Who's letting you here, Troll? Oh, yes, Alan is. Sad.Quote:
Originally Posted by Rob.Singer
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Who's letting you here, Troll? Oh, yes, Alan is. Sad.Quote:
Originally Posted by Rob.Singer
Lol, from the guy with 3,374 or so posts. And not a valid argument or theory among them.Quote:
Originally Posted by Rob.Singer
What's your iq? Which test? Just curious.
R. Singer (like a bird) is essentially harmless. Not the same thing as AP's who think that they ARE mathematicians/physicists.Quote:
Originally Posted by kewl
Yes two dice were thrown and we are told at least one is a 2. The answer then depends on one die which has six faces.Quote:
Originally Posted by kewl
All of this doesn't matter in answering the original question. Sorry. It's always been simple.Quote:
Originally Posted by kewl
No!Quote:
Originally Posted by Alan Mendelson
You get confused by the statement at least one is 2. Its not like saying the red/blue is 2. It is one OR the other. That makes possible either one is. Twice the more combos. On every single throw.
There are only two dice. It doesn't make a difference which die is the 2 because the answer is always 1/6 for any die in a two dice problem when you only consider one die.
Your 1/11 is a Rube Goldberg answer.
Sorry there is no reason to consider 11 combinations. You have a 2. To get 2-2 you need 1/6 faces on another die.Quote:
Originally Posted by kewl
OK--"it's one or the other". So let's say "one" is a 2. How many faces are on the "other"? Envelope please----6. Chance of a 2--1 of 6.Quote:
Originally Posted by kewl
Let me go over it again, to emphasize what regnis told you:Quote:
Originally Posted by kewl
Once a 2 has been identified on either die you only have to consider six faces on one die. And if a 2 is on both dice it is still the same: you only have to consider the six sides on one die.
No one said anything about counting up the faces on two dice and adding together the possible combinations -- this is needless. Not only is it needless but it's an exercise that is not required to answer the original question.
Not on every single throw. Single throws are the one-2 rolls by column or row but not both, or 2-2.Quote:
Originally Posted by kewl
Absolutely correct.Quote:
Originally Posted by Alan Mendelson
Either way, we agree on this.
OneHit, you've definitely grown on me. I think pretty much anything I had to say on this is in better hands. Thanks, and good luck.
I'll limit my future comments to (1) the personal value this topic has had for me and (2) the irony that some math folks are blind to their competency limitations with language, but that they get all snarky pointing out the math limitations of others. Math has its own language, but language has its own math, in a sense. Language is not the first language of many of these folks.
I don't understand why is this so hard to comprehend.
You have a bag with 36 balls - 25 red and 11 blue. One of the blue balls has a smiley face on it.
You draw a random ball out of the bag. If its red you disregard it and put it back in the bag, cause it is the blue ones you're after.
And every time you draw a blue one, the chances you have drawn the one with the smiley is 1/11.
Or,
You find a magical pair of dice which rolls perfect distribution of all 36 possible throws one after the other.
You roll them 36 times and you get 11 2-x and exactly one 2-2.
Do it on your own. Roll dice many times. Count. Report the results here. And be shocked.
Do you understand what "OR" means? You can't say "its one that is 2" without accounting there are twice as much combos for that "one". Subtract one combo (2-2) so you don't double count it and there you go.Quote:
Originally Posted by regnis
In order for you to "identify" a die you need to include the chances it (the die) has to become a 2. You throw two dice. Either die has equal chance to become a 2. And it will fulfill it's chances - half the time one will become 2 with 1/6 frequency and half the time the other. During each half the other die will be a non deuce 5/6 of the time.Quote:
Originally Posted by Alan Mendelson
kewl you just don't understand? This is not about a bag with 36 balls or any combination of 11 dice with a 2.
This is about TWO DICE and we know there is a 2 on at least one of them.
And the question is what would it take for us to have 2-2 showing on both dice? Really -- in plain, simple English that's the question.
And if we have 2 on one die, the other die in the problem has a 1/6 chance of also being a two.
That's all this question asked. You guys got all wrapped up in your math and statistics and probabilities and forgot to look at the simple question that has a simple answer.
And as proof I asked the WOV forum to show videos of your thought process with two dice. And there it was for everyone to see how you compounded the question and came up with the wrong answer.
Yes, tell the Wizard he's wrong too.
this is the most ridiculous comment I've read yet.Quote:
Originally Posted by kewl
Don't you understand that since either die can be your "2", you now have twice as much chances to see a 2-x, compared to what you would have if you were only looking for a specific die (only the red or only the blue die) ?Quote:
Originally Posted by Alan Mendelson
No I don't understand. When one die is a 2, there are 6 chances on the other die. It doesn't matter which die is the 2. And if both dice are showing a 2 it was still a 1/6 chance that either die is showing a 2.Quote:
Originally Posted by kewl
I encourage you to go ahead and bet real money on your believe that it is 1/6.Quote:
Originally Posted by Alan Mendelson
Roll the dice and count the results. It might come as a shock to you, but if you do that you will actually find out empirically that the odds are indeed 1/11.
Then post here your thoughts.
Yes, I know.Quote:
Originally Posted by Alan Mendelson
Go ahead and roll the dice. It will give you your proof eventually.
Correct. And there are 11 unknown faces that could be up, only one of which is a 2!Quote:
Originally Posted by Alan Mendelson
You made a bald, unsupported claim (regarding horizontal exclusive-or vertical). Well, not only is it unsupported - it's wrong. And I made that bald counter-claim.Quote:
Originally Posted by OneHitWonder
Here are all the ways you could have "at least one two" under the cup:
http://www.imageurlhost.com/images/k...vfowiygkmn.jpg
You are told by the peeker:
- "At least one of the dice is a 2." (verbatim)
- Not "the green die is a 2."
- Not "the purple die is a 2."
- Not "both dice are 2's."
- Not "one of the dice isn't a 2."
- Not "neither die is a 2."
- Not "one die is a 2, but the other one isn't."
Explain how you can eliminate any of those combos, from potentially being under the cup. Explain how you can restrict the possibilities to "ones with a green deuce." Explain how you can restrict the possibilities to "ones with a purple deuce." Remember, good buddy, you can't see them.
Look you guys are nuts. Are you trying to sell us snake-oil? Because that's what it sounds like when you come up with hocus pocus about there being 11 possible faces when we know that at least one die is showing a 2.
I am going to tell you this point blank. When either die is showing a 2 you can't count the other five faces on that die. End of discussion.
We deal with reality on this forum. Now go back to the WOV and bad mouth me there.
Alan (or anyone whom believes the answer to the question/bet as proposed in the first post is 1/6) could you please explain what is wrong with OnceDear's Excel spreadsheet?
I know there are a number of simulations out there…but I feel most should understand Excel.
I’m aware that Alan wrote…
But I’m not seeing how it’s not responding to the question. The spreadsheet picks two numbers from 1 to 6 (aka roll two dice). Then, it shows that when at least one of them is a 2 that both of them will be a 2 about 1/11 of the time. This IS the question.Quote:
OnceDear's spread sheet and all those other spreadsheets looked at the combinations of 2 dice that contained at least 1 two, and from that they said the answer is 1/11. They did not respond to the question.
So please, explain how his spreadsheet does not represent the question/bet.
Because Alan says so, plain and simple. The simulation is wrong and the question you wov-ers are asking is wrong and all is wrong. The only right thing here is Alan. And the answer is 1/6. Point blank.Quote:
Originally Posted by Zedd
There you go.
The spreadsheet does not represent the question. The spreadsheet applies to any two dice and shows the various combinations of any two dice.Quote:
Originally Posted by Zedd
In the question we have isolated that at least one die is a 2. So -- picture that die frozen as a 2. It does not change.
Given that at least one die is a 2 there can only be six faces on the second die used to answer the question. That's the 1/6 answer.
We've heard various excuses and reasons from the 1/11 "camp" about how we can't freeze one die as a 2. Well it's going to be treated as 2 because it's a question involving two dice -- and at least one them is a 2. Deal with it.
Zedd, welcome to the forum, and thanks for posting.
But,but,but the spreadsheet does isolate one die is a 2?Quote:
Originally Posted by Alan Mendelson
See the column where it says "Count of "At Least One Deuce" Showing" ? That's isolating one die as a two right there?
You're welcome, but thank you. Many of these sorts of "paradoxes" I have already thought about a lot before now in one form or another.Quote:
Originally Posted by redietz
Sometimes, we have "it" in the (sequential) language of our thoughts, and must struggle to put "it" into math; sometimes, the other way around. Sometimes, "it" isn't about language at all; but our (random-access or simultaneous) spatial senses instead.Quote:
Originally Posted by redietz
I leave you with an anecdote. One of the curious things, discovered upon examination after his death, about Einstein's brain was that the two lobes - lingual left and spatial right - were closer together than normal. This could have facilitated his ability to perform accelerated calculations. Princeton had hired him just for that purpose later on in his life. It was noted as well that he had the brain of a much younger man without the usual plaque of someone of 75.
The only spreadsheet I want to see shows one die as a two with the following options for the second die in this two dice problem: 1, 2, 3, 4, 5, 6.
The spreadsheet doesn’t necessary show all the various combinations of two dice though. All it does is simulate the rolling of two dice. If at least one of them is a two then the wager is on. He even paid 8 units for deuce-deuce and lose 1 unit for when it’s not deuce-deuce.Quote:
Originally Posted by Alan Mendelson
The spreadsheet simulates exactly the bet you described in post one.
This is not a rolling of two dice problem. It is separate from the bet with the Wizard. Frankly, I concede the bet because the way the Wizard and I agreed to our own bet was a limited number of rolls.
So the bet is not an issue. Don't make it an issue.
The issue is the original question.
The reality is the two sides will never come to an agreement. We who say the answer is 1/6 read the question one way, and you who say the answer is 1/11 do some sort of picking 36 balls out of a bag, or looking at charts of 36 dice combinations, or looking at a graph showing 11 combinations with a 2 and come up with your answer.
Go in peace. And thanks for posting.
Don't you guys know that Alan is just trolling? Let it go...
So then are you acknowledging that the Wizard bet has a probability of 1/11 to pay out? And it’s a different question that we are answering?Quote:
Originally Posted by Alan Mendelson
No. I am not saying it has a probability of 1/11. But I am saying he will probably win.Quote:
Originally Posted by Zedd
Are you aware of the specifics of the bet that I have with the Wizard? I tried to meet up with him in Vegas a couple of weeks ago when I was in town to meet with a client but I could not make it in time.
First of all none of you in the 1/11 camp has yet to read the "original problem" in the same way that those of us who say it's 1/6 read the problem.
You guys can't even tell me which die is showing the 2. Those of us who say it's 1/6 say it doesn't matter which die is the 2.
So, when you guys tell me it doesn't matter which die is the 2, OR, if you can tell me which die it is, then we can discuss a different bet.
Zedd you came to this forum late in the discussion. Early in the discussion I said the original problem resembled this event at a craps table:
Two dice are thrown and one die immediately comes to rest showing a 2. The second die becomes a spinner and spins like a top. What are the chances that the "spinner" will also land on a 2?
If you don't view this as the same as the original question, we are going in different directions on the same road.
This is clearly not the same as the original question.Quote:
Originally Posted by Alan Mendelson
The answer to your question is obviously 1/6, no doubt about that and everyone here will agree.
However, the question remains, how is it possible that you (and a few others obviously) can read the original question, then transform it in their mind to become your question, and even more so, boldly claiming it is the rest of us who don't understand the original question?
So kewl what do you think is the meaning of the original question? Do you think there are 36 dice in a bag, or eleven dice combinations showing a 2 and you have to pick from those 11 combinations?
Sorry. I don't buy it.
Alan, if it isn't obvious by now I'll give a helping hand: these guys all had no choice but to step in the same trap the wiz did when he jumped to a theoretical, more complicated and weakly supportable conclusion, as he rushed to be the first mensa to throw up an answer. All this group is doing is saving face for their mentor.
The question of course has a 1in6 answer when taken verbatim along with a picture in one's mind. But the intent of the question was to see who'd be the first to over-think it so they could incorporate as much theory as possible, thereby impressing the many members over there who only read math books and do not actually gamble. Revenge of the nerds, so to speak. But they never calculated on their slide rules that they'd ever run into folks here who are " Too smart for words, Too hot for nerds, Too tough for turds, & Too high for birds".
If the meaning of the question isn't clear to a fair number of the reading audience, and the question is intended for that reading audience, then the fault lies with the author of the question, not the reading audience. The responsibility to be clear lies with the author.
What about this is so difficult to grasp?
If the question can be misinterpreted, there is a problem with the question as written. Period. It doesn't take an IQ of 140 to understand this. It may take an obsession with one's IQ of 140 to ignore it.
I’m aware of the specifics of the bet…which, as far as I know, is basically what you had described in your original post. These are the same specifics that OnceDear used in his spreadsheet.Quote:
Originally Posted by Alan Mendelson
I’m also aware of the original question. And you’re right; it can be read in different ways. The original question is ambiguous because it’s unclear how the condition of ‘at least one deuce’ was satisfied. Therefore, we must make assumptions on what sample space to use in our calculations. With this particular problem—one assumption produces an answer of 1/6 while another assumption produces 1/11.
BUT…this ambiguity is eradicated in your original post. We now know that the condition is satisfied only when EITHER one of the two dice is a deuce. This is different from when a specific die is a two—which is a more difficult condition to meet. Therefore, we must look at all two-dice combinations with either dice being a deuce (there are 11 of them).
1/11 is the absolute, no doubt about it, correct answer to the question in your OP.
The meaning? Its a very straightforward question, there is no hidden meaning to it.Quote:
Originally Posted by Alan Mendelson
You roll pair of dice. Mind you, they both come to rest peacefully, no spinners, no nothing.
Given that at least one of them shows 2, what are the odds they are both 2.
What is your proof the odds are 1/6? Its the fact that a die (the other die lol) has 6 faces.
Now, go ahead and show this "proof" to any person you trust is good in probabilities.
No, scratch that , I don't trust your judgment.
Show this "proof" to any person which has a proven record in dealing on at least mid level with probabilities.
You don't have a proof (you can't have one, cause your conclusion is incorrect, very incorrect). You are only running your mouth here, honestly. You have no idea how do pair of dice work in reality.
It is not a matter of ambiguous wording or misunderstanding the question. No. As I earlier quoted a perfectly worded question you answered the same - 1/6.
The odds of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2 are?
You do understand that the Rawtuff's bet was word by word based on the original question, right? What made you disregard it as a proof to the answer then?
You understand that the simulation with the OnceDear's spreadsheet does exactly like the bet would've and what the results will look like after 50 000 trials, right?
What made you say it is not applying to the question then?
Listen this is the last time I am going to say it:
IT DOESNT MATTER WHICH OF THE TWO DICE IS A TWO. If Die #1 is a 2, then Die #2 has a 1/6 chance of being a 2.
If Die #2 is a 2, then Die #1 has a 1/6 chance of being a 2.
If both Die #1 and Die #2 show a 2, then the chance that either Die #1 or Die #2 was a 2 was still 1/6.
1/11 does not answer the question.
The spread sheet does not matter.
The graph of 36 dice combinations does not matter.
The graph showing all 11 combinations of dice with at least one die showing a 2 does not matter.
The bet only tried to monetize the problem but it doesn't matter either because anything can happen when you throw two dice.
So, it's over. And it has been over since the day the Wizard showed that video with him rotating the die with the 2 to show the 1/11 answer.
Alan, it's been over for a long time. Your denial does not change reality. You can continue to deny the facts forever just like you always do. It doesn't change a thing. 1-11 is the right answer and will always be the right answer.
Alan! Eureka! You're Right!
You had the key. Ignore the impossible faces.
I now have undeniable evidence and proof that your evaluation - one out of six chances (1/6) - is the correct answer to the pair-of-deuces puzzle.
http://www.imageurlhost.com/images/d...b1j3qw7vvc.jpg
Dotted cubes (dice) are utterly impossible, when that cube shows a Deuce.
My picture shows that you have 2 winners out of 12 combinations.
So pay attention, folks.
2 out of 12 is exactly the same as 1/6. Alan was right; is right; always will be!
You did it again, Indignant. You used FOUR sets of dice instead of just TWO.
We caught you. And you're not going to get it by us.
It's still 1/6 no matter how many pretty pictures you post.
Give it a rest here and go back to WOV. Thanks for coming by.
What? I'm agreeing with you.
I show a Red Deuce, with all its six partners.
I show a Blue Deuce, with all its six partners.
I'm only showing exactly one Red Deuce (with five red impossible ghosts, not real).
I'm only showing exactly one Blue Deuce (with five blue impossible ghosts, not real.)
The extra cubes are for illustration purposes only. (What coulda been, but was not in reality).
Dotted cubes (impossible) cannot show up.
Solid cubes (possible) are only to show what could have happened. (The solid ones ain't real either.)
Did you just react to my name? Or did you focus and read the post?
I'm only showing ONE RED DEUCE, and the six-ways going along with it.
I'm only showing ONE BLUE DEUCE, and the six-ways going along with it.
Of course, as you say, it doesn't matter which die is showing the deuce.
The only real combinations, possible, are at the ones at the ENDS OF THE ARROWS.
I didn't show 4 sets of dice.
I show 24 total cubes. If anything, that's TWELVE pairs of dice.
Can't you notice the difference between theoretical, versus actual?
And that's exactly right.Quote:
Originally Posted by Alan Mendelson
So, to visualize:
Die #1 will be a deuce 1/6 of the time. Die #2 will be a deuce 1/6 of the time also.
Now if we were only interested in scenarios where Die #1 is a deuce we would have Die #2 be also a two 1/6 of those scenarios, thus giving us a 1/6 chances we see 2-2 when, and only when Die #1 shows 2.
And if we were only interested in scenarios where Die #2 is a deuce we would have Die #1 come as deuce 1/6 of those scenarios, thus giving us a 1/6 chances we see 2-2 when, and only when Die #2 shows 2.
But we are not interested in those separate scenarios. It indeed does not matter which die is a 2.
So, we now have both dice equally likely to show a deuce (1/6 each) and we are interested in all of those scenarios where either of them is 2.
That makes twice as many chances we see a 2-x --> 1/3 of the time. Well almost.
We need to subtract the one of the cases where Die #1 is deuce and Die #2 is deuce, or where Die #2 is deuce and Die #1 is deuce because we've just double counted the same roll. Since each of these rolls has 1/36 chance to happen we subtract 1/36 from 1/3.
That gives us: 1/3-1/36=11/36.
In other words the chances we see 2-x are exactly 11/36 of the time.
But we didn't increase our chances to see exactly 2-2. They remain 1/36 for each throw.
The fact that it doesn't matter which die is a 2 gives as (almost) twice the odds on seeing 2-x, while changing nothing else.
So when we see 2-x that means that either Die #1 is a 2 or Die #2 is a 2 and we still have 1/36 chance of seeing exactly 2-2 (both dice come up 2).
Well, simplifying our results gives us 11/36 chances for 2-x versus 1/36 chance for 2-2, or one time out of eleven 2-x we will see 2-2.
The very fact that it doesn't matter which die is a 2 makes it 1/11. If we'd cared which die is 2 and were only interested in one particular die to show 2 first and only then examine the second die we would have 1/6 chance that the other die is also a 2.
Kewl when you roll two dice ONE TIME you'll have to choose ONLY ONE of the three scenarios. You can't have two and you can't have three.
We have all our chances decided before we threw the dice. Not after the throw. Remember this.
What are you saying?Quote:
Originally Posted by kewl
Two dice have been thrown. You know at least one die is a two. The second die in the problem has a 1/6 chance of also being a 2. That's it. Finished.
Let me give you an analogy.Quote:
Originally Posted by Alan Mendelson
How do you know that upon throwing a single die the chances it will come up as 2 (or whatever) is 1/6?
You've derived it from your knowledge in probability that when there is one outcome you are interested in and six possible outcomes total, your chance to get a particular outcome is 1 divided by 6.
You don't throw the die one time and conclude from the outcome anything - its impossible.
Therefore, you have all the information about your chances based on how many outcomes total are there and how many outcomes (combos) you are interested in.
Two dice have been thrown. You know at least one die is a two.
Same thing.
You have such and such ways to arrive to at least one two and such and such ways to arrive to 2-2.
You don't come midstream and conclude since the outcome is 2-x you can base your odds now on one die.
Both the bet from post#1 and the simulations are based verbatim on the original question. Yet you refuse to acknowledge that.
Kewl let me ask you a question: are black helicopters circling your house?
No, not ones I can see at least. Why, you seeing any?Quote:
Originally Posted by Alan Mendelson
I was thinking - this post here:
http://vegascasinotalk.com/forum/showth...5054#post25054
does indeed have some good points. Well put :), admittedly some vulgarity was used, but nevertheless it really reflects what you are doing here (and over there in that thread).
Kewl just consider who wrote it.
That would be MickeyCrimm, the world-class drunk who was banned from also the Wizard's -reluctantly I might add - because there seemed no subject or manner of phrasing thereof, or ridiculous claims, off limits to him. It was becoming so obviously pathetic that the Wizard didn't want the intellectually challanged over there spotting it and mistaking the Wizard's mentality for the same. Why would the Wizard be afraid of such? I don't get it, lol.Quote:
Originally Posted by Alan Mendelson
Anyway, Kewl, you had better spend at least one a day a week in the casinos making that $100,000 or $200,000 free and clear a year. Or the Wizard might have to ban you as well.
Lol, dude, you come of as a totally inadequate poster to me, what the hell are you trying to say?Quote:
Originally Posted by OneHitWonder
Are you by any chance assuming I am "KewlJ", the member on WoV? Cause I'm not lol.
I'm really having a hard time truing to decipher what are you saying/your point actually.
Most of the time it sounds like you actually farted instead of say something meaningful, no offence lol.
Alan's very good at being a troll.
I got warning:
"careful
No need to use a word like farted. Thanks"
Okie dokie :), I thought farted is considered mostly harmless and humorous lol
In my country when someone is making a moot point or is incorrect or ambiguous we use to say " Did you say something or did you fart?"
kewl will be taking a week off to think about his use of certain words here especially after being warned about it.
Good time to lock this thread up. It's run it's course, and with the WoVers starting to claim that they're not the same anonymous superstars they are over there, let's see what other topics they'd like to be taken to the woodshed on.
There is one question I had which I don't think anyone has answered: why does it matter which of two dice is initially showing the 2?
Because the WOVers (and I think Arci too) said we don't know which of the two dice is showing a 2 that their answer of 1/11 is the correct answer.
I have always said that it doesn't matter which of the two dice shows a 2 because it's only a two dice problem. I said if Die A shows a 2, then it's Die B which has the 1/6 chance -- and vice versa. And I even offered that if both Die A and Die B show a 2 then the chance for either die being a 2 was also 1/6.
Yet, no one ever explained their position why my position was wrong.
In fact their insistence that we don't know which of the two dice shows a 2 is the foundation for their 1/11 answer as the Wizard showed in his video and as miplet showed in his video.
Quote:
Originally Posted by Alan Mendelson
- I can see now, that my diagram was unclear, confusing, and erroneous.
- You're right... Any picture must use 2 dice only.
- I have made a new diagram - two dice only - to prove you're correct.
http://www.imageurlhost.com/images/p...gjvjova99g.jpg
- You can use 1-of-6 on top... one fixed red deuce against six blue faces.
- You can use 1-of-6 on bottom... one fixed blue deuce against six red faces.
- Or you can use 2-of-12 combinations - the whole enchilada - on one pair of dice.
I have yet to see the Wizard post anywhere else.Quote:
Originally Posted by Rob.Singer
Indignant99 I still don't understand your graphic.
It doesn't matter. What matters is how the question as written is properly interpreted. The mathematical interpretation of this must match its plain English gist. Believe me, real mathematicians use about one word a paragraph. What we have "drummed into us" over at the Wizard's is a bunch of math talk. The reason it's not very logical if you look closer.Quote:
Originally Posted by Alan Mendelson
Short answer is that it matters the dice are taken as two separate dice in a specific roll of the stated condition; or as taken together in single cards from a deck with the condition on each card (for both dice).Quote:
Originally Posted by Alan Mendelson
Because it's not wrong? I agree with you that this isn't about 1/11; but I think that you haven't distanced yourself far enough away from the 1/11.Quote:
Originally Posted by Alan Mendelson
This is about a specific roll, so as far as 1/11 and your way of saying 1/6 say the same thing, anyway, yours is the better part of the dispute then.
Simulating his own interpretation, doesn't prove that the Wizard's is the correct one, let alone the only possible correct one. He proved that 1/11 by calculation is 1/11 by simulation. A strange error in thinking for a self-styled math genius.Quote:
Originally Posted by Alan Mendelson
P.S. I will bring this question before a real math forum pending more nonsense from over there. More time to try to sum this up later.
Quote:
Originally Posted by Alan Mendelson
- The red die is on the left. I merely showed all six of its faces.
- The blue die is on the right. I merely showed all of its six faces, too.
- The top arrows connect a fixed red deuce, coupling it with any of the six blue faces.
- The bottom arrows connect a fixed blue deuce, coupling it with any of the six red faces.
Well done.Quote:
Originally Posted by OneHitWonder
Quote:
Originally Posted by indignant99
I'm trying to be real super-clear here. (Since I wasn't super-clear before.)
http://www.imageurlhost.com/images/0...xstjirbd4c.jpg
This is what I did to the Red Die. I did the same to the Blue Die.
Thanks, Indignant99. I take your word for it, but I still don't understand all the lines going back and forth.
OnHitWonder this comment of yours had a very big impact on me: "Simulating his own interpretation, doesn't prove that the Wizard's is the correct one," and that says a lot.
From the get-go the Wizard immediately jumped on the 1/11 answer and resisted at all times any idea that the question itself might have been flawed or that following the information in the question might cause you to come up with a different answer.
Even now, on the WOV forum there is discussion about another problem with two dice using the numbers 3,5 and pointing to how the answer inolves 11 dice combinations. And while the correct answer is 2/11 it is because there the question lacked the specifics in the question we discussed.
Frankly, it shows me how far they will reach to protect their own flawed thinking.
People have explained why your position is wrong over and over again. You ignore the explanations. Why should anyone repeat what has been done many times?Quote:
Originally Posted by Alan Mendelson
They're just arrows, showing how a Deuce can be combo-ed. (Gotta have a two-dice result, right?)Quote:
Originally Posted by Alan Mendelson
Red Deuce ---> Blue Ace, Blue Deuce, Blue Three, Blue Four, Blue Five, Blue Six
Blue Deuce ---> Red Ace, Red Deuce, Red Three, Red Four, Red Five, Red Six
Arc what was wrong was the understanding of the original question. How many more times do you have to hear about it?Quote:
Originally Posted by arcimede$
I explained the two possible interpretation long ago. I also explained why your interpretation was wrong. You ignored the comment. Face-palm.Quote:
Originally Posted by Alan Mendelson
I have an alternate diagram. I separated the arrows, so they're not all on top of each other. I'm not sure this is any clearer. What do you think (about clearer)?Quote:
Originally Posted by indignant99
http://www.imageurlhost.com/images/z...fjb6xbj9cv.jpg
And I would still say that once you show me the blue deuce, the other 5 faces on the blue die are eliminated.
To me the original question was no different than throwing two dice down a craps table and one coming to rest on a 2 while the other die is a spinner.Quote:
Originally Posted by arcimede$
And therein lies what remains mysterious to the math people. This is the 1in6 interpretation in its entirety. This is the way the original question was meant to be interpreted. Once wizard took the deep-thinker's interpretation to arrive at 1in11, his people and a few others had no choice but to follow him down his yellow brick road.Quote:
Originally Posted by regnis
This is all it is.