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

The question was intentionally worded that way, just as the GRE logic portion was intentionally written so as to provide a verbal maze and just as Alan's law exam used invented language to provide obliqueness.

The question was written to divide the reading audience into those tricked and those not tricked.

I despise when the best and brightest spend their time with these kinds of intellectual masturbations. I mean, really? This is the best use of your time, energy, skills and logic? To write convoluted opaque crap in efforts to demonstrate cleverness?

The WoV crew is a math club, not some group of gambling experts. If they want to be spend their days being clever by writing purposefully torturous prose, let them. As far as actual gambling goes, that entire forum appears woefully overrated.

Quote:

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Originally Posted by redietz

The question was written to divide the reading audience into those tricked and those not tricked.

I despise when the best and brightest spend their time with these kinds of intellectual masturbations. I mean, really? This is the best use of your time, energy, skills and logic? To write convoluted opaque crap in efforts to demonstrate cleverness?

---

I readily admit I, myself, was tricked by the question into believing the answer was 1/6 for a few minutes after Alan first presented it here. It was only after a more careful look did I become a 1/11 convert. There's no shame for me in admitting I was wrong when I first looked at the question. I actually liked the question! It indirectly provides a valuable life lesson about how the beautiful subtleties of mathematics can fool the instinctive mind in a practical application.

I'm not so impressed. Our minds are designed to prioritize personal experience and anecdotal evidence over the abstractness of probability because, well, we don't really have a "probability sense." So bad judgements are made every day by all of us all of our lives. The last thing we need is for those who can do so to manipulate language to cause further confusion.

Quote:

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Originally Posted by Alan Mendelson

OMG Arc you can't use the same die twice to solve the problem.

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I used the red die once and the blue die once. That is obvious from my description. And that is obviously the meaning of "at least one die". If they meant just one die then that is what they would have said. The fact you intentionally lie about this tells me you are not being honest in this entire discussion. Bad Alan.

Arc, why are you accusing me of lying? You said I "intentionally lie" -- and what did I lie about?
Where does the word "lie" come from?

Alan, you lied when you said "OMG Arc you can't use the same die twice to solve the problem." Since I never used the same die twice, your statement is an outright LIE.

And no, this statement is NOT TRUE ... ""At least one" means that one or both of the dice in the problem can be a 2." ... What it means is what I told you above and you keep denying because it destroys your silly logic. It means EITHER die can be a two (or both). Now, if you would quit lying and redefining what other people are saying maybe we could have an intelligent discussion. However, you continue to lie and misquote others. Quit it, Alan.

I just tried to post and somehow the posts got screwed up with what I think were Arc's posts. So I deleted what I posted and if it deleted what Arc wrote, I am sorry.

Quote:

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Originally Posted by arcimede$

And no, this statement is NOT TRUE ... ""At least one" means that one or both of the dice in the problem can be a 2." ... What it means is what I told you above and you keep denying because it destroys your silly logic. It means EITHER die can be a two (or both).

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Arc I am going to disagree with you about the meaning of "at least." When you have two dice and tell me that at least one is a 2 it means that one or both. It doesn't mean "either." Now if you want to say that it's either, we have a problem. Because "either" means only one and it can't mean both. "At least" can include "both" dice.

Consider this:

Idioms
6.
at least,
at the lowest estimate or figure:
The repairs will cost at least $100.
at any rate; in any case:
You didn't get a good grade, but at least you passed the course.

And the word "lie" is not appropriate. I have not lied nor have I msquoted. You quit it.

I invite everyone to read thru redietz' posts. The original problem was a play on words meant to be taken in two different ways depending on if your life is driven by conflict or if you're at peace in general. Math people tend to get a special chill up their leg when they smell an argument. And no one is lying here, that's what's obvious. Arci hasn't much tolerance these days for views opposite his, and for him to be calling someone a liar....now THAT'S special.

As I said earlier, the way wizard presented his challange, it's a 1in11 proposition. I'd still be inclined to take on his 1in9 offer, simply because it's similar to playing -EV video poker, and everyone knows, anything can and will happen in a short term application of long term theory. But the way Alan interpreted the question--which in my opinion is the more grounded of the two, he is absolutely correct about the question strongly suggesting that the 2 shown "on at least one of the dice" was there to stay. In short, the geek who proposed the original question wasn't the super deep thinker he thought he was. He simply got outsmarted by his own question, the wizard then went all in with another of his bad choices, and the math club had no option but to follow and persist. When you think about it, and how they will never admit their error in judgement, it's quite entertaining.

An intentionally confusing framing device is being used to point out math or logic errors by readers, but the authors of the framing device seem to think their ability to be confusing should be applauded. This is like anti-teaching.

Again, Alan, myself, and Rob all have writing backgrounds. We avoid this kind of Don King trickeration at all costs. It gives me the willies when people intentionally do this. If nuns were teaching technical writing and you did this, your knuckles would be smacked.

Quote:

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Originally Posted by Alan Mendelson

Arc I am going to disagree with you about the meaning of "at least." When you have two dice and tell me that at least one is a 2 it means that one or both. It doesn't mean "either." Now if you want to say that it's either, we have a problem. Because "either" means only one and it can't mean both. "At least" can include "both" dice.

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There you go again. I said " It means EITHER die can be a two (or both)". And you reply with "you want to say that it's either ... it can't mean both" ..,. Did you fail to see my words "or both"? Looks like you are lying once again. You just quoted me wrong. It happens continually.

Look I already said this above. The red die could be a 2 or the blue die could be a 2 or they could both be 2s. That is what "at least" means.

Quote:

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Originally Posted by Alan Mendelson

Consider this:

Idioms
6.
at least,
at the lowest estimate or figure:
The repairs will cost at least $100.
at any rate; in any case:
You didn't get a good grade, but at least you passed the course.

And the word "lie" is not appropriate. I have not lied nor have I msquoted. You quit it.

---

You've done both, repeatedly. Now, I cannot tell if this is intentional since I can't read your mind. But, your typed words are lies. That is not in doubt.

Hold it Arc, I admit I am totally mixed up and confused.

Let me start at the beginning with the term "at least" and your red and blue dice.

At least means the red die can be a 2 or both the red and blue dice can be a 2.
At least also means the blue die can be a 2 or both the blue and red dice can be a 2.
BUT at least does not mean you can consider the blue die to be a two and then switch it to the red die to be a 2.

Do we agree?

Quote:

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Originally Posted by Alan Mendelson

Hold it Arc, I admit I am totally mixed up and confused.

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No kidding.

Quote:

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Originally Posted by Alan Mendelson

Let me start at the beginning with the term "at least" and your red and blue dice.

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Oh, so now you are actually going to read what I wrote. How interesting.

Quote:

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Originally Posted by Alan Mendelson

At least means the red die can be a 2 or both the red and blue dice can be a 2.
At least also means the blue die can be a 2 or both the blue and red dice can be a 2.
BUT at least does not mean you can consider the blue die to be a two and then switch it to the red die to be a 2.

Do we agree?

---

You can't switch it on any given throw but the sum of all the DIFFERENT throws can have either the red or the blue die being a 2.

You throw the dice ... it has a red 2 and a blue 3. That is one possibility.
You throw the dice again ... it has a blue 2 and a red 3. That is another possibility. (Do you agree?)

No one "switched" anything. They are two distinct possibilities.

Now, you keep on doing this and you find out you can have 11 distinct possibilities. A red 2 with a blue 1,3,4,5,6 .... a blue 2 with a red 1,3,4,5,6 ... and a red 2 and a blue 2.

All of these are based on different throws of the dice when "at least" one of the dice has a 2.

On this we can agree or you can continue to ignore reality.

Yes, Arc, I understand what you are saying. But please put it in the context of the original question because that is what I am willing to discuss. And I will continue to discuss it.

Now there was some discussion that we didn't know which of the two dice contained a "2" and I explained it didn't matter. Choosing either die as the 2 gives you six other options. Choosing both dice as 2 also gives you six other options. But when you say Dice A is a 2 and then say Dice B is a 2 to justify the 1/11 answer -- that is plain wrong.

Quote:

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Originally Posted by Alan Mendelson

Yes, Arc, I understand what you are saying. But please put it in the context of the original question because that is what I am willing to discuss. And I will continue to discuss it.

Now there was some discussion that we didn't know which of the two dice contained a "2" and I explained it didn't matter. Choosing either die as the 2 gives you six other options. Choosing both dice as 2 also gives you six other options. But when you say Dice A is a 2 and then say Dice B is a 2 to justify the 1/11 answer -- that is plain wrong.

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No one said " Dice A is a 2 and then say Dice B is a 2" on the same throw. We are talking about odds, probabilities ... do you understand what that means? As I stated above,

One possible throw has a red 2 and a blue 3.
Another possible throw has a blue 2 and red 3.

Both cases meet the definition in the original question. At least one 2 is present.

These are two separate possible throws. As such, they make up the part of the list that needs to be considered when discussing probabilities. Once you list all the possibilities you can then compute the odds. Since there are 11 possible throws as I listed above, with only one of them being 2-2, you end up with the odds being 10:1 (or 1 in 11). You aren't "choosing" any die to be a 2. You are considering the total number of possible throws where either die (or both) can be a 2.

This gets back to what redietz has been saying all along. It's a trick question. You have been misinterpreting/redefining the question to fit you initial erroneous answer. You were wrong. Admit it and move on.

EDIT:Alan, the question you are answering would need to be stated differently. To get 1-6 the question would need to ask when the die that was viewed is a 2 what are the odds the other die is a 2.

Well said blount2000. And indeed, for every roll you've seen with at least a 2 in it, you probably saw a 2-2 1/11 of the time (or close to 1/11).

Interesting that Alan agrees that it satisfies the original question. But then he follows up saying the answer is still 1/6. I'm not sure if that's really funny or really sad....

RS___ I am not saying that when you see a 2 playing craps that you will see another 2 1/11 times. I said when you see a 2 playing craps, the chance of seeing 2/2 is a 1/36 event.

Here's what I said before:

Quote:

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Originally Posted by Alan Mendelson

Your statements 1 and 2 are correct, blount2000. Whether or not you saw 2-2 when at least one of the two dice is a 2 one-sixth of the time doesn't matter. Frankly, when you roll two dice the chance of getting 2-2 is 1/36.

The original question and the question to be answered is simply this: if you have at least one 2 showing, what are the odds for the second die to be a 2 as well. And that is the same as rolling one die with a 2 showing 1/6 times.

If you are not seeing 2-2 with one 2 showing one-sixth of the time you're just like me. Like I said, that would be a 1/36 chance.

---

Rolling two dice at the craps table was not the question either. And I guess it will never be settled. Thanks for stopping by.

Quote:

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Originally Posted by arcimede$

No one said " Dice A is a 2 and then say Dice B is a 2" on the same throw. We are talking about odds, probabilities ... do you understand what that means? As I stated above,

One possible throw has a red 2 and a blue 3.
Another possible throw has a blue 2 and red 3.

Both cases meet the definition in the original question. At least one 2 is present.

These are two separate possible throws.

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Thanks Arc you made that clear.

One die is a 2. For both to be a 2, the other has to be a 2. What are the odds of that----1 of 6. The one that was a 2 can't suddenly become a 1,3,4,5, or 6 to create more combinations. I once got beaten pretty badly for having dice like that!!!!

Once a 2, always a 2. Arci's dice are chameleons.

Quote:

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Originally Posted by regnis

One die is a 2. For both to be a 2, the other has to be a 2. What are the odds of that----1 of 6. The one that was a 2 can't suddenly become a 1,3,4,5, or 6 to create more combinations. I once got beaten pretty badly for having dice like that!!!!

Once a 2, always a 2. Arci's dice are chameleons.

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That's interesting. But I think your comments are more fitting for the videos posted by the WOV folks. They both had the dice changing.

They're still talking about me over on the WOV forum and I am really surprised at what they're saying.

First from miplet who did one of the videos:

"I think Alan's missing the fact that once you roll two dice at the same time until there is at least one 2, there are only 11 possible outcomes. Just one of those outcomes is a pair of 2s. If he does understand this, I don't know how he is getting 1 in 6."

I'm sorry miplet but once you do that roll with two dice and you know "there is at least one 2" there are only 6 possible outcomes -- which are the six faces on the other die.

And from the Wizard:

"I'm sure deep down Alan knows he is wrong. Otherwise, he would be betting thousands of dollars as opposed to lunch money. However, I truly believe he doesn't understand the flaw in HIS argument. I tried to explain it in my video. He is simply can't get past a die having six sides."

That's right, I can't get past a die having six sides.

And it gets funnier. From Ibeatyouraces:

"12 faces on the dice. One is known, but since we can't know exactly which one, you CANNOT discount the other 11.

I'm not a "math guy" and I can clearly see this."

No, I can't discount all of the other 11. But I can eliminate from consideration the five other faces that are on the die with the 2.

Alan, they're still talking about and making fun of you on WoV because they indeed have recognized you exposed a flaw in both the original question wording and how they all quickly & very unwisely followed the wizard down the tunnel of mis-interpretation with their blinders firmly on. They now have no choice but to dig in. And the fact that RS keeps showing up here on orders from Shack just to repeat the same verse, is pure gold, if you know what I mean. :)

I think the original question refers to one throw and one peek. It's part of the verbal illusion.

You know, Shackleford making that "Alan can't get past a die having six sides" comment is pretty funny. Maybe it's deserving of multiple threads and 300 responses and loads of debate. Was Shackleford being unbelievably stupid or was he making a cogent comment about Alan's psychology? That line could be taken either way.

Perhaps we "writing guys" should repeat ad nauseam how intellectually superior we are to anyone who could stupidly write something that could be interpreted, at face value and in isolation, as really, really dumb.

Naaawww, we don't have the time or inclination.

The show continues. Now the participants on the WOV are arguing about the rules and the conditions in the original question:

"Ayecarumba" says: What I am proposing is that the peeker would call out whatever was on the observed die.

"Indignant" says: You're having a problem, son. The scenario does not have "the observed die." It has "an observed pair of dice." Never is the peeker restricted to observing only one die. He's obligated to see both, and if any deuce manifests, he's further obligated to say "at least one of the dice is a two."

Let me remind everyone on the WOV what the original problem said:

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?

I don't think the peeker was obligated to say anything but the truth. And when he said "at least one of the dice is a 2" it could have been that he saw one 2 or a pair of deuces.

Quote:

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Originally Posted by Alan Mendelson

I said when you see a 2 playing craps, the chance of seeing 2/2 is a 1/36 event.

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THIS is gold.

Quote:

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Originally Posted by regnis

One die is a 2. For both to be a 2, the other has to be a 2. What are the odds of that----1 of 6. The one that was a 2 can't suddenly become a 1,3,4,5, or 6 to create more combinations. I once got beaten pretty badly for having dice like that!!!!

Once a 2, always a 2. Arci's dice are chameleons.

---

Yes, my dice are chameleons. They can be either red or blue to help folks understand the problem. On any given throw the red die can be 2 or the blue die be a 2 (or both). If you ask what are the odds of the blue die being a 2 when the red die is a 2 the odds are 1-6. Or, if you ask what are the odds of the red die being a 2 when the blue die is a 2 the odds are 1-6. But, those are NOT the question. The question allows EITHER die to be a 2 which means all those situations could occur.

Arc I think the point is this: I understand what you are saying now. You are just using colors instead of saying, for example, the die on the left or the die on the right. You are not saying that the numerical value will change.

For example on roll #1 if you say the Red=6 and Blue=5 there will not be a change for roll #1.

Do I have it correct now?

The problem was that others have suggested changing dice "values" to justify their 1/11 claim, and they have used "colors" to illustrate what they are saying -- and I think that is what caused confusion. I know it caused confusion with me.

Quote:

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Originally Posted by Alan Mendelson

For example on roll #1 if you say the Red=6 and Blue=5 there will not be a change for roll #1.

Do I have it correct now?

---

What does that even mean, Alan?
On any roll of the dice, any slamming of the cup, there will be no change of anything. The outcome would be the outcome. The probability of that outcome, until it's revealed, of being a pair of deuces, does not change. When it's been revealed what the full outcome is, by lifting the cup, then the probability of the dice being a pair of deuces will either be 0% (if it's not a pair) or 100% (if it is a pair). But until the great reveal, the probability of the pair is 1/11.
Are you now at the Eureka moment when you realise that the original question has the one correct answer of 1/11?
If so, what persuaded you, and are you grateful?

OnceDear we'll have to wait for Arc to respond. Thanks for dropping by. And I hope you collect on your bet with the Wizard. Send my regards.

And welcome back to the forums Rob. Unbanned how often now?
Gotsta say Alan is a generous host.

(:o)

Quote:

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Originally Posted by OnceDear

And welcome back to the forums Rob. Unbanned how often now?
Gotsta say Alan is a generous host.

(:o)

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That would be "forum" once dear. If you haven't heard or can't tell by how often my name appears on the forum now owned by online casino activists, I'm too feared over there to be "welcomed". BTW I saw where the collection of genius moderators banned mickeycrimm. You mean they finally figured out his stories of gambling-prowess were nothing more than what one would get when visiting Scotty of Scotty's Castle lore? A little slow for the self-proclaimed Walter O'briens of the world, wouldn't you say??

Quote:

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Originally Posted by Rob.Singer

That would be "forum" once dear.

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I stand corrected Rob. Oh, and it's Oncedear, all one word.
But I respectfully withdraw my welcome because of your hostile tone. 'Gotsta' was bad English too, do you wish to pull me up on that? Maybe I misused an apostrophe somewhere.

I know that there is history and hostility twixt you and that other forum. Not my history: Not my hostility.

I wouldn't say anything about Scotty or Walter, because those names are not familiar to me and I'm not inclined to google them. My opinion on the banning of MC is all by the by.

Wouldn't all 1/6-ers be persuaded by the fact that if they put money on a bet which follows the OQ exactly, i.e. roll dice,peek and announce if at least one deuce is showing and if yes,then bet on 8 for 1 odds for a pair of deuces, they will lose in long term?

Doesn't that mean that the real,physical dice concur that indeed the probability of a pair of deuces when at least one deuce is showing is 1/11 in the real,physical world?

Or do they still believe the bet will win them money?
Cause it was proved and demonstrated (and even the 1/6-ers know it) in multiple ways that the 11 combos of showing at least one deuce are "in play" before any particular throw and they WILL come out 11/36's of the time. And only 1 out of those 11/36's will there be a pair.

Quote:

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Originally Posted by kewl

Wouldn't all 1/6-ers be persuaded by the fact that if they put money on a bet which follows the OQ exactly, i.e. roll dice,peek and announce if at least one deuce is showing and if yes,then bet on 8 for 1 odds for a pair of deuces, they will lose in long term?

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What Kewl, you mean like when I offered to stake my entire and not unsubstantial pension fund? I got no takers for that. Wife is grateful.

When I roll two dice I get 2-2 1/36 of the time -- not 1/11 and not 1/6. When I know one of two dice is a 2 the chance of another two on the other die is 1/6. What bet do you want?

Quote:

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Originally Posted by Alan Mendelson

When I roll two dice I get 2-2 1/36 of the time

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Yep.
And you get at least one deuce 11/36 of the time, yes?
And no deuce 25/36 of the time, yes?

So, since you'll be interested only in those 11/36 of the time when at least one deuce is under the cup, there will be only one 2-2 for those 11 times,yes?

OnceDear,

Yeah, I'll bite. In my opinion, misusing an apostrophe is usually a more egregious intellectual error than getting the dice problem "wrong." What makes you think your dice problem trivia is somehow more important?

This is like a practical joke on you Alan. All these people know it's 1 of 6 but they are playing a joke trying to get you to switch.

This is how they get convictions here in Chicago Alan. You know you didn't do it but they will keep telling you that you did it until you can't take it anymore and give in and confess.

Stay strong Alan!!!! On my dice the 2's don't disappear and reappear. Once you roll a 2, it's a 2. The other die only has 6 options--1 to 6. I've got the proof to exonerate you.

And welcome back mr. Singer!!!!!!!

The bet is as follows:

Roll 2-Anything, lose.
Roll 2-2, get paid 9:1.
Roll anything else, bet pushes.

Only a moron would take this bet!

Regnis go post the common sense of how a 2 is always a 2 on a die that no longer moves, on the WoV forum, then watch as the super duper math conspiracy theorists all attack you at the same time. We already know of their thin skin as shown by the gamblers wizard sent over here for his own CYA needs.

Oncedear, Scotty was our most famous made-up story teller of all time. Walter O'brien was a true genius. From here--not the UK.

Quote:

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Originally Posted by regnis

This is like a practical joke on you Alan. All these people know it's 1 of 6 but they are playing a joke trying to get you to switch.

This is how they get convictions here in Chicago Alan. You know you didn't do it but they will keep telling you that you did it until you can't take it anymore and give in and confess.

Stay strong Alan!!!! On my dice the 2's don't disappear and reappear. Once you roll a 2, it's a 2. The other die only has 6 options--1 to 6. I've got the proof to exonerate you.

---

But if you follow the original question word by word you'll get the inevitable 1 in 11 ratio, so how can it be?

Let's make it even more childish.

Imagine a pair of dice which ,for the purpose of better visual explanation, always rolls a perfect distribution of all 36 combinations on every 36 rolls.
So that, when you roll them 36 times you get all 36 combos one after the other, no repeats. I'm guessing we can all agree that is not how it is in reality, but it is a simple representation of what the distribution is expected to be in infinity since its derived from the pure fixed probability for each and every die.

So you roll those dice 36 times and you get 25 "no deuce"'s and 11 "at least one deuce"'s. With exactly one "deuce-deuce" among those 11 "at least one deuce"'s, of course.

And again, new 36 rolls and the result is again - 11 "at least one deuce"'s and exactly one 2-2.

So, if you are to bet $1 on those 11 "at least one deuce"'s ,which the peeker meticulously announces, at odds of 8 for 1 that a pair of two's will be under the cup , you'd have lost $2 total.

So how can that be?
How can it be that a 1 in 6 probability of "the other die" being a 2 actually yields an 1 in 11 actual results?

Doesn't this tell us that the dice themselves actually agree with the 1/11 bunch?
Doesn't this mean that 1 in 6 is actually the answer to a different question?

I'd like to offer my sincerest thanks to Regnis and Rob.

Quote:

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Originally Posted by kewl

But if you follow the original question word by word you'll get the inevitable 1 in 11 ratio, so how can it be?

Let's make it even more childish.

Imagine a pair of dice which ,for the purpose of better visual explanation, always rolls a perfect distribution of all 36 combinations on every 36 rolls.
So that, when you roll them 36 times you get all 36 combos one after the other, no repeats. I'm guessing we can all agree that is not how it is in reality, but it is a simple representation of what the distribution is expected to be in infinity since its derived from the pure fixed probability for each and every die.

So you roll those dice 36 times and you get 25 "no deuce"'s and 11 "at least one deuce"'s. With exactly one "deuce-deuce" among those 11 "at least one deuce"'s, of course.

And again, new 36 rolls and the result is again - 11 "at least one deuce"'s and exactly one 2-2.

So, if you are to bet $1 on those 11 "at least one deuce"'s ,which the peeker meticulously announces, at odds of 8 for 1 that a pair of two's will be under the cup , you'd have lost $2 total.

So how can that be?
How can it be that a 1 in 6 probability of "the other die" being a 2 actually yields an 1 in 11 actual results?

Doesn't this tell us that the dice themselves actually agree with the 1/11 bunch?
Doesn't this mean that 1 in 6 is actually the answer to a different question?

---

Don't help them.. you'll cost us money!

Quote:

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Originally Posted by jbjb

Don't help them.. you'll cost us money!

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You helped them too jbjb. Even I helped them. We live and learn.
Now with cold hard cash at stake, I'm grateful to anyone who helps demonstrate that Alan is correct and to anyone encouraging him to 'stick it to' those maths guys from WOV.
I'm on Alan's side now.

$;o)

Quote:

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Originally Posted by kewl

Yep.
And you get at least one deuce 11/36 of the time, yes?
And no deuce 25/36 of the time, yes?

So, since you'll be interested only in those 11/36 of the time when at least one deuce is under the cup, there will be only one 2-2 for those 11 times,yes?

---

Regarding the cup question: when one die shows a 2 the other five faces are not eligible. That leaves 1/6.

Quote:

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Originally Posted by Alan Mendelson

Regarding the cup question: when one die shows a 2 the other five faces are not eligible. That leaves 1/6.

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Way to go Alan. Of course they're not. You stand your ground mate!

Quote:

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Originally Posted by OnceDear

You helped them too jbjb. Even I helped them. We live and learn.
Now with cold hard cash at stake, I'm grateful to anyone who helps demonstrate that Alan is correct and to anyone encouraging him to 'stick it to' those maths guys from WOV.
I'm on Alan's side now.

$;o)

---

https://vine.co/v/b9IXzIqe9F2

Quote:

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Originally Posted by Alan Mendelson

Regarding the cup question: when one die shows a 2 the other five faces are not eligible. That leaves 1/6.

---

But the question was:

So, since you'll be interested only in those 11/36 of the time when at least one deuce is under the cup, there will be only one 2-2 for those 11 times,yes?

I don't see how you answering this question?
So, yes or no?

Also please consider my reply to regnis from above.
What are your thoughts about the questions raised?

Quote:

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Originally Posted by kewl

https://vine.co/v/b9IXzIqe9F2

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Nice. But seriously, we have to admit that Alan has been right all along..... If I'm to collect on my advantage play.
:D:D:D:D:D:D:D:D

Quote:

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Originally Posted by kewl

But the question was:

So, since you'll be interested only in those 11/36 of the time when at least one deuce is under the cup, there will be only one 2-2 for those 11 times,yes?

I don't see how you answering this question?
So, yes or no?

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I think the bet you are proposing and has been proposed before is a good bet. But it involves the variance of dice so I could lose.

Since you're so sure the answer is 1/11 why don't you offer me even better odds?

Better yet-- let's skip rolling two dice and just set one as "2" and roll the "other die"? But wait-- you folks have been saying you can't tell which one the "other die" is.

I'm proposing even "better" bet - zero variance.
We set a pair of dice in all 36 possible combos one at a time and then start all over. Every time there is at least one 2 you bet and if there is 2-2 you win 8 for1 (alright , I'm giving you 9 for 1, I feel generous). If not 2-2 - you lose your bet (its a flat betting though,no cheating).

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Originally Posted by Alan Mendelson

I think the bet you are proposing and has been proposed before is a good bet. But it involves the variance of dice so I could lose.

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Indeed, one session of this game would involve a lot of variance.
Whoever claims victory could hardly call it a proof. I'm already relying on that variance factor to pull me back from the brink....Of p1551ng myself laughing

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Originally Posted by kewl

I'm proposing even "better" bet - zero variance.
We set a pair of dice in all 36 possible combos one at a time and then start all over. Every time there is at least one 2 you bet and if there is 2-2 you win 8 for1 (alright , I'm giving you 9 for 1, I feel generous). If not 2-2 - you lose your bet (its a flat betting though,no cheating).

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That's just silly Kewl. This is a two dice problem.

But there is a way that Alan could eliminate variance while still using fair and random throws of the dice. He could hedge his bets.

Regarding your bet: if one die shows a 2, eliminating five faces, how can you believe 2-2 is still 1/11? I want you to think carefully before betting.

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Originally Posted by Alan Mendelson

Arc I think the point is this: I understand what you are saying now. You are just using colors instead of saying, for example, the die on the left or the die on the right. You are not saying that the numerical value will change.

For example on roll #1 if you say the Red=6 and Blue=5 there will not be a change for roll #1.

Do I have it correct now?

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There's never a change after the dice are rolled. The result is the result. Not sure where you are going with this.

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Originally Posted by Alan Mendelson

The problem was that others have suggested changing dice "values" to justify their 1/11 claim, and they have used "colors" to illustrate what they are saying -- and I think that is what caused confusion. I know it caused confusion with me.

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No one has suggested changing dice values on any given throw. The only changes that occur are the result of more throws. Every throw is a unique and independent event just like playing one hand of VP.

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Originally Posted by Alan Mendelson

I want you to think carefully before betting.

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I'm trying my best, but hey who could I blame for my losses?
So, you agree on my proposed - zero variance - bet then, eh?

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Originally Posted by Alan Mendelson

Regarding your bet: if one die shows a 2, eliminating five faces, how can you believe 2-2 is still 1/11? I want you to think carefully before betting.

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You didn't respond on my request to examine this:

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Originally Posted by kewl

But if you follow the original question word by word you'll get the inevitable 1 in 11 ratio, so how can it be?

Let's make it even more childish.

Imagine a pair of dice which ,for the purpose of better visual explanation, always rolls a perfect distribution of all 36 combinations on every 36 rolls.
So that, when you roll them 36 times you get all 36 combos one after the other, no repeats. I'm guessing we can all agree that is not how it is in reality, but it is a simple representation of what the distribution is expected to be in infinity since its derived from the pure fixed probability for each and every die.

So you roll those dice 36 times and you get 25 "no deuce"'s and 11 "at least one deuce"'s. With exactly one "deuce-deuce" among those 11 "at least one deuce"'s, of course.

And again, new 36 rolls and the result is again - 11 "at least one deuce"'s and exactly one 2-2.

So, if you are to bet $1 on those 11 "at least one deuce"'s ,which the peeker meticulously announces, at odds of 8 for 1 that a pair of two's will be under the cup , you'd have lost $2 total.

So how can that be?
How can it be that a 1 in 6 probability of "the other die" being a 2 actually yields an 1 in 11 actual results?

Doesn't this tell us that the dice themselves actually agree with the 1/11 bunch?
Doesn't this mean that 1 in 6 is actually the answer to a different question?

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Why is this happening when the probability for "the other die" is 1/6?

Alan I swear you're a fucking retard....

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Originally Posted by RS__

Alan I swear you're a f****** r*****....

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Are you taking bets on that RS. Can you give me odds?

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Originally Posted by RS__

Alan I swear you're a f------ retard....

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What I think is happening is Alan doesn't truly read any of the comments. Even the ones he responds to. He just skims them at best. Hence, he doesn't have a clue what you said. Since he's already made up his mind I think he just assumes it is a waste of his time. The reason I said he was lying last night was kind of a slap in the face. I wanted to get his attention. It did for a brief moment and you could see from his response he hadn't read my previous comment at all ... and then he went right back into ignore mode.

So, I can see where you get that impression. You really are talking to a wall. In general, wall's aren't real bright.

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Originally Posted by arcimede$

So, I can see where you get that impression. You really are talking to a wall. In general, wall's aren't real bright.

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When watching someone trying to train a donkey to count, who's the least intelligent? Donkey? Teacher? Cheering onlooker?.... or the bookmaker taking bets on the Donkey

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Originally Posted by arcimede$

There's never a change after the dice are rolled. The result is the result. Not sure where you are going with this.



No one has suggested changing dice values on any given throw. The only changes that occur are the result of more throws. Every throw is a unique and independent event just like playing one hand of VP.

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Thanks Arc I just wanted to be sure I understood what you said. Walls do listen.

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Originally Posted by kewl

I'm trying my best, but hey who could I blame for my losses?
So, you agree on my proposed - zero variance - bet then, eh?

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No. And how many reasons do you need?

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Originally Posted by RS__

Alan I swear you're a fucking retard....

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Can you say this on the WOV forum? You can't say it here either. Don't do it again.

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Originally Posted by Alan Mendelson

No. And how many reasons do you need?

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I don't know, you didn't give any. What are your reasons?

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Originally Posted by kewl

I don't know, you didn't give any. What are your reasons?

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I'd venture that would be far too many dice. A different question.
When will you get back to the original question as asked? Alan's trying his best to get idiots like me to understand. If I can see the light, there is hope for you too.

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Originally Posted by kewl

I don't know, you didn't give any. What are your reasons?

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Here's one: when a 2 appears on a die it can't be changed to another number. If that were possible I would never seven-out.

Yes, on charts and graphs the answer is 1/11. With two real physical dice that are not rotated for convenience the answer is not.

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Originally Posted by OnceDear

When watching someone trying to train a donkey to count, who's the least intelligent? Donkey? Teacher? Cheering onlooker?.... or the bookmaker taking bets on the Donkey

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Touche. I'm thinking donkey is the least intelligent.

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Originally Posted by Alan Mendelson

Can you say this on the WOV forum? You can't say it here either. Don't do it again.

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My apologies. I will try to not do it again. It's just that sometimes when you post, it makes me say things I normally would not say [see one of my previous posts].

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Originally Posted by OnceDear

I'd venture that would be far too many dice. A different question.
When will you get back to the original question as asked? Alan's trying his best to get idiots like me to understand. If I can see the light, there is hope for you too.

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But its just one pair of dice. We set them one combo at a time as if we are throwing a pair of perfectly-distributed-dice and examine if there is a two showing. And then the bets are on.
I'm trying to see the light as well and if Alan can really state his point clearly maybe I'll convert too, who knows..."cough" doubtful "cough"

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Originally Posted by Alan Mendelson

Walls do listen.

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Only with simple straight forward comments. If the comment requires you to think you just dismiss it. So, no, the Alan Wall does not listen.

Quote:

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Originally Posted by Alan Mendelson

Here's one: when a 2 appears on a die it can't be changed to another number. If that were possible I would never seven-out.

Yes, on charts and graphs the answer is 1/11. With two real physical dice that are not rotated for convenience the answer is not.

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But I'm not claiming the 2 can be changed to another number.
Please read this and comment on the questions raised if you care:

Quote:

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Originally Posted by kewl

But if you follow the original question word by word you'll get the inevitable 1 in 11 ratio, so how can it be?

Let's make it even more childish.

Imagine a pair of dice which ,for the purpose of better visual explanation, always rolls a perfect distribution of all 36 combinations on every 36 rolls.
So that, when you roll them 36 times you get all 36 combos one after the other, no repeats. I'm guessing we can all agree that is not how it is in reality, but it is a simple representation of what the distribution is expected to be in infinity since its derived from the pure fixed probability for each and every die.

So you roll those dice 36 times and you get 25 "no deuce"'s and 11 "at least one deuce"'s. With exactly one "deuce-deuce" among those 11 "at least one deuce"'s, of course.

And again, new 36 rolls and the result is again - 11 "at least one deuce"'s and exactly one 2-2.

So, if you are to bet $1 on those 11 "at least one deuce"'s ,which the peeker meticulously announces, at odds of 8 for 1 that a pair of two's will be under the cup , you'd have lost $2 total.

So how can that be?
How can it be that a 1 in 6 probability of "the other die" being a 2 actually yields an 1 in 11 actual results?

Doesn't this tell us that the dice themselves actually agree with the 1/11 bunch?
Doesn't this mean that 1 in 6 is actually the answer to a different question?

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You are correct, Alan, this is a trick question. I want to explain why it's a trick question. It's something I think everyone is missing the point on. The question asks, "If you roll the dice, and at least one is a 2, what are the odds that BOTH dice rolled a 2?".

That's the wording of the original question, and it's worded pretty different than normal speech. Normally, you would ask the question, "If you roll the dice and at least one of the the die is a 2, what are the odds the OTHER die also rolled a 2?" You wouldn't ask about both dice, after explaining that one is already a 2, you would just ask about the OTHER die.

This is why it's a trick question. In the original question, the answer is 1/11, the odds of BOTH dice rolling a 2 is 1 in 11. And, that's what the question asks. It's not asking the odds of the other die rolling a 2, it's only asking the odds for BOTH dice rolling 2's.

In the second question, the one you would ask in normal speech, the chances of the OTHER die also rolling a 2 is 1/6. The odds of 1 particular die rolling a 2 is 1 in 6. And that's what the 2nd question asks. It doesn't ask the odds for BOTH dice rolling 2's, only the "other" die.

This is why the question is a trick question. It asks the question in a different way than you would normally speak, and by doing this it gives a different answer than would normally be given.

I think a lot of these math guys don't understand why it's a trick question, most would probably, incorrectly, give the same 1/11 answer to both wordings of the question. It just so happens, in this case they are correct, the question asks the odds of BOTH dice rolling a 2, that's 1/11. If the question was asked about the "other" die, you would be right with 1/6, and I'm guessing most of the math guys would give the wrong answer of 1/11. But, this is why it's a trick question.

If you do the experiment itself, you'll see that BOTH dice will roll 2's 1 in 11 times. And, the "other" die will have a 1 in 6 chance of rolling a 2, just like in your video. So, depending on whether you say BOTH dice, or if you say the OTHER die, you'll get 2 different answers.

The original question asks for BOTH dice, it does that on purpose, to get the 1/11 answer, because if it asked about the "other" die, it would get a totally different answer of 1/6 (and this would trick a lot of people who think it's 1/11). Hope this helps explain why this is such a tricky question for you.

Quote:

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Originally Posted by kewl

But its just one pair of dice. We set them one combo at a time as if we are throwing a pair of perfectly-distributed-dice and examine if there is a two showing. And then the bets are on.
I'm trying to see the light as well and if Alan can really state his point clearly maybe I'll convert too, who knows..."cough" doubtful "cough"

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I don't think Alan quite understands how dice even work. Really.

There was a discussion before on the WOV forum about how if a dice controller could change the house edge. Alan's opinion was that no -- you could not change the edge, because every number has a 1/36 and 2/36 chance of being rolled, simply because each die has 6 faces x 6 faces.

Even after I gave an example, like, what if you could throw a 7 1/10 times and a 6 1/4.5 times and an 8 1/5 times....would the edge change on the 6 and 8? Alan said no, it does not, because there are 5/36 ways to roll a 6, 5/36 ways to roll an 8, and 1/6 [6/36] ways to roll a 7.


The one game Alan thought he was a master of (craps)....he really knowing nothing about.

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Originally Posted by PharoahsWin

The original question asks for BOTH dice,

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Exactly and you know one of them is a two. So we are not interested in both dice anymore, just 'the other one'
I see where Alan's coming from. It's the same question. It has to have the same answer.

Wait wouldn't it be 1/36 for both dice to be a deuce? :claps_hands:

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Originally Posted by RS__

Wait wouldn't it be 1/36 for both dice to be a deuce? :claps_hands:

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It's ALWAYS 1/36.

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Originally Posted by OnceDear

It's ALWAYS 1/36.

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But the chance of rolling a 2 is 1/6.

(1/6) * (1/36) = 1/216


So the answer is really 1/216. Any1 interested in a side wager?

Quote:

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Originally Posted by RS__

But the chance of rolling a 2 is 1/6.

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I think you are wrong.
Take two dice.
Look at those dice.
Each die has 6 faces
That's 12 faces all together
So that's twelve different ways that they can land. Count them.
1 of those ways is Deuce Deuce.
So you can throw Deuce Deuce ON AVERAGE 1/12 of the time.
1/36 is just a trick played by the casino to maximise its edge.