said, however, there is a problem with it as presented that leaves open two interpretations. I leave it to those with doctorates in technical writing to have at it, because I was lost in the middle of the second sentence.

All I will contribute is this. Looking at the dice under the cup is, in reality, a sequential act. An actual human being's eyes process one die before the other by darting from area to area. One die will be processed before the other. This makes it a sequential act. An actual person is not, technically and actually speaking, drawing the conclusion about what is on which die simultaneously. It is sequential. Therefore, when a person sees a 2 under the first die, he has very possibly not yet processed the second die. If he immediately reports that he has seen a 2, that does not necessarily mean he has seen the second die. If he processes the first die and it is not a 2, then he processes the second die. Mickey's experiment does not technically fulfill the wordage of the trick question because he is reporting the sighting of the dice as if it's simultaneous, not sequential.

Math people assume the writing is easy to grasp because, what the hell, it's only writing. Do you see how arrogant and self-referential that is?

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Originally Posted by redietz

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Originally Posted by quahaug

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Originally Posted by redietz

Arci, good to see you! I appear to have inherited your mantle as Rob's most beloved poster, only he wants me to put my driver's license online. That'll be the day. At least he believed you were you!

Anyway, arci and I disagree on this, as it is the responsibility of the writer to make clear the math of the answer. If the way a question is written leads a large number of reasonably educated readers to the wrong answer, then the writer is more at fault than the reader. Math may be about finding the correct answers, but writing is about making the correct answers understandable.

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The question couldn't be clearer. You say it's poorly written. Please explain why that is. And please don't count Alan in as a reasonable person.

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First of all, your last sentence reminds me of everybody with a doctorate who thinks they can write. Just because something is clear TO YOU doesn't mean it's clear to a general population. Write a different variation of it that makes it clear and doesn't allow a good chunk of the readership to draw an erroneous conclusion. That's the writer's job. YOU are not the general population. As a writer, your job isn't (1) to be tricky, (2) to show how smart you are, or (3) to dichotomize the readership into those who get it and those who don't.

When this was first posted two years ago, one of the posters described the rules of writing concerning conditionals and tense uses. I am not going to go back and review it because (1) I'm not an expert on it and (2) I remember my ex-girlfriend (who is an editor at Penn State) describing the same rules. From what he said, however, there is a problem with it as presented that leaves open two interpretations. I leave it to those with doctorates in technical writing to have at it, because I was lost in the middle of the second sentence.

All I will contribute is this. Looking at the dice under the cup is, in reality, a sequential act. An actual human being's eyes process one die before the other by darting from area to area. One die will be processed before the other. This makes it a sequential act. An actual person is not, technically and actually speaking, drawing the conclusion about what is on which die simultaneously. It is sequential. Therefore, when a person sees a 2 under the first die, he has very possibly not yet processed the second die. If he immediately reports that he has seen a 2, that does not necessarily mean he has seen the second die. If he processes the first die and it is not a 2, then he processes the second die. Mickey's experiment does not technically fulfill the wordage of the trick question because he is reporting the sighting of the dice as if it's simultaneous, not sequential.

Math people assume the writing is easy to grasp because, what the hell, it's only writing. Do you see how arrogant and self-referential that is?

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I've got to step in here. There is no place on this Forum for such an intelligent and well thought posting.

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Originally Posted by arcimede$

Alan has already admitted that if the question is worded one way the answer is 1:11. Hence, the argument here is what is the wording of the question. Alan needs to show everyone the precise difference in wording that would yield 1:11 vs. 1:6 and explain why.

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Hey arcimede$ how have you been?

Here's the original question to which Alan says the answer is 1/6:

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?

And here's Alan's version of a question that he says would lead to a 1/11 answer:

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Originally Posted by Alan Mendelson

For those of you who want the answer to be 1/11 here is your question:

Given two dice, with a 2 showing on at least one of the two dice, what are the odds that both dice will show 2-2?

The answer to this question is 1/11.

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I am not smart enough to see the difference between these 2 scenarios. :(

Redietz: I'm not going into a long dissertation over this, but it still doesn't matter at all whether your partner peeked under the cup and looked at the dice sequentially or not.

Yes, some of the time it would only be necessary for the partner to look at one die and other times it would be necessary for the partner to see both dice in order to say, "At least one of the dice is a 2."

The fact that the partner does not reveal to us that s/he had to look at one or both dice truly keeps the odds at 1/11.

If the partner explicitly reveals to us that he only looked at one die would the odds become 1/6. The whole point of this exercise is that he does NOT say this to us.

Moreover, if the partner DID tell you that it was necessary to look at both dice in order to say, "At least one of the dice is a 2", the odds of both dice being a 2 given that information would become:

https://www.youtube.com/watch?v=2V3CfD8TPac

This whole dice problem is a beautiful illustration of conditional probabilities based on limited information. It does not need to be worded any differently at all.

Alan--I guess you are going to have to change your line to "Alan simply can't get past a die having 12 sides".

It's really amazing that intelligent people can't see the actual wording of the question and interpret it accordingly.
They are reading a question and then responding to a different question, and I can't believe that so many are doing it.

Look guys, would you all agree that if I threw two dice down a table and one die came to rest on a two and the second die was a spinner, that the odds of the spinner settling on a 2 were 1/6?

Do we all agree on that?

Well, that's the closest analogy to the original wording of the question. If you do not believe the "spinner analogy" is the closest to the original question, what is your analogy?

Are you telling me that the original question tells you to rotate both dice to see 11 possible results containing a 2 on at least one die? If so, how do you reach that conclusion?

(thank you regnis and redietz.)

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Originally Posted by Count Room

Redietz: I'm not going into a long dissertation over this, but it still doesn't matter at all whether your partner peeked under the cup and looked at the dice sequentially or not.

Yes, some of the time it would only be necessary for the partner to look at one die and other times it would be necessary for the partner to see both dice in order to say, "At least one of the dice is a 2."

The fact that the partner does not reveal to us that s/he had to look at one or both dice truly keeps the odds at 1/11.

If the partner explicitly reveals to us that he only looked at one die would the odds become 1/6. The whole point of this exercise is that he does NOT say this to us.

Moreover, if the partner DID tell you that it was necessary to look at both dice in order to say, "At least one of the dice is a 2", the odds of both dice being a 2 given that information would become:

https://www.youtube.com/watch?v=2V3CfD8TPac

This whole dice problem is a beautiful illustration of conditional probabilities based on limited information. It does not need to be worded any differently at all.

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I didn't say it mattered. What I did say is that mickey's demonstration starts from the presupposition that it does not matter, and since the physical processes for the two things are different, the demonstration presupposes the conclusion. The experiment does not recognize that the processes are different.

If you don't immediately see the difference in the two processes, you have a blind spot when it comes to running a proper experiment. It's also possible you have blind spots regarding word usage. Should I now say, "Gotcha? You're an idiot!"

If the writing leads a large chunk of readers to the wrong conclusion, then it's a bad piece of writing. This whole "gotcha" mentality is like some Saturday afternoon Mensa get-together. It serves no purpose. Just rewrite the damn thing so it's clear what's going on. But that wasn't the purpose. It was a word game to generate a gotcha. I saw this thing 20 or 25 years ago. I know the proper "answer." It's a stupid exercise in trickeration of which Don King would be proud.

There's probably a reason the disparate group of Mendelson, Singer, regnis, and Dietz all found fault with the exercise. Unless I miss my guess, these are probably the folks who have done the most writing for the public or for use in court.

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Originally Posted by Alan Mendelson

Look guys, would you all agree that if I threw two dice down a table and one die came to rest on a two and the second die was a spinner, that the odds of the spinner settling on a 2 were 1/6?

Do we all agree on that?

Well, that's the closest analogy to the original wording of the question. If you do not believe the "spinner analogy" is the closest to the original question, what is your analogy?

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I have a question for you, Alan. If your partner is looking at a resting die with a "2" and the other one is a spinner, does he show you this or tell you this fact? Again, this situation would be exactly the same as if your partner telling you s/he only needed to look at one die to tell you "At least one of the dice is a 2". Yes, at that point the odds truly become 1/6.

It's not a good analogy for the original problem, though.

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Originally Posted by redietz

I didn't say it mattered. What I did say is that mickey's demonstration starts from the presupposition that it does not matter, and since the physical processes for the two things are different, the demonstration presupposes the conclusion.

If you don't immediately see the difference in the two processes, you have a blind spot when it comes to running a proper experiment. It's also possible you have blind spots regarding word usage. Should I now say, "Gotcha? You're an idiot!"

If the writing leads a large chunk of readers to the wrong conclusion, then it's a bad piece of writing. This whole "gotcha" mentality is like some Saturday afternoon Mensa get-together. It serves no purpose. Just rewrite the damn thing so it's clear what's going on. But that wasn't the purpose. It was a word game to generate a gotcha.

There's probably a reason the disparate group of Mendelson, Singer, regnis, and Dietz all found fault with the exercise. Unless I miss my guess, these are probably the folks who have done the most writing for the public or for use in court.

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Redietz: OK, fair enough. How would you re-word this problem to satiate the masses, yet keep the information limited enough to have the same answer as originally intended (1/11)?

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Originally Posted by Alan Mendelson

Look guys, would you all agree that if I threw two dice down a table and one die came to rest on a two and the second die was a spinner, that the odds of the spinner settling on a 2 were 1/6?

Do we all agree on that?

Are you telling me that the original question tells you to rotate both dice to see 11 possible results containing a 2 on at least one die? If so, how do you reach that conclusion?

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This isn't close to the original wording of the question at all.

You do not see any of the 2 dice. Your partner does.

Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

In your analogy, you KNOW for certain which of the dice is a 2.

Let's say we use coloured dice. One is red and one blue. If you see that the red die is 2, of course you can eliminate all of the combinations where the blue die is 2. That would be 1/6

But you do not know which die is 2, so you cannot eliminate any of the combinations.

Therefore any of the 11 possible combinations where "At least one of the dice is a 2." could be under that cup. Hence 1/11.

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Originally Posted by Count Room

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Originally Posted by redietz

I didn't say it mattered. What I did say is that mickey's demonstration starts from the presupposition that it does not matter, and since the physical processes for the two things are different, the demonstration presupposes the conclusion.

If you don't immediately see the difference in the two processes, you have a blind spot when it comes to running a proper experiment. It's also possible you have blind spots regarding word usage. Should I now say, "Gotcha? You're an idiot!"

If the writing leads a large chunk of readers to the wrong conclusion, then it's a bad piece of writing. This whole "gotcha" mentality is like some Saturday afternoon Mensa get-together. It serves no purpose. Just rewrite the damn thing so it's clear what's going on. But that wasn't the purpose. It was a word game to generate a gotcha.

There's probably a reason the disparate group of Mendelson, Singer, regnis, and Dietz all found fault with the exercise. Unless I miss my guess, these are probably the folks who have done the most writing for the public or for use in court.

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Redietz: OK, fair enough. How would you re-word this problem to satiate the masses, yet keep the information limited enough to have the same answer as originally intended (1/11)?

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I have no idea, but you bring up an interesting question. Why make this a word game? Why not just a paragraph lecture on conditional probability so everyone learns something? The only reason for the dice example is to create a gotcha. Unless the dice example has some real world application of some kind, which I doubt.

I don't know why the information has to be limited or what reason there is to format the information into a word game. Cleverness isn't everyone's raison d'etre. Dr. Michael Starbird at Ut-Austin does probability lectures for the masses. I'd ask him.

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Originally Posted by slingshot

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Originally Posted by Alan Mendelson

In this case the dealers have a better idea of the actual problem. And they are not using one die to have multiple values. In other words, they have a better grip on the reality of the question.

You "1/11 guys" still don't understand how you are misinterpreting simple English.

I guess when you take probability classes they forget to teach how to understand the English language.

I gave you the question that would have an answer of 1/11.

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One die is a fixed parameter, the other is variable.

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This is true... for the partner who peeks under the cup. We don't know which die is "the fixed parameter" so we must consider all possible combinations.

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Originally Posted by Alan Mendelson

It's really amazing that intelligent people can't see the actual wording of the question and interpret it accordingly.
They are reading a question and then responding to a different question, and I can't believe that so many are doing it.

Look guys, would you all agree that if I threw two dice down a table and one die came to rest on a two and the second die was a spinner, that the odds of the spinner settling on a 2 were 1/6?

Do we all agree on that?

Well, that's the closest analogy to the original wording of the question. If you do not believe the "spinner analogy" is the closest to the original question, what is your analogy?

Are you telling me that the original question tells you to rotate both dice to see 11 possible results containing a 2 on at least one die? If so, how do you reach that conclusion?

(thank you regnis and redietz.)

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And if I asked what's the probability of rolling a 12 with 2 dice, would you jump and scream saying, "What part of the question has you looking at both dice, looking at all 6 sides on each, to come to your '1/36' conclusion? Heresy! It's clearly 50/50 -- it either happens or it doesn't!"

NOTE: I actually wouldn't be surprised if you said that.

Honestly, I think the "if people are drawn to many different conclusions, the wording is bad" argument is a bit silly.

Many people would get the monty hall problem wrong.


Of course it's a "gotcha" question. Monty hall is the same way. But it's not due to the wording. It's due to the fact the answer is different than what you'd likely think it is.

Same thing with the "what's more likely, rolling a both a 6 and an 8 or rolling two 7's first?" Or the poker one, I don't remember exactly, something about picking 3 cards, then if any of those 3 cards (ranks, not including suits) shows up on the flop, the other guy loses.

Same with many riddles.

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Originally Posted by RS__

Honestly, I think the "if people are drawn to many different conclusions, the wording is bad" argument is a bit silly.

Many people would get the monty hall problem wrong.


Of course it's a "gotcha" question. Monty hall is the same way. But it's not due to the wording. It's due to the fact the answer is different than what you'd likely think it is.

Same thing with the "what's more likely, rolling a both a 6 and an 8 or rolling two 7's first?" Or the poker one, I don't remember exactly, something about picking 3 cards, then if any of those 3 cards (ranks, not including suits) shows up on the flop, the other guy loses.

Same with many riddles.

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Exactly. It is a trick question and most people will get it wrong. However, the words were clearly chosen carefully so that people with good logical skills will come to the right answer. It's perfectly OK to get it wrong. It is not perfectly OK to deny the correct answer once the details of the problem are explained. Those still pushing the 1:6 answer are showing the effects of a closed mind and/or massive ego.

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Originally Posted by RS__

Honestly, I think the "if people are drawn to many different conclusions, the wording is bad" argument is a bit silly.

Many people would get the monty hall problem wrong.


Of course it's a "gotcha" question. Monty hall is the same way. But it's not due to the wording. It's due to the fact the answer is different than what you'd likely think it is.

Same thing with the "what's more likely, rolling a both a 6 and an 8 or rolling two 7's first?" Or the poker one, I don't remember exactly, something about picking 3 cards, then if any of those 3 cards (ranks, not including suits) shows up on the flop, the other guy loses.

Same with many riddles.

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I gave a perfect example to this puzzle using people instead of dice. The answer is the same.

P.S. Singer believes the Monty Hall problem is 50/50. Look on Alan's Best Buys site for his article.

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Originally Posted by arcimede$

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Originally Posted by RS__

Honestly, I think the "if people are drawn to many different conclusions, the wording is bad" argument is a bit silly.

Many people would get the monty hall problem wrong.


Of course it's a "gotcha" question. Monty hall is the same way. But it's not due to the wording. It's due to the fact the answer is different than what you'd likely think it is.

Same thing with the "what's more likely, rolling a both a 6 and an 8 or rolling two 7's first?" Or the poker one, I don't remember exactly, something about picking 3 cards, then if any of those 3 cards (ranks, not including suits) shows up on the flop, the other guy loses.

Same with many riddles.

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Exactly. It is a trick question and most people will get it wrong. However, the words were clearly chosen carefully so that people with good logical skills will come to the right answer. It's perfectly OK to get it wrong. It is not perfectly OK to deny the correct answer once the details of the problem are explained. Those still pushing the 1:6 answer are showing the effects of a closed mind and/or massive ego.

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+100

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Originally Posted by redietz

All I will contribute is this. Looking at the dice under the cup is, in reality, a sequential act. An actual human being's eyes process one die before the other by darting from area to area. One die will be processed before the other. This makes it a sequential act. An actual person is not, technically and actually speaking, drawing the conclusion about what is on which die simultaneously. It is sequential. Therefore, when a person sees a 2 under the first die, he has very possibly not yet processed the second die. If he immediately reports that he has seen a 2, that does not necessarily mean he has seen the second die. If he processes the first die and it is not a 2, then he processes the second die. Mickey's experiment does not technically fulfill the wordage of the trick question because he is reporting the sighting of the dice as if it's simultaneous, not sequential.

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The original scenario/question can be reduced to one simple sentence "if you roll two dice and at least one of them is a two then what is the probability that both dice are 2's." The "friend" looking under the cup is all fluff. It doesn't matter who looks under the cup and it doesn't matter which die you see first, the result of the dice throw is still the same.

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Originally Posted by Alan Mendelson

If you do not believe the "spinner analogy" is the closest to the original question, what is your analogy?

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Neither dice is spinning when I look under the cup.

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Originally Posted by RS__

Or the poker one, I don't remember exactly, something about picking 3 cards, then if any of those 3 cards (ranks, not including suits) shows up on the flop, the other guy losees.

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This was a hustle I seen around the poker tables years ago. The hustler would bet you that either a six, seven or eight, would appear on the flop. There are at least 12,000 combinations out of 20,000 that have a six, seven or eight in them. It's a pure sucker bet but some people fell for it.

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Originally Posted by arcimede$

the words were clearly chosen carefully so that people with good logical skills will come to the right answer.

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As I pointed out on the WOV forum two years ago it was obvious it was a trick question when you looked at the "answer" in the spoiler. Did anyone besides me happen to read the "spoiler"?

And then did you happen to see how the original poster on the WOV set up his question in attempt to fool everyone to think the wrong way -- and they did?

But putting that aside, redietz has the whole thing perfectly explained: we are as humans unable to process two dice simultaneously. In theory (which is how the 1/11 answer is developed) you can consider two dice and 12 faces. But in the real world you can't.

We use logic and reason to know that when at least one die is showing a 2, that die cannot show any other number. And in a two dice problem, when at least one die is showing a 2, we know that there will forever be one die showing that 2 and the only variable is the second/other die.

In the real world it doesn't matter which of the two dice is showing a 2 because as humans we logically understand that if there are two dice with one showing a 2 the variable is that second die.

To not accept that human condition you must have a closed mind and/or massive ego.

This is why I asked the Wizard to demonstrate the solution using a cup and two dice, just as I did in my video. The first time he did a video similar to mine (I can't find it any longer on YouTube) he actually took the die showing a 2 and rotated it to show 11 possible faces.

If you think a die showing a 2 can be altered then yes, go with your 1/11 answer. That is your reality.

http://www.alanbestbuys.com/id362.html

Scroll down about 1/3 the way to Monty Hall vs The Wizard vs Rob Singer.


Alan, if something is true in theory, then it's true in reality. If one is claimed to be true and the other false, then either one of those claims is untrue or the premise is different for the two claims. Otherwise, theory would be completely unnecessary and a waste of time. I certainly hope we can all agree it is not.

Wow, RS__, for a minute I thought you really had me when I read this:

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Originally Posted by RS__

Alan, if something is true in theory, then it's true in reality.

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But then I did a little research. Just a little research. And I just could not find a REAL dice game that allowed you to change the face of a die after the die came to rest.

So... maybe in theory you can change the face of a die after it comes to rest, but in reality you can't change the face of a die that comes to rest. At least, I couldn't find a game that allowed it. Can you tell me a game that allows it? You were a dealer in Vegas... did you ever allow the face of a die to be changed after it came to rest?

Getting back to our original question: It seems you are basing your answer of 1/11 on a fantasy game -- a game that allows you to change the face of a die after it comes to rest.

I came up with the answer of 1/6 on reality -- what everyone knows about every dice game which is you roll the dice and they come to rest on a number.

Gosh, had you told me that shaking two dice in a cup was part of a fantasy game that allowed you to change the faces after the dice came to rest I guess I might have come up with the answer of 1/11 all by myself. But no... I didn't know we were talking about a fantasy game. I thought we were talking about reality.

I never said anything about changing the face of a die after it comes to rest. Smells like a strawman to me.

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Originally Posted by RS__

I never said anything about changing the face of a die after it comes to rest. Smells like a strawman to me.

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In order to get 1/11 you must rotate the die showing a 2 to show other faces... otherwise the answer is 1/6. Are you telling me this is news to you?

This is how close-minded my massively egotistical brain was over this two dice problem. When I first read the question it was something I had never thought about before and my initial reaction was "It's a no brainer. It's 1 in 6." But when I seen that some people were saying it was 1 in 11 I thought "What the hell? Those people are saying that the great Mickey Crimm is wrong?" I wasn't about to put up with that bullshit. No doubt about it I was out to prove them wrong. I went to work on the math of the problem so I would be thoroughly correct when I explained to them just how wrong they were. Damn whippersnappers. I'll show them.

But alas! In doing the math of the problem my close-minded and massively egotistical mind came to the realization that....that....that....God forbid.....I was.....was....I hate to admit this.....dammit....I WAS WRONG! Just think of all the pain and humiliation my close-minded and massively egotistical brain suffered through when I discovered that the correct answer is 1 in 11. I still can't believe that I was wrong about something. You talk about a bruised ego. I'll never live down the shame.

In theory 2 + 2 = 4. And in the real world 2 + 2 = 4.

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Originally Posted by jbjb

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Originally Posted by RS__

Honestly, I think the "if people are drawn to many different conclusions, the wording is bad" argument is a bit silly.

Many people would get the monty hall problem wrong.


Of course it's a "gotcha" question. Monty hall is the same way. But it's not due to the wording. It's due to the fact the answer is different than what you'd likely think it is.

Same thing with the "what's more likely, rolling a both a 6 and an 8 or rolling two 7's first?" Or the poker one, I don't remember exactly, something about picking 3 cards, then if any of those 3 cards (ranks, not including suits) shows up on the flop, the other guy loses.

Same with many riddles.

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I gave a perfect example to this puzzle using people instead of dice. The answer is the same.

P.S. Singer believes the Monty Hall problem is 50/50. Look on Alan's Best Buys site for his article.

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It feels really good when my mind grasps a different concept like the Monty hall problem which I had much more trouble seeing than the dice problem. It's like ah ha I see it now but you have to be willing to think out of the box which not everyone can do.

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Originally Posted by mickeycrimm

This is how close-minded my massively egotistical brain was over this two dice problem. When I first read the question it was something I had never thought about before and my initial reaction was "It's a no brainer. It's 1 in 6." But when I seen that some people were saying it was 1 in 11 I thought "What the hell? Those people are saying that the great Mickey Crimm is wrong?" I wasn't about to put up with that bullshit. No doubt about it I was out to prove them wrong. I went to work on the math of the problem so I would be thoroughly correct when I explained to them just how wrong they were. Damn whippersnappers. I'll show them.

But alas! In doing the math of the problem my close-minded and massively egotistical mind came to the realization that....that....that....God forbid.....I was.....was....I hate to admit this.....dammit....I WAS WRONG! Just think of all the pain and humiliation my close-minded and massively egotistical brain suffered through when I discovered that the correct answer is 1 in 11. I still can't believe that I was wrong about something. You talk about a bruised ego. I'll never live down the shame.

In theory 2 + 2 = 4. And in the real world 2 + 2 = 4.

---

And if I realize that you are right and I am wrong, my sense of being wrong might start to spread and I will fall into an abyss of wrongness where I will disintegrate and become nothing and die!

Here are 2 dice. One shows a 2. What are the odds the other is a 2?

I don't know--third base





Answer to follow later

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Originally Posted by Alan Mendelson

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Originally Posted by RS__

I never said anything about changing the face of a die after it comes to rest. Smells like a strawman to me.

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In order to get 1/11 you must rotate the die showing a 2 to show other faces... otherwise the answer is 1/6. Are you telling me this is news to you?

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You don't need to have dice, let alone rotate them, to figure out the answer. Do you think you need to look through a 52 card deck to figure out proper VP strategy?

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Originally Posted by RS__

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Originally Posted by Alan Mendelson

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Originally Posted by RS__

I never said anything about changing the face of a die after it comes to rest. Smells like a strawman to me.

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In order to get 1/11 you must rotate the die showing a 2 to show other faces... otherwise the answer is 1/6. Are you telling me this is news to you?

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You don't need to have dice, let alone rotate them, to figure out the answer. Do you think you need to look through a 52 card deck to figure out proper VP strategy?

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Dumbass doesn't know the difference between a "certain" die and "at least one" die.

Again, I gave a perfect example using people instead of dice in the same scenario. Only an idiot can't see it's 1 in 11.

Name calling and swearing etc. a clear sign of ignorance. Just sayin.

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Originally Posted by jbjb

Dumbass doesn't know the difference between a "certain" die and "at least one" die.

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At least one of the dice is a 2, what are the odds that both are showing a 2? In this example, the answer is still 1/6.

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Originally Posted by RS__

You don't need to have dice, let alone rotate them, to figure out the answer.

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If you did have two dice and you didn't rotate them (because you cant rotate dice after they come to rest) you'd see you couldn't have an answer of 1/11.

If you want the answer of 1/11 I gave you the question that would yield that answer.

Let's color code the dice. One die is RED and the other is GREEN.

Combinations with at least a two in them:

Red 1, Green 2
Red 2, Green 2
Red 3, Green 2
Red 4, Green 2
Red 5, Green 2
Red 6, Green 2
Red 2, Green 1
Red 2, Green 3
Red 2, Green 4
Red 2, Green 5
Red 2, Green 6

Problem #1.

Our friend looks under the cup and says "At least one die is a two." What is the probability that both dice are 2's?

Why is the answer 1 in 11 and not 1 in 6? Because we don't have enough information to determine it is 1 in 6. All we know is that one of the dice is a 2. With that limited information all we can do is calculate the number of combinations that have a 2 in them. There are 11 and they are listed above. There is only one combination where both the dice are 2's. With the limited information we have we cannot eliminate any of the 11 combinations from possibility. So the correct answer can only be 1 in 11.

Problem #2

Our friend looks under the cup and says "The RED die is a 2." What is the probability that both dice are 2's?

What can we do with the information given here? Well, we can do a lot more with it than we can with the information given in problem #1. We can eliminate all the combinations that don't have a RED 2 in them. They are:

Red 1, Green 2
Red 3, Green 2
Red,4, Green 2
Red 5, Green 2
Red 6, Green 2

That leaves only six possible combinations. In one of those combinations both the dice are 2's. So the correct answer can only be 1 in 6.

Problem #3

Our friend looks under the cup and says "The GREEN die is a 2." We can do the same here and eliminate the combinations that don't have a GREEN 2 in them. So the correct answer here is also 1 in 6.

That's the best I can do in explaining it, folks.

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Originally Posted by mickeycrimm

Let's color code the dice. One die is RED and the other is GREEN.

Combinations with at least a two in them:

Red 1, Green 2
Red 2, Green 2
Red 3, Green 2
Red 4, Green 2
Red 5, Green 2
Red 6, Green 2
Red 2, Green 1
Red 2, Green 3
Red 2, Green 4
Red 2, Green 5
Red 2, Green 6

Problem #1.

Our friend looks under the cup and says "At least one die is a two." What is the probability that both dice are 2's?

Why is the answer 1 in 11 and not 1 in 6? Because we don't have enough information to determine it is 1 in 6. All we know is that one of the dice is a 2.

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You still don't get it, do you?

That's all the information you need to know to figure the answer is 1/6. There are only two dice. If at least one is a six you know that 2 dice - 1 die = 1 die and one die has six faces on it, one of which is a 2. Is that really so difficult?

Really, is it?

Regnis gets it. Maybe he can explain it better than I do.

Alan, the only information you have to make your determination is that at least one of the dice is a two. There are 11 possible combinations with "at least a two" in them. Could you show us from the list below which five combinations you eliminated from possibility to get to the answer of 1 in 6.

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Originally Posted by mickeycrimm

Combinations with at least a two in them:

Red 1, Green 2
Red 2, Green 2
Red 3, Green 2
Red 4, Green 2
Red 5, Green 2
Red 6, Green 2
Red 2, Green 1
Red 2, Green 3
Red 2, Green 4
Red 2, Green 5
Red 2, Green 6

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No Mickey. If one of the dice is showing a two, then the other die has one out of six sides. It's that simple.

Why are you making it complicated? Why are you considering eleven possible combinations with a 2?

We know that one die can be excluded. Why are you including it again?

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Originally Posted by Alan Mendelson

No Mickey. If one of the dice is showing a two, then the other die has one out of six sides. It's that simple.

Why are you making it complicated? Why are you considering eleven possible combinations with a 2?

We know that one die can be excluded. Why are you including it again?

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So 5 combinations just disappeared into thin air? Dice are pretty tricky, Alan. Which die is showing the 2? The red die or the green die? That is the crux of the problem. You don't know which die is showing the two. So none of the 11 possible combinations can be eliminated.

Mickey if you want the answer to be 1/11 CHANGE THE QUESTION.

Ask: given all of the combinations of two dice that include the number 2, how many combinations can be 2-2?

That answer is 1/11.

My question is the correct question to get the answer 1/11.

You guys are not answering the question that was asked.

Did you ever play Jeopardy? They give you the answer and the contestants have to come up with the correct question.

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Originally Posted by mickeycrimm

Which die is showing the 2? The red die or the green die? That is the crux of the problem. You don't know which die is showing the two. So none of the 11 possible combinations can be eliminated.

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It doesn't matter which die is showing a two.

Let me repeat that: given the problem, that there are two dice in the cup, it doesn't matter which of the two dice in the cup is showing a 2. This is why the dice are in a cup.

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Originally Posted by Alan Mendelson

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Originally Posted by mickeycrimm

Which die is showing the 2? The red die or the green die? That is the crux of the problem. You don't know which die is showing the two. So none of the 11 possible combinations can be eliminated.

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It doesn't matter which die is showing a two. Let me repeat that: given the problem, that there are two dice in the cup, it doesn't matter which of the two dice in the cup is showing a 2. This is why the dice are in a cup.

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As I said, Alan, dice are tricky. I know I'm not going to change your mind. I'm not even trying to. This thread has had 1800 views in just four days. That means a lot of people are watching it, not just those who have participated in the discussion. I've put up the best argument I can for 1 in 11. It's for their benefit. They need to know there is a counter argument to your argument that it is 1 in 6. They've been given information by you. And they have been given information by several of us in the 1 in 11 crowd. They'll make up their own minds about it. The world won't quit turning no matter what they decide. Let the chips fall where they may.

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Originally Posted by regnis

Here are 2 dice. One shows a 2. What are the odds the other is a 2?

I don't know--third base

The answer is 0, because the effin dice don't flip and the other die is a 1. The point being that once one die is a 2, that can't change. You have the necessary information to determine that the odds of the other die showing a 2 would be 1/6, except in my example.




Answer to follow later

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Quote:

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Originally Posted by regnis

Here are 2 dice. One shows a 2. What are the odds the other is a 2?

I don't know--third base





Answer to follow later

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The answer is 0, because the effin dice don't flip and the other die is a 1. The point being that once one die is a 2, that can't change. You have the necessary information to determine that the odds of the other die showing a 2 would be 1/6, except in my example.

Quote:

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Originally Posted by regnis

The answer is 0, because the effin dice don't flip and the other die is a 1. The point being that once one die is a 2, that can't change. You have the necessary information to determine that the odds of the other die showing a 2 would be 1/6, except in my example.

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They'll never understand because somewhere along the line someone told them that whenever you have a dice problem you have to look at all of the combinations on the dice. Never were they told to use common sense or to consider the conditions of the problem and in this case the condition of the problem is that there are two dice in a cup and only two dice and the dice don't move after they are slammed down on the table.

You do not consider the various combinations of dice that include a two. You consider only the problem in front of you and that the observer TRUTHFULLY told you that AT LEAST one of the dice is showing a two.

How friggin hard is it? I guess for people who can't interpret the question it is very hard.

A few pages back in this thread I asked Alan to ask the craps dealers how many combinations add to 10 when you are rolling three dice. Alan quickly wrote it off as "bullshit questions about three dice." As Alan has said, a die has six sides. But the fact that it has six sides has led some people to draw some wrong conclusions about dice. The question of how many combinations add to 10 when you roll three dice played a roll in the history of the study of the probabilities of dice throws. It's the story about the dice gambler and the dice hustler.

In very early 17th century Italy a dice gambler got introduced to a new gambling game played with three dice. I call the person that introduced him to the game the "dice hustler." It didn't matter who rolled the dice. They were betting on totals. And the only two totals that mattered were the nine and ten. All other outcomes were no action.

The dice gambler bet the nine and the dice hustler bet the ten. When the total added to nine the hustler paid the gambler. When the total added to ten the gambler paid the hustler. The dice gambler had streaks where he did pretty good....but over time he was slowly but surely losing all his money to the dice hustler. He couldn't figure it out. He thought he had the best number. He thought he should be winning. After all, the dice are six sided....three times six is eighteen....half of eighteen is nine....that's the mean number....it should be the best number to have. But he kept losing and losing and losing.

Finally he had had enough. He quit the game. But he still wanted to know why he lost. He had a friend, a noted scientist and mathematician by the name of Galileo. He took the problem to Galileo.

Galileo studied the dice and the game for a couple of days then gave his friend an answer. He told his friend that there were 216 total combinations with three dice (6X6X6). Twenty-seven of those combinations added to 10, but only twenty-five combinations added to 9.

Because of this event Galileo was credited with being the first scientist/mathematican to work out the probabilities of dice throws. But you can bet your ass the dice hustler knew the reason the dice gambler was losing long before Galileo did.

Quote:

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Originally Posted by a2a3dseddie

How about this:

In a standard game of craps considering only the rolls where "At least one of the dice is a 2." What are the odds of being paid on a Hard 4 bet?

Let's see, you CAN'T have an "easy 4" since you know "At least one of the dice is a 2." so 1,3 or 3,1 have been ruled out.

So, your only loss comes if a 7 appears... 2,5 or 5,2

The odds of a 5 appearing on a single die is 1/6 (the same as another 2 appearing).

Do you think it's an even money bet by disregarding the possibility of a 5,2? And what wording in the original question would make you disregard the possibility of 5,2?

Or are the odds actually 2-1 against you? (2,5 or 5,2 versus 2,2)

---

Alan/regnis

What are were your thoughts on this previous post?

Quote:

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Originally Posted by a2a3dseddie

Quote:

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Originally Posted by a2a3dseddie

How about this:

In a standard game of craps considering only the rolls where "At least one of the dice is a 2." What are the odds of being paid on a Hard 4 bet?

Let's see, you CAN'T have an "easy 4" since you know "At least one of the dice is a 2." so 1,3 or 3,1 have been ruled out.

So, your only loss comes if a 7 appears... 2,5 or 5,2

The odds of a 5 appearing on a single die is 1/6 (the same as another 2 appearing).

Do you think it's an even money bet by disregarding the possibility of a 5,2? And what wording in the original question would make you disregard the possibility of 5,2?

Or are the odds actually 2-1 against you? (2,5 or 5,2 versus 2,2)

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Alan/regnis

What are were your thoughts on this previous post?

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I pass.

I am not a mathematician. I can just tell you what happens when I look at two dice.

Quote:

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Originally Posted by regnis

Quote:

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Originally Posted by regnis

Here are 2 dice. One shows a 2. What are the odds the other is a 2?

I don't know--third base.Answer to follow later

---

The answer is 0, because the effin dice don't flip and the other die is a 1. The point being that once one die is a 2, that can't change. You have the necessary information to determine that the odds of the other die showing a 2 would be 1/6, except in my example.

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Trump changed the name of fake news to very fake news. Your answer is not only wrong it is very wrong. It doesn't make you a bad person though. Just very wrong. The only information you have is that "at least one die is two." Remember, the dice are under a cup and only your friend see's them, you don't. You don't know which die is the unknown die. Not one of the 11 possible combinations that have "at least a two" in them can be ruled out. You don't have enough information to do that.

Quote:

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Originally Posted by Alan Mendelson

I am not a mathematician.

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I absolutely agree. See there, we CAN agree on something.

Quote:

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Originally Posted by Alan Mendelson

Mickey if you want the answer to be 1/11 CHANGE THE QUESTION. Ask: given all of the combinations of two dice that include the number 2, how many combinations can be 2-2. That answer is 1/11. My question is the correct question to get the answer 1/11.
You guys are not answering the question that was asked.

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This is the question, the original question, that I answered. "your friend looks under the cup and says "at least one of the dice is a two. What is the probability that both dice are 2's?" This is the question I answered. It is a clear and concise question. I totally understand the question. I DID NOT, repeat, DID NOT, repeat, DID NOT, change the question. And I answered it clearly and concisely. When the only information you have is that "at least one die is a two" you cannot eliminate any of the 11 possible combinations from possibility. Both the dice are 2's in only one combination. Therefore, the answer is 1 in 11.

And that is why, Alan, when you finish a VP session and return another day, the machines will pick up right where you left off. The millions of hands others play will not count, so short term, goal oriented play is useless, and only AP play wins.:rolleyes::rolleyes::rolleyes::rolleyes:

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Originally Posted by mickeycrimm

Quote:

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Originally Posted by Alan Mendelson

Mickey if you want the answer to be 1/11 CHANGE THE QUESTION. Ask: given all of the combinations of two dice that include the number 2, how many combinations can be 2-2. That answer is 1/11. My question is the correct question to get the answer 1/11.
You guys are not answering the question that was asked.

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This is the question, the original question, that I answered. "your friend looks under the cup and says "at least one of the dice is a two. What is the probability that both dice are 2's?" This is the question I answered. It is a clear and concise question. I totally understand the question. I DID NOT, repeat, DID NOT, repeat, DID NOT, change the question. And I answered it clearly and concisely. When the only information you have is that "at least one die is a two" you cannot eliminate any of the 11 possible combinations from possibility. Both the dice are 2's in only one combination. Therefore, the answer is 1 in 11.

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Mickey-mine was a trick question to show the absurdity of some of this. In my example, you can see both dice. The picture didn't transfer over to where I gave the answer but the original picture showed one die as a 2 and one as a 1. My odd sense of humor was not meant to to anything except highlight the way that the phrasing of the question may lead to this mess.

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

Quote:

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Originally Posted by regnis

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

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That's an excellent question regnis. It reminds me of a scene in the Martin Scorsese movie "Casino".

It involves a pair of cheaters playing blackjack. In the scene, one player... the "big player" is betting $1000s on each hand. His partner "the spotter" is seated at a table directly behind the dealer of the "big player's" table.

The dealer is "weak" and lifts his hole card too high and the spotter can see what it is. The "spotter" betting just the table minimum then uses an electronic signaling device to notify the "big player" whenever the dealer has a poor card in the hole.

Using this information, the "big player" can stand on stiff hands like 12 through 16 against a dealer upcard of 10 when he knows the dealer doesn't have a pat hand hoping he draws a card and busts. This increases their edge significantly and the pair win several hundred thousand before Robert De Niro figures it out and gives them the hammer.

The "big player's" optimal move minus any additional information is to play basic strategy blackjack. The "spotter" increases their odds of winning by obtaining hole card information and getting his partner to deviate from basic strategy where appropriate.

In the dice question, AFTER the roll, the partner peeking under the cup has information that the other guy does not. For the partner, part of the roll has already been resolved. Using mickeycrimms coloured dice example, the partner KNOWS which dice (red or green) is already a 2. For him, if the remaining die is "spinning" as Alan puts it, relative to him the odds are 1/6 for it landing on 2.

For the guy who has not seen any of the dice, no part of the roll has been resolved relative to him and he does not know which dice (red/green) is a 2 and which one is still "spinning". Relative to him, any of the 11 combinations of two dice where at least one die is a 2 could be under that cup.

Quote:

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Originally Posted by regnis

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

---

Warning: if you are ever in court and the other side is represented by regnis, you lost.

(You are brilliant, regnis.)

Quote:

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Originally Posted by regnis

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

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Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

Quote:

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Originally Posted by regnis

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

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If all he has seen is one die and he can't distinguish one die from the other then "one die is a two" is all the information he has. Consequently it's 1 in 11. If the dice are color coded, such as one is red and one is green, he has more information and can eliminate five combinations from possibility, putting his chances at 1 in 6.

Note: The dice don't necessarily have to be color coded. If one die has some sort of distinguishing mark on it that he is aware of then he has the same information as if they were color coded.

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Originally Posted by arcimede$

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Originally Posted by regnis

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

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Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

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How right you are, if the peeker sees two twos The odds are 100% there is another two. If he sees any other number the odds are 0% there is a another two, but that has nothing to do with the original question Regnis' picture notwithstandig. And I don't think I want him representing me against the IRS.

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Originally Posted by Alan Mendelson

Arc, I've done the work. When I have a 2 showing and throw a die, one out of six times it's another 2.

That's the work because that's the question.

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No Alan, that's not the question.

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?

Using a craps analogy, let's say you throw the dice down the felt and one of them lands on a 2. The other bounces off the table.

Does the crew let you throw just 1 die in order to complete the roll, add it to the 2 and resolve all the bets on the table?

No. They call it a "no roll" and make you roll 2 dice again. You know why that is right?

While it's still 1 in 6 that you could end up "sevening out" it also becomes 1 in 6 that you could end up with a 3, hard 4, 5, 6 or 8 instead. The chances of you winning a bet on any of those numbers goes up dramatically if you are allowed to just roll 1 die!

By keeping both dice in play for every roll, any of the 36 combinations could come. The chances of a seven out remain at 1 in 6 but the odds for the other numbers 3, hard 4, 5, 6, 8 revert back.

With regards to the original question, both dice are in play for you, but NOT for the peeker. The peeker knows which dice is a 2. You do not.

The question:

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


would elicit different correct responses from the peeker and from the person who has not seen any of the dice.

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Originally Posted by arcimede$

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

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Absolutely false statements.

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?

It comes down to this -- what happens if neither die is a 2?

Or in other words, does he always truthfully say, "At least one die is an X."? In this case, assuming he picks a random die to express the value (either the one closer to him, first one he saw, etc.), then the answer would be 1/6.

But if he can only say this when one of the dice is a 2...and doesn't say anything if neither die is a 2, then it's 1/11.


Can we agree on that, Alan?

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Originally Posted by RS__

It comes down to this -- what happens if neither die is a 2?

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If neither die is a 2 then the peeker cannot TRUTHFULLY say "at least one die is a 2."

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Originally Posted by Alan Mendelson

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Originally Posted by RS__

It comes down to this -- what happens if neither die is a 2?

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If neither die is a 2 then the peeker cannot TRUTHFULLY say "at least one die is a 2."

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Well yeah, duh. That's not quite what I was asking though.....

If you are looking at a graph of dice, or if you are examing two dice, YES there are 11 different combinations of the two dice that include a 2, and only ONE of those 11 combinations will show 22. 1/11

But the question asks us to consider what happens when two physical dice are put into a cup and rolled and it is determined that at least one of those two dice is showing a 2 when it comes to rest. That particular question asks us to consider the second die in the problem: one out of six faces will also have a 2. 1/6

As Ive been saying over and over again... the questions are not the same. And the 1/11 people are answering the question about the two physical dice that have been rolled with the graph or thinking that would be used when looking at a graph of dice results or examing two dice.

Is that a little bit easier to understand now?

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

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Absolutely false statements.

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?

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Silly nonsense. If the die he sees when he sees only one die is not a 2 and the die he does not see is a 2, then he is not answering the question "correctly". Face-palm.

Using the red/green dice, when "at least" one die is a two, there are ELEVEN possible faces for the other one because the NON-PEEKER does NOT know which two is showing.

Either 1g, 2g, 3g, 4g, 5g, 6g, 1r, 3r, 4r, 5r, 6r OR 1r, 2r, 3r, 4r, 5r, 6r, 1g, 3g, 4g, 5g, 6g.

Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY. I say "at least one is an ace, what is the probability the other is an ace?" Any of the 11 cards could be the "other one." Same 1 in 11 as the dice problem and people problem I mentioned earlier

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Originally Posted by Alan Mendelson

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Originally Posted by regnis

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

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(You are brilliant, regnis.)

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My cross to bear Alan. yet the beat goes on here.

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Originally Posted by jbjb

Using the red/green dice, when "at least" one die is a two, there are ELEVEN possible faces for the other one because the NON-PEEKER does NOT know which two is showing.

Either 1g, 2g, 3g, 4g, 5g, 6g, 1r, 3r, 4r, 5r, 6r OR 1r, 2r, 3r, 4r, 5r, 6r, 1g, 3g, 4g, 5g, 6g.

Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY. I say "at least one is an ace, what is the probability the other is an ace?" Any of the 11 cards could be the "other one." Same 1 in 11 as the dice problem and people problem I mentioned earlier

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JB--As soon as you say one is an ace, the other cards in that suit are disqualified. If it is the ace of spades, there can't be 2 aces of spades, so the other spades are out. That only leaves the 6 hearts. It is the same as the dice.

Eddie took a shot at my question about the peeker, albeit a failed shot. I'm waiting for someone to explain how the same dice can have different odds.

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Originally Posted by arcimede$

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Originally Posted by Alan Mendelson

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Originally Posted by arcimede$

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

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Absolutely false statements.

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?

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Silly nonsense. If the die he sees when he sees only one die is not a 2 and the die he does not see is a 2, then he is not answering the question "correctly". Face-palm.

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Arc, now you're going to start playing word games? Your response was to regnis who was talking about one die being a 2. I was challenging your use of the English language and what "at least one" means. Of course if the peeker sees two 6s he cannot truthfully say at least one die is a 2 which sets up the original question.

Take your face palm and slap yourself.

Regnis: too bad you don't have this gang in court. They would be pleading for a deal. They don't know what hit them.

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Originally Posted by Alan Mendelson

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Originally Posted by regnis

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

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Warning: if you are ever in court and the other side is represented by regnis, you lost.

(You are brilliant, regnis.)

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I'm sorry, but I just have to say I am marshalling a significant portion of my personal restraints not to have an undiplomatic response here.

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Originally Posted by jbjb

Using playing cards. I take the ace through 6 of spades and ace through 6 of hearts. In their own piles. Mix them up separately and pull one from each pile and look at both SIMULTANEOUSLY. I say "at least one is an ace, what is the probability the other is an ace?" Any of the 11 cards could be the "other one." Same 1 in 11 as the dice problem and people problem I mentioned earlier

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Very good, jbjb. If your "question" allows all 12 cards to be in one pile, then the answer is 1/11 because you have picked and identified the ace of spades and all of the other 11 cards are still "in play."

But if there are two piles -- spades in one pile and hearts in a second pile -- and the Ace of spades is the only card you can choose from the Spades pile, then there are 6 hearts that remain "in play." Of those 6 hearts one is the ace of hearts.

(pause)

My goodness, I just reread your question and you said there are two piles. So this means you got the answer wrong to your own question. If the ace of spades has already been chosen from your spades pile and only the hearts pile remains the answer can only be 1/6.