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There are TWO 2's right? Yes count them.Quote:
Originally Posted by OnceDear
Chance of rolling a 2 on the first die is 1/12.
Now there are 11 [eleven] faces remaining. How many 2's? 1? Chance of a 2? 1/11.
Chance of rolling a 2-2 with 12 faces is 1/(11*12) = 1/132
Quote:
Originally Posted by RS__
I'm Beginning to See the Light...
http://www.reactiongifs.com/wp-conte...mg/amazing.gif
Would Archbishop Desmond Tutu (TWO-TWO) be the consultant of last resort on this matter?
AgreedQuote:
Originally Posted by RS__
No. Once you rolled the first die, 5 of those faces became inelligableQuote:
Originally Posted by RS__
Chance for second die is 1/6
1/12 X 1/6 = 1/72
Which is exactly half of 1/36. Coincidence? I think not.
Helleluia.Quote:
Originally Posted by kewl
If you say (and I quote Pharoah here), "If you roll [present tense] the dice, and at least one of the dice is a two [present tense, as far as I can tell], what are [present tense] the odds that BOTH dice rolled [past tense] a two?"
People do not speak like this. People do not write like this. This would be edited in a technical writing assignment.
Let's match up the tenses, which is the way people speak, and the way non-quantum reality works. "If you rolled the dice, and at least one of the dice was a two, what were the odds that BOTH dice rolled a two?"
Mixing tenses (in a way that would get you edited in any journal) just to create confusion is childish. The lack of reference to the specific number of dice rolls has two effects. Math folks will interpret the lack of explication as "any number of rolls" or some abstract scenario. Non-math folks will tend to assume it refers to one roll of one pair of dice with no reference to probability in general.
In addition, the use of the present tense implies (in my mind) that one is talking about more than one roll. The use of past tense implies that it's more likely that one is talking about a single roll.
Quote:
Originally Posted by redietz
And what were the odds? If you think 1/6 cause the die has 6 faces, then why is that when you bet on that 1/6 event at a 8 for 1 odds you still lose?
At no point have I said the odds are/were (I can be as tricky as you boys) 1/6. Perhaps you should actually read my posts and test yourselves for reading comprehension as you would have others tested for math comprehension.
This trick question has been around a long time, decades I think, so don't believe you've discovered some great brain teaser.
Do tell what do you think the odds are then. Please.
I should probably be making fun of your reading comprehension. My take on this entire waste of space has been as crystal clear from the start as your trick question. Some people (arci, Rob) got it. Some people didn't.
Gee, I must have divided the reading audience into those who got it and dullards. Mercy me, what a rush (yes, Sheldon, that was sarcasm). What a clever boy I am.
Hi PharoahsWin and thanks for joining and for posting.
I agree it's a trick question. But let's go back to the actual, original question:Quote:
Originally Posted by PharoahsWin
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?
The "trick" is really in how you decide to solve the problem. Clearly the two sides (1/11 vs 1/6) are divided in their methodology.
The 1/6 camp looks at the simple, in-your-face condition: one die shows a two, so what are the odds that the other die will also show a 2? That's 1/6.
The 1/11 camp doesn't look at the simple, in-your-face condition and looks at all of the different combinations of dice containing a 2 and then the single combination that has 2-2 which is 1/11.
I said it early on that it was a trick question and if you want to ignore the conditions of the question the answer is 1/36.
If you accept the condition of the question with one die already showing a 2 it's 1/6.
But if you don't accept the condition of the question and start questioning it by saying things like we don't know which of the two dice is a 2 then you will choose the 1/11 answer.
I accept that out of the two dice one is showing a 2. I don't care which of the two dice it is and with one die showing a 2 the only choice for me are the six sides on the other/second die.
Indeed how you read and interpret the question will define your answer. And because interpretations can be different there can be four answers:
1. 1/36
2. 1/11
3. 1/10
4. 1/6
What I said early on was that the question was written to throw a curve ball. I think the question was written to make you look for the 1/11 answer because of this "set-up line" that preceded the question:
Along the vein of the Two Coin Puzzle,
It was the two coin puzzle that asked you to look for the combinations of two coins, without one coin being "set" so you would get HH, TT, HT, TH.
But ignoring the line Along the vein of the Two Coin Puzzle, I accept that one die is set as 2 which removes the other five faces on that die and leaves only the six faces on the second die. When I start reading comments like we don't know which of the two dice is the 2 I just throw my hands up in disbelief. For heaven's sake -- there are only two dice and 2-1=1.
Clearly both the Wizard and "miplet" when they did their videos explaining the two dice problem they did exactly opposite what the original question stated as a condition and they rotated both dice to show the different combinations. Had they left one die frozen showing a 2 and did not touch that die they would be left with only one die to rotate for an answer of 1/6.
I've asked this over and over again: why rotate the dice known to be showing a 2? The original question told you "at least one of the dice is a 2" so why change that condition? And by rotating that die to explain the 1/11 answer they corrupted the question.
So, it was the 1/11 folks who were tricked. They were tricked into not following what the question asked. Whether or not that was the intent of the author of the question I don't know. But perhaps it was unintended because as redietz who is also a writer pointed out, the question was written very poorly. Very poorly. And as I said early in the discussion, had this been part of any national standardized test like an SAT or a LSAT it would have been thrown out just as another confusing SAT question was thrown out back in the 1960s.
Understanding that this is a trick question, look at what you posted.
The bold "other" in the quote there, is exactly what I was trying to explain to you in my first post. You're clearly saying that you feel that if one of the dice is a 2, then the "OTHER" die has a 1 in 6 chance of being a 2 as well. This is 100% correct, but the question is asking the odds of "BOTH" dice showing a 2, not the "OTHER" die.Quote:
Originally Posted by Alan Mendelson
You understand, that there's 1 in 11 combinations that can produce a 2-2, when at least 1 die is a 2. This is, of course, before the dice are actually rolled. 11 combination with a 2, only 1 that is a 2-2, before you roll the dice. I know you understand this because you've said so on other posts. So, before the dice are rolled there's a 1 in 11 chance of 2-2. Then, the dice are rolled, the peeker peeks, and the question asks what are the odds "BOTH" dice rolled 2's. Tricky, but the answer is still 1 in 11.
As, the quote above shows, you change the question to "We know 1 die is a 2, so the "OTHER" die is 1 in 6 to be a 2. Which is absolutely correct, and understandable, if one die is a 2 the "OTHER" has a 1 in 6 chance of being a 2. But, this does not answer the question of the odds "BOTH" dice rolled 2's. There's a 1 in 11 chance that "BOTH" dice rolled 2's. 1 in 6 the "other" die rolled a 2.
I understand it's tricky, but if you look at all your responses, including the response to my post, you are answering what the odds of the "OTHER" die rolling a 2 is. This is your arguement, that the "OTHER" die is 1 in 6 to roll a 2. The question, however, clearly asks what the odds are "BOTH" dice rolled a 2.
This is why the wizards bet will lose you money. Why all the tests show 1 in 11. And, why if you try the experiment yourself you'll get 2-2 1 in 11 times, while the "other" die always has a 1 in 6 chance of being a 2.
The thing that really tripped you up, Alan, was taking that bet. The bet will lose you money because the answer is 1 in 11. Otherwise, 1 in 6, was a valid arguement, if you could get people to see the question the "other" way. But, taking that bet shows that you really didn't understand the question, because it would cost you and everyone else who takes the bet a lot of money.
You can test the bet out and see for yourself. 2-2 comes up 1 in 11 tries, even though the "other" die is 1 in 6 to throw a deuce. That's just never going to change, and it will always cost you money. Misunderstanding or not.
As I have said over and over again: there are two dice and at least one of them is a 2. If you have two dice and at least one of them is a 2 whatever die you decide is a 2 has no other odds or options or choices. It is a 2 -- period.Quote:
Originally Posted by PharoahsWin
Even if you think "at least one of the dice is a 2" means that both dice are showing a 2, you still would consider one of them and say "if dice A is a 2, dice B only has a 1/6 chance of being a 2 also."
Look math guys -- there are two ways to attack this problem: using rational, common sense, or creating the blue print for a Rube Goldberg machine.
And stop with your "test" of 1/11. Why don't you try MY TEST: Set one die as a 2 and then roll the other die and see if a 2 shows up 1/6 times?
The reason you have to do the 1 in 11 tests, is because that's the bet you made with the wizard. LOL. Honestly, I'm not trying to trick you, that's the bet you said you'd accept with the wizard. And, that bet will lose you a load of money, because the answer to that question is 1 in 11.
If you would have made a bet that the "other" die rolls a 2, 1 in 6 times, I would have left this whole thing alone. LOL. It's honestly what is tripping your whole arguement up. Unless you just like to lose money for some reason.
Alan, your test, asks the tester to roll just 1 die, not "BOTH" die. That 1 die, of course, being the "other" die. As I said, the "other" die is always 1 in 6 to be a deuce. Roll "Both" die, and you lose money.
Just IMAGINE this scenario.
I have 60 dice and roll them all. I remove up to 10 of the dice that show a 2. If there are less than 10 deuces, I remove them all. If there are more than 10 deuces, I only remove 10. You do not know I did this, since you're in another room.
You walk in, can't see any of the dice. I tell you "pick one of the dice, any die you want" (they're all under Dixie cups, sorry forgot to mention). And now I ask you "Okay, what's the chance there's a deuce under that cup?"
From your perspective (without the info that I removed the "up to 10 deuces"), would say " 1/6 chance it's a deuce". Which is correct with the information you have.
We make our bet, I say I'll pay you 9-to-1 if it's a deuce, and you accept. I win your $1. Then I tell you, "oh yeah, I removed a bunch of the deuces". And you would (hopefully) say " well that's not fair!"
Does this illustration make sense to you? I'm not saying it's the same as the 2-dice problem. However, it is very similar.
Thoughts?
PharoahsWin: Yeah, well it was a good natured bet with a limited number of trials that I never expected to win and was limited to buying lunch because I wouldn't mind having lunch with Mike. Actually the bet was proposed by others in an attempt to monetize the debate. So, why not?
As I said on the WOV I'd be very happy to buy lunch for the Wizard and enjoy the time.
As we all know anything can happen rolling two dice but that's not the real issue. The real issue is how to read the question, and that is the great divide.
No I can't imagine that scenario. Try to imagine this one: You have two dice and at least one of them is a 2. Knowing that at least one of them is a two what does the other die have to show to have the combination 2-2? How many faces are on that other die? How many of those faces have the number 2?Quote:
Originally Posted by RS__
Okay, then I'm gone. May I suggest asking the math guys... When at least one 2 is rolled, what are the odds the "other" die also rolled a 2?Quote:
Originally Posted by Alan Mendelson
I bet they give your argument in reverse. I bet they'll say that the odds the "other" die rolls a 2 is 1 in 11, because if the "other" die rolled a 2, that would be a 2-2, which had a 1 in 11 chance of happening. So, they'd say, the "other" die had a 1 in 11 chance of being 2. They would be wrong, and you would be right saying 1 in 6.
Pharoah when will you guys get a grip on reality: two dice have a total of 12 faces, right? When you have one die showing a 2 the other five faces on that die no longer count or matter or can be used. That leaves the six faces on the other die.
Your mathematical gymnastics of switching the 2 from one die to another sounds good for math theory, but when you have two physical, real dice, you can't do that.
So let me suggest this: There is no cup. This roll of two dice is done in the open.
You roll two dice. You observe "at least one of the two dice is a 2." Now tell me, can you really consider 1/11 as your answer?
It seems to me the only way you can possibly consider 1/11 as the answer because you don't actually see one die with a 2. But you don't really have to see one die has a two. You are told in the question that at least one die has a two.
And thanks for stopping by.
LOL. Alan, I wasn't trying to trick you there. LOL. Hasta.
Really Alan, you can't imagine that? What a poor imagination you must have...
I'm one of said Maths guys. I'm an English graduate, in England. English is my first language. I'm familiar with some American English.Quote:
Originally Posted by PharoahsWin
I'd go with 1/6 to that latter question.
Thanks.
Thanks for trying to persuade Alan, but give it a rest now, as I have money at stake :o)
Alan, I'm being serious. Try to imagine that scenario I presented above (where I roll a bunch of dice [60] and remove some of the 2's).
Are you being stubborn or ignorant? You seem very uncooperative in all of this. Why? Try to think outside the box a bit, instead of disallowing any other scenario.
That's just the kind of maths guy ' overthinking' that led to all this trouble. He won't fall into that trap. He's not stupid. Ah! This forum doesn't support strikethrough font.Quote:
Originally Posted by RS__
One more time:
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?
What do the 1/6-ers say?
Its still 1/11. If you ask the question before you throw the dice and not after seeing which die has a two. And disregard all non 2 throws.Quote:
Originally Posted by PharoahsWin
Sorry but for a guy with avid writing skills you really said nothing of substance here. What a waste of space.Quote:
Originally Posted by redietz
A quote from ME from the WoV forum:
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"
No humans,intentions or any unintended ambiguity. Though IMHO the original question is written very well and not tricky or poor writing, or anything. The 1/6-ers just can't grasp the concept. Instead they're accusing all the rest in trickery,overthinking and what not. Its not like that at all. Its a simple probability question from our very real and physical world, a little counter intuitive but that's all.
Your answer people?
If you want me to say "I'm wrong" you are going to have to demonstrate the 1/11 answer using two dice, with one die showing a 2. That is what I asked in the video. Neither miplet or the Wizard showed the 1/11 with one die showing a 2. Both showed a 2 and then proceeded to alter the faces on the die that showed a 2 in order to reach their 1/11 result.
I am not questioning the 1/11 result when you are asking how many combinations of two dice include at least one 2, and of those how many would result in 2-2. That would be 1/11.
But given that at least one of the dice has settled on 2 -- SHOW ME -- using two dice that the chance of 2-2 is 1/11 and not 1/6.
Please, you have to use two physical dice.
I don't WANT you to do anything. It'd be cool though if you and the rest of 1/6 supporters could see the question and the real probabilities for what they are.Quote:
Originally Posted by Alan Mendelson
My demonstration is as follows:
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 $2 total.
So how can that be?
How can it be that a 1 in 6 probability of "the other die" being a 2 actually yields an 1 in 11 actual results?
Doesn't this tell us that the dice themselves actually agree with the 1/11 bunch?
Doesn't this mean that 1 in 6 is actually the answer to a different question?
Or, you could roll a pair of dice a statistically significant amount of times,disregard all the non 2 rolls and count all the 2x outcomes in relation to all 2-2 outcomes.
What that does is - demonstrate and prove that whenever you have at least one 2 showing the odds for a pair of 2's is 1/11.
I'm sorry but I am going to tell you what the problem is: this is not a question of probabilities the way you have been presenting it. This is more a problem of arithmetic. It's really very simple -- almost mechanical. But you have turned it into a question of probabilities.
If it were a question of probabilities the answer would clearly be 1/11. But we have more information that cannot be overlooked and cannot be dismissed:
We are told that there are only two dice, and at least one of those two dice shows a 2. And that takes this problem out of the realm of probabilities and brings it down to earth as a question of simple arithmetic. One die is a 2 -- and how many other faces on the other die would create 2-2?
As I have said so many times, you math guys have over-thought a simple problem and by overthinking you came up with an impossible answer. Really -- 1/11 is impossible when you know what one of the two dice is and when you clearly understand the question.
So if you can possibly take two dice and show me the answer is 1/11 without the flip flopping of faces which is what the Wizard and miplet did, then I will say it's not a simple matter of arithmetic. But right now that's all it is. And I am very tired of you and me and the others going over the same arguments so many times.
Show me a video.
try a semi-glossQuote:
Originally Posted by arcimede$
Ok, a video(a virtual one, like in my mind) -Quote:
Originally Posted by Alan Mendelson
I hold a pair of dice in my hand and I start to think:
What are the odds I'll roll 2-2? 1/36? Correct.
Now, what are the odds I'll roll at least one 2? 11/36? Correct.
Now, I'm not interested in all other rolls. I only care about the "at least one 2" and a"2-2".
So, (keep thinking and holding the dice in my hand) when I'm rolling at least one deuce, 1 out of those 11 times I'll have rolled a 2-2.
How come,then, when I actually throw the dice, the 1 in 11 chance of rolling a 2-2 when at least one 2 is rolled, somehow collapses to a 1 in 6 chance?
Can you explain that for me?
In the OQ I'm told at least one of the dice is a 2. That is exactly the same as what I am thinking before the actual throw - I have one out of eleven chance to roll 2-2 given that I roll at least one 2.
So,how does the probability collapse to 1/6?
Edit: Sorry,where it says "collapse" it should read "expand" or "rise". I'm not a math guy.
You use a poorly written trick question to have intellectual orgasms over other people's failings to grasp your cleverness. It's pathetic, really. I'm hoping none of you folks are in teaching professions.Quote:
Originally Posted by kewl
Glad to hear you think I have avid writing skills. I've never made such a claim, and never will, but it's good someone thinks I'm competent, at least.
Kewl,Quote:
Originally Posted by kewl
Don't you get it? As Alan has pointed out :-
Alan forgot to capitalise the T' after the colon.Quote:
Originally Posted by Alan
I'm Just saying, because these things are important to we English.
One of the issues I'm having with the attitude of the WoV posters is that they're attempting to denigrate a public person, namely Alan, while (A) using language they wouldn't use face to face with any one of us and (B) maintaining anonymity. Academic journals don't publish papers from the anonymous. Step out, sign your real names, and stop being a high school math clique.
God knows, arci and I have gone head to head with Alan tooth and nail over many math issues, and I have been sarcastic and direct on many occasions. But you guys go way, way beyond the bounds of semi-formal civility and set records for being snide. So instead of these weirdly cliquish narcissistic responses, why not just sign your names and pretend there's an actual person facing you?
That'll never happen red. These guys are making so many personal attacks, first because they're not used to anyone contradicting them or telling them why they're wrong (Heaven FORBID!!), and secondly because the admins on their forum are such big wussys that anyone and everyone will get banned for hurting their feelings in the slightest. Face is a penniless non-gambler, Mission saves up all year to make that big trip to LV to play quarters and lose, beachbum wishes he were gay, and Shack gets torn between using that stupid "nuclear option" & going into a dark corner to cry if someone gently insults his social or gambling beliefs. So they come over here to let off steam, but with wizard's blessing. And you have to remember, most of the "geniuses" over there gamble very little or not at all because they much prefer the safety of theory over reality.Quote:
Originally Posted by redietz
If you want to know their real names you'll have a long wait indeed. I do know one of these guys since I've trained him. But there's so much sensitivity with the group overall that all you'll get is more insults the more you ask.
Because that is not the problem that was asked. Why do you continue to deny the English language?Quote:
Originally Posted by Alan Mendelson
Great one! And that rant above coming from the insult king! But agreed, there shouldn't be any.
I've been kicked out of more casinos than you all have ever been in, COMBINED! Last thing I'd do is reveal my identity anywhere!
One last thing. You COULDN'T PAY ME to be "trained" by Singer!! That's asinine!!
Huh? I use nothing to have nothing dude. I just love the "oh it's that then, wow that was more satisfying then I thought,I really get it now,plus it will help me make much better decisions from now on" moments in my own experience and is good to see more of these around.Quote:
Originally Posted by redietz
I know this thread is long and one can easily be confused, but you either mistake me for someone else or are just indeed a very strange dude. Nowhere did I try to act as if I'm(or we, or whatever) superior or stg. Pointing errors doesn't necessarily come with intellectual orgasms, get over yourself.
As for your writing skills I should have made it clear that it is you and Alan and others who point to them so I referred to them like in quotes and in the end you didn't say anything after all.
This is not your usual trick puzzle, when you know why the answer is 1/11 you really get stg from it. Stg substantial.
I know, I'm trying to demonstrate it is a question of probabilities cause I thought I can come up with one or two better examples and mind experiments then what were given so far, but alas , they don't work either so far...Quote:
Originally Posted by OnceDear
Do you agree, that if I show you a video of someone rolling dice 10 000 times and counting the number of times when at least one die is 2 and when both dice are 2 the ratio is very, very, likely to be close to 1/11?Quote:
Originally Posted by Alan Mendelson
Sure thing Mr Redeitz.Quote:
Originally Posted by redietz
Actually, I could agree with you that we 'maths guys' are becoming more and more snide, deriding Alan almost as much as we are deriding his stubborn and absurd argument.
It wasn't always so. I came to this forum to try to be as patient as possible in guiding him to the enlightenment of the correct answer to what could be considered a subtly tricky question. Not a trickster with a trick question.
But he's worn down any attempts at patience. I think he was first to describe someone as 'nucking futs'.
So, his forum, his rules and his precedents shall prevail.
Now there's another good one. "I've been kicked out of more casinos......blah blah blah" as a reason to hide the identity. jbjb, care to present written proof on that statement? You know, like when I was banned IN WRITING for winning at vp at Bellagio, then published the letter in my Gaming Today column? With my real name in it. Did the world cave in? Nope. And you fear revealing your true name HERE!? As I said, these WoVers for the most part do not gamble with any significance if they do at all, and instead make believe they're the world's most feared "AP's" who claim to win. Theory = safety. Don't want to end up like Face or Teddy, right?Quote:
Originally Posted by jbjb
Personally, I have great respect for anyone who's been kicked out of more casinos than I've visited. He must be doing something right. Once you've been booted from the 85th place, you not only have to be a really good gambler. You've gotta be a master of disguise.
I subscribe to the James Grosjean method to winning in a casino. Yes disguises are a part of this. For sure, it's not a badge of honor being in Griffin and OSN.
I don't know, nor do I care. I've been playing craps for more than 20+ years and I've seen all kinds of strange things happened with dice and dice combinations. I've seen players bet the 12 for $100 for 40 rolls in a row and never hit it. But after their $4000 was gone the next shooter rolled 12 three times in a row. And I saw a player lose $100,000 at the table making only one bet: hopping the 7 on the come-out with $100 on each of the 7s -- and the dice went around the table all night long and no one had a natural on the come out roll.Quote:
Originally Posted by kewl
What you show me when you roll two dice only shows me what happens when you roll two dice and it doesn't answer the question.
I'm not familiar with craps, but.....Quote:
Originally Posted by Alan Mendelson
Here's a challenge I will respond to. I've been booted from NYNY, MGM and Bellagio for allegedly being a dice mechanic. Yes, I was winning and they didn't like that I set my dice and had "controlled throws" -- but I just said I was lucky. So, I have 3. How many for you?Quote:
Originally Posted by jbjb
If you are referring to the $40,000 -- yes I had too many zeros. Fixed it.Quote:
Originally Posted by OnceDear
Devil's in the detail. :oQuote:
Originally Posted by Alan Mendelson
What is interesting here is that very few are acknowledging the confusion in the question and they treat the question only in the way that matches the answer of 1/11.
Personally I don't see the need to discuss this anymore. The way I read the question the answer is 1/6. I also know that when I use real, physical dice to present the question which means I cannot rotate any dice showing a 2 to show a different number, the answer is still 1/6.
I cannot accept the Wizard's video or miplet's video as a true representation of the original question -- because each of them changes the value of a die showing a 2. I still maintain that if at least one die is showing a 2 that die's value cannot be changed which eliminates 5 of the 11 possible dice combinations.
I have listened to all of the reasoning why rotating the die showing a 2 is valid -- including the reason that we don't know which of the two dice under the cup has the 2 on it. And that reason in particular makes no sense to me when there are only two dice in the problem.
I also asked you to show a video representing the problem without a cup and leaving the die with a 2 frozen showing that 2 and still maintain how the answer is 1/11 -- but no one has done it.
Ironically, even the 1/11 camp says if we saw that one die was a 2 and rolled just one die (the "other die") then the answer would be 1/6. Well, that's what I interpreted the original question to ask.
So, I really have nothing else to say. I am not closing the thread because I don't believe in closing threads. For those of you who still want to expound on your theories and your beliefs please knock yourselves out. But I can't imagine what is left to be said?
I see...Quote:
Originally Posted by Alan Mendelson
Well, I thought that rolling two dice is exactly what answers the question, but you seem to be adamant in your decision to try and answer the question (or is it A question) by setting one die and rolling only the other. The way I read the OQ it involves rolling two dice and then, having some information on the outcome, try and determine a probability. But that's just me.
I think this statement of yours really crystalizes the difference of opinion. Yes, rolling two dice does answer the question. And that's what the question asks you to do. And when you roll two dice and at least one shows a 2 the chance that the other die shows a 2 as well is 1/6. Now in order for you to come up with your 1/11 answer -- and this was illustrated by the videos of both the Wizard and miplet -- you must switch the die that shows a 2. Don't believe me? Look at the videos they did. If either one of those videos kept the die showing a 2 on the 2 the only solution you could come up with would be 1/6.Quote:
Originally Posted by kewl
Don't believe me? Take two dice and roll them. When at least one die comes to rest showing a 2 is there a 1/6 chance that the other die shows a 2 or is there a 1/11 chance that the other die shows a 2?
Based on the original question, when you are told that one die is a 2, why would you go through the exercise of considering either dice as being the 2 which is what Wizard and miplet did in their videos? Why wouldn't you accept that one die is a 2 and just consider the 6 possibilities on the second die? That is the wording of the original question as I interpret it.
I can't possibly interpret the original question to consider your methodology of considering all of the 11 combinations of dice showing a 2 when I am given the information that one die is a 2. And when you say "at least" one die is a 2 that means thee is a two on one of those dice. And if there is a 2 on one of those dice it stays a 2 -- you don't get to flip it around.
And by the way -- when I did my video I kept the die showing a 2 as a 2. I didn't flip that die around. I flipped around the second die to show it had 6 options for also showing a 2.
And please tell your buddies over on the WOV forum that I noticed my error about the $40,000 before I saw the post pointing out the error. I am in my office working and had not yet read through the thread to see the responses. If they want to make a big deal out of the typo they can. It's typical of what they do.
I just wanted to thank all of the WoV folks for their practical contributions to a Las Vegas forum. My next excursion to Las Vegas will undoubtedly be much better thanks to them.
I learned a lot from this thread. WoV has a gentlemen who has been tossed by somewhere between 100 and 200 casinos, which has to be a record. I'm not sure if it's a record for persistence or having really bad disguises or what, but it's undoubtedly a record. I learned that writing in mixed tenses can be really helpful in creating verbal illusions. I had never really considered that, but then a friend in law enforcement said tense changing in an interview is a dead giveaway and an obvious red flag for obfuscation, and I was somewhat embarrassed I hadn't immediately realized that. I learned that being on the math department grad basketball team at Penn State was poor preparation for touting math skills; I should have joined WoV.
I hate to ask for one more insight, but I'm hoping one of the honored guests from WoV can explain what "stg" stands for. I will forever be in your debt.
Thanks.
Hopefully this will die down with the slow death that it deserves. But clearly, Alan did a number on these geniuses because not only are they now arguing with each other--the discussion dragged in MathExtremist, who loathes getting involved in circle jerks.
Meanwhile, they're still making believe Alan's still arguing in real time, probably so they can keep insulting him. I even saw them take another shot at me by claiming I live in an RV park, etc.! It is just too funny, what you've done to them. And have you EVER seen so many Brits turn away their delicious gourmet nightly dinners in favor of spending that precious time one-upping each other in their posts? :)
You were told about this possible difference in interpretation long ago ... many times. No one is switching dice. We are talking about multiple independent throws. No different than multiple VP hands. You went off in la la land. That is your problem.Quote:
Originally Posted by Alan Mendelson
You've been told many times your interpretation is wrong by numerous folks.Quote:
Originally Posted by Alan Mendelson
Silly nonsense. All that was stated is one die was a 2 on the independent throws in question. No one said that same die has to be a 2 on future throws. That is your own silly BS.Quote:
Originally Posted by Alan Mendelson
Arc well before you ever saw this thread on THIS forum I was the one who brought up on the WOV forum about how the question was worded and the difference in interpretation. You came late to the party.
And yes, dice are being switched and dice values are being changed and I saw two videos showing it being done to justify the answer 1/11.
The dice values change because they are from different throws. Duh. You've been told this fact over and over again.Quote:
Originally Posted by Alan Mendelson
There's no coming late to this party. It's been going on for weeks.
Arc stop. The Wizard and Miplet both in their videos are rotating the dice showing the 2 to justify their 1/11 answer. That is to say: on the same simulated roll, they have to rotate the die with the 2 to simulate how there can be eleven different combinations -- and that is not showing different rolls. They are doing that on the same roll. Look at the Wizard's video again. Look at miplet's video again.
One roll. One die shows a two. Yet both of them rotate that die with the two to justify their claim that the odds of 2-2 are 1/11.
And for the record, this "party" was going on the WOV forum long before it ever came here.
If Pharoah's version of the question is indeed the original question, it is not at all clear that there are multiple throws. If this were a well-written experimental protocol, the question would actually state that there are multiple throws and how many.
This is starting to get stupid. Now we are arguing about who said what first and what the original question is? Here is the original question, copied from the WOV forum:
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?
There is only ONE roll of the dice. For the Wizard and for miplet to rotate the die with a 2 to justify their 1/11 claim when there is only ONE roll of the dice and one of the dice rolled is showing a 2 is just WRONG.
WHEN do you determine a probability? Before the roll or after?Quote:
Originally Posted by Alan Mendelson
After the roll is made there is no probabilities, there is just an outcome. Don't you get it?
In your video you made a blatant mistake. You rolled the dice and THEN you start examine faces and determine odds. There is no such thing as 1 in 6 at that point.
Read the basics of probability and then comment things.
Here we go. Kewl now admits that the 1/11 folks are ignoring the question because he says "when do you determine a probability? Before the roll or after?" And then he goes on to say after the roll is made there is no probabilities, there is just an outcome." EUREKA -- THEY HAVE FOUND IT!!
They have been making this error the entire time. They are talking probability to come up with the 1/11 answer and those of us who say it's 1/6 are looking at the actual problem.
Kewl: this question is not going to be answered by imagining the different combinations of two dice containing a 2. The question is answered by the conditions being presented: you have one die showing a two so what will it take to have 2-2?? That is 1/6 and that is determined by the conditions of the question.
Part of the trick is that in the version of the question referenced by Alan, you have a description of an event in present tense. The use of present tense lends itself to a possible interpretation that the event in question is just one of many such events. That, however, is not stated, so deciding that this is one event of many requires drawing an unmerited conclusion.
All we do know, based on this description, is that there was one event.
If the question had been put in past tense ("rolled the dice" and "peeked"), then it would be more likely to be interpreted by most readers (and editors) as a single event.
The bottom line is that if the author wanted to be clear, and intended to describe multiple throws of the dice, one would assume he would say exactly that. Instead, he does not, which leaves the verbiage open to multiple interpretations if you're trying to be tricky. Most people are not trying to be tricky, and editors do not like this kind of being tricky.
From the question as stated, we can infer there may have been or be more than one throw of the dice, but that's speculative. All we definitely know is that there was one throw.
You guys keep saying it's a trick question. I'm not sure why you think that.
[QUOTE=redietz;28167]From the question as stated, we can infer . . .[QUOTE]
Guys, Guys, Guys,
The thread has flown off course.
Please, let's stop all this bickering about some question from long ago. Look at the title of this thread. Look at post 1 and look at post 5.
A real wager has been offered where Alan believes the odds are 1/6 and where the Wizard (and I) would pay out at 1/9.
Doesn't that get anyone's interest? The amount you can stake is massive and by implication, you can wipe out the Wizard, and if you wish, you can wipe out my pension fund.
Not one taker on that bet. Really? I'm not counting Alan, because he would only bet Lunch and has not yet materialised that bet.
Alan only asks you to agree or disagree whether the bet is a good one. No need to mess around with arguments or insults.
From post 5...
Does this forum support polls? Maybe after all the debate, someone could create one with three options such as "No. This is a bad bet for the player, Yes. This is a good bet for the player, Yes.This is a good bet for the player and I've contacted Wiz or Oncedear to arrange taking it up for serious money.Quote:
Originally Posted by Alan Mendelson
Let me return to the wager: I said no. And the reason why I said no is that I do not make bets against individuals. Yes, I play in casinos -- but my parents taught me not to make bets against individuals, friends, relatives, co-workers, etc. because they can lead to bad feelings.
I did, however,agree to bet the Wizard lunch of up to $50. I did try to meet the Wizard to have this bet a couple of weeks ago but I got into Vegas too late to hook up with Mike and I had to leave in the morning for a meeting with a client and then the drive back to LA. I am more than happy to have the meeting again and the lunch wager.
Now, there were some others who said they wanted to make more substantial bets with the Wizard, but they dropped out for whatever reason.
The bet I have with the Wizard calls for a limited number of dice rolls and frankly I don't expect to win because of the way the bet is structured -- but it doesn't matter I think it will be fun to have lunch with Mike.
But frankly the bet is a separate issue. What's more important is the truth about the actual question and the actual solution. You 1/11ers have corrupted the question and manipulated the information to come up with your answer. Sorry -- just gotta tell it like it is.
I asked everyone to do a video explaining your methodology and the videos PROVE your methodology corrupted the question. This should not have been treated as a question of probability -- but only as a simple math problem: You have one die showing a 2 was fact number one, and you have a second die with six sides which was fact number 2. You only had six sides to choose from -- and not eleven. As soon as you turned that die showing a 2 to expose five more faces you corrupted the question. Shame on you.
It's not exactly a trick question. But the question was framed to mislead, and then the way the question was actually worded, left open all sorts of interpretations. Let's face it -- not only are we arguing over whether the answer is 1/11 or 1/6 but over on the WOV forum they're arguing over whether or not the "peeker" has seen one or both dice.Quote:
Originally Posted by RS__
If you wanted a question about probability it should have been written as a question about probability. If you wanted a question about dice combinations, it should have been written as a question about dice combinations. And when you frame the question as it relates to the "two coin problem" it immediately raises the issue of how you are supposed to come up with the answer.
So, it's the confusing nature of how the question was written that is "tricky."
About polls -- yes, if you start a new thread you can choose the poll option.
Because you brought it up. ?Quote:
Originally Posted by Alan Mendelson
Aaaaaand...probability and the number of combinations are the same thing (well, they lead to the same answer).Quote:
Originally Posted by Alan
You have yet to answer how the two-coin problem is significantly different than the 2-dice puzzle.
No. It's people who get the answer wrong then later on blame their wrong answer on "the question being tricky". You (and your rob-singer friends) got the answer wrong and are making an excuse.Quote:
Originally Posted by Alan
We can find the answer to this question quite easily -- via simulation. However, the anti-math people tend to naysay simulations and their results.
Quick question (naysayers will likely get it wrong or refuse to answer) -- If I do a simulation (say 100 million rounds) of dice throws/tosses and gather a sufficient sample size, and one of the events in the simulation occurs 1 out of 11 times. Now on my own, not part of the simulation, I throw the dice once and they meet the standards in the simulation (ie: at least one die a deuce), what is the probability or chance that both dice are deuces?
RS__ I didn't use the word "tricky." Redietz did. I was offering an explanation on how the question was problematic. There is no doubt the question is problematic. You are also problematic and I am tired of you and your ridiculous comments -- not to mention your cursing which left a very bad taste in my mouth and makes me not want to communicate with you. Go back to the WOV forum -- you fit better there.
I'm sorry. Did it taste salty?
I told you Alan, with these people you have to interpret their brain teasers exactly as they do. When you don't and you clearly explain why by exposing a geek-fault, the agitation sets in and the results are vulgarity & insults. They do that over there and the sensitive admins immediately suspend them. You also experienced what occurs when someone outside the gay math club brings up an important point about a posted problem that none of them thought of before you. Who'd of thought to solve this issue by using situational reality, when WoV wisdom demands you use probabilities & theory??
"It's not whether you made a 'good' bet; it's whether you win or lose"
I am sure that those of us who came up with 1/6 got that answer because we visualized two dice. The "probabilities crew" probably visualized that chart of 11 dice combinations. This is why I asked them to use real, physical dice because if they had (and didn't alter the conditions of the problem) they would have seen for themselves the answer was 1/6. But, lo and behold, they manipulated the dice just to fit their probability chart again.Quote:
Originally Posted by Rob.Singer
To compound their corruption, first the Wizard and then one of the WOV visitors on this forum starts talking about the 36 dice combinations of which only one is 2-2 and they start using that for some evidence of whatever.
And they make up their own rules such as: we don't know which of the two dice is a 2 so let's figure the problem by switching the 2 from one die to another. And that's the only way they can come up with 1/11. Anyone who looks at two dice with one die showing a 2 knows there are only six combinations for that 2. But at the WOV they have their own rules, don't they?
Meanwhile over at the WOV they're still fighting among themselves over what the problem says and they're still quoting what's posted here.
I'm offended by your childish name calling.Quote:
Originally Posted by Rob.Singer
Exactly right. They keep tossing in so many varying events of probabilities because when frustration takes over, obfuscation results. And that constant arguing among themselves now is "pure gold"....
The wiz should be shutting it down soon. It's becoming an embarrassment for him.
And I, even more so, by you spelling MATERIALIZED incorrectly. Please use the host country spelling.Quote:
Originally Posted by OnceDear
I'll spell materialise and colour, etc. any way I damned well choose. I owe you no courtesy, Rob.
I suggest you create that Poll. It would carry your authority and none of us 'Maths' people could bias it with 'overthinking' or 'trickiness'Quote:
Originally Posted by Alan Mendelson
Why don't you write it up and PM me. My first concern is that the poll questions might not be clear. But let's give it a shot.Quote:
Originally Posted by OnceDear