Statistics aren't the issue. The issue is what are the odds that one die is showing a particular number, and in this case a 2?
Another video: https://www.youtube.com/watch?v=uCT-5Rye9bk
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Statistics aren't the issue. The issue is what are the odds that one die is showing a particular number, and in this case a 2?
Another video: https://www.youtube.com/watch?v=uCT-5Rye9bk
If it is one particular die then it is 1 in 6, if it is either die, it is 1 in 11. How many times does this need to be repeated?Quote:
Originally Posted by Alan Mendelson
You will have to explain to me why, in a problem involving only two dice, it matters which of the two dice shows the 2.Quote:
Originally Posted by arcimede$
If, for example, die A is showing a 2 then isn't it a 1/6 chance for die B?
And if die B is showing a 2 isn't it a 1/6 chance for die A?
And if we don't know if its die A or die B that is showing a 2 but one of them is showing a 2, then isn't it always 1/6 for the other die whichever it is?
And just in case both die A and die B are showing 2s then it was still a 1/6 chance that either A or B had a 2 showing, right?
<<;)....chuckle....;)>>Quote:
Originally Posted by arcimede$
OK--one last try at this. Then mums the word--this has gotten tiresome.
If the original question asked: "If I have 2 dice in a cup and I slam them down, what are the odds that at least one is a 2", the correct answer is 1/11. However, once you added the parameter that we know one of the dice is a 2, only one die is left and that is always 1/6.
As soon as the question took one die out of the equation, 1/11 bit the dust. If the 1/11ers can't see or understand this then I give up.
So again, take the peeking out of the original question. Just make it what are the odds that at least one die will show a 2. Voila--1/11. The question is in no way confusing--it is clear--1/11.
Now do it again and peek and tell me one is a 2. Voila--1 of 6 that the other is a 2.
I rest my case.
But Alan, if you take the time to explain everything clearly, then you don't get to separate the reading audience into those who are "geniuses" and those who are not, which takes all the fun out of it for the "geniuses."
Geez, get with the program.
If I have 2 geniuses in a cup and slam them down, what are the odds that at least one of them will get some sense beaten into them and understand this.
I think you missed something here - the odds of at least one two showing up is NOT 1/11, it's 11/36. How do you figure the odds of getting a two by rolling two dice as LOWER than the odds of rolling a two with one die? I agree, this should not be confusing in any way. I might expect you to (incorrectly) estimate the odds of rolling at least one two with two dice to be 1/3, but can't fathom how you think you're less likely to get at least one two when you're rolling more dice.Quote:
Originally Posted by regnis
Yes, regnis misspoke about 1/11 and synergistic is correct that out of 36 dice combinations a 2 will show up 11 times for 11/36.
I wrote this before: the answer "1/11" would apply to a question such as this: "the combination 2-2 on two dice shows up in how many combinations of two dice with at least one die that shows a 2?" The answer to that question is indeed 1/11.
And obviously, that is not the original question.
Unfortunatley the "1/11 crowd" is using that 1/11 answer to respond to the actual question being asked. And I can understand why they are doing it. They are looking at all of the combinations of two dice -- and there are 11 of them -- showing at least one two. And they are correctly identifying that only one combination of the 11 dice that show a 2 is the single combination 2-2. BUT THAT IS NOT THE QUESTION AND THAT IS NOT THE ANSWER TO THE ORIGINAL QUESTION.
As I said early on -- this is a matter of reading comprehension. The math that says 1/11 is correct if you ask a different question.
In fact, this nonsense about which die is showing a 2 is part of the convoluted thinking to justify the 1/11 answer. And again, it doesn't matter which die shows a 2 when there are only two dice to consider in the ORIGINAL QUESTION. But (and this is very important) if the question were worded differently and you looked at all of the dice combinations with a 2 then you could justify that you need to know which of the two dice shows a 2.
It's really become silly. The "non math" people "get it" because they are able to look at the question and use their common sense. It's the "math people" who insist on the convoluted mathematical thinking which gets the wrong answer because it doesn't answer the actual question.
I've given up on the WOV forum because they are all math guys.
yep-sorry--hand quicker than the brain--especially in my case
No, it was the original question but that is not how you interpreted it. So what. We all misinterpret things once in awhile. You now understand exactly what the 1/11 crowd is thinking. You are just unwilling to admit you got it wrong.Quote:
Originally Posted by Alan Mendelson
Arc I just don't see where the original question asks me to consider 11 dice combinations. Why don't you explain why 1/11 is the answer to this problem? Many have said it is but never explained it. I even asked the Wizard to explain it and he didn't though he said he tried to do a video. So explain why we must use 11 dice combinations?
Over on WOV they've noted my comment about giving up on "math guys." Not noted were my comments about the lack of reading comprehension. Is that more proof?
So, I just posted the following on the WOV forum:
Putting aside all of your personal attacks let me ask you to explain WHY you are using eleven dice combinations to answer what I (and some others) believe is answered with one die?
I've asked this several times and I have posted a video. The Wizard said he tried to post a video but he said he wasn't happy with it. Still, no one has answered. Why eleven dice combinations instead of one die with six faces?
The closest explanation given so far is that we don't know which of the two dice is showing the 2. You're going to have to explain that even more in a two die problem because there are a lot of others who also say it doesn't matter when there is a two dice problem.
Those of you who have an answer to my question and are a member of my forum please post your answer there, because I am sure the others on my forum who say the answer is 1/6 will be interested in what you have to say.
=============
And regarding your personal attacks: please don't take comments out of context -- you're smarter than that, and I'm smarter than that.
Regarding DI -- yes, I believe it is possible and in a random game it doesn't hurt to try. Can you argue against trying? Or do you deliberately play craps with the intention of losing?
There is nothing wrong with win/loss goals if they make your play more enjoyable and comfortable for your budget. I don't have an unlimited gaming budget so I always use win/loss goals.
And if I am the poster boy for Total Rewards, why isn't Caesars paying me?
Latest exchange on the WOV forum:
Quote: Ibeatyouraces
I think you guys read it as if you look at on die first, determine if there is a 2 or no 2 then separately look at the other. In that case I'll say 1/6. Now if we see BOTH at the exact same time, we know the full outcome so now it's 1/11. Take two blank dice. Now put a 2 on only one of them. How many blank faces are left? 11 correct? Only one of these can have a 2 but all 11 of the faces of both dice could possibly show up. Not just 6.
Anyway, who cares. It's not a matter of life or death. Now I'm going back to "work." :-)
And my response:
Here's the disagreement. When we see a 2 on one die, we no longer consider the die with a 2 on it, and that eliminates FIVE faces. And that leaves only the six faces on the "other die."
Sorry, answer is still 1/6.
My comment would be that no human sees two dice at the exact same time. One is always seen before the other, or pieces of each are seen in a sequence whereby the totality of one die is always processed before the other. There is no simultaneous both-dice perceptual process.
I thought you might be interested in this exchange I just had over on the WOV. In bold are the comments from the WOV 1/11 members and in italics is what I said:
Okay, I will be a good sport, and I will respond to your responses:
Quote: AlanMendelson
Here's the disagreement. When we see a 2 on one die, we no longer consider the die with a 2 on it, and that eliminates FIVE faces. And that leaves only the six faces on the "other die."
Sorry, answer is still 1/6.
Quote: RS
And that is where you are wrong. That's not how math works.
But that's how reality works. In this case the reality is we are told that there is at least one die showing on two dice under a cup.
So why don't you do this: take two dice and have at least one of them show a 2 and tell me what are the chances that you can have 2-2 showing on both dice? In fact, do a video for me with your cell phone while you are looking at the two dice and explain to me your thought process. I really want to know how -- when knowing that at least one die is a two -- that you can come up with the answer of 1/11 ?? Please pick up the dice and illustrate your reasoning as you discuss it. Post it here or on YouTube with the link. Don't be bashful, this is not a Hollywood audition. I just want you to explain your reasoning in a "show and tell" style -- like I did.
Moving on to the next response:
Quote: Ibeatyouraces
Quit looking at one at a time. You can't eliminate the other five faces because you have no clue which die has the 2 showing, so you have to include them in the probability. That's where you are hung up.
You can do this exact same thing with 12 cards. Ace through 6 of two different suits.
I am going to ask you to do the same thing, Ibeatyouraces: take two dice and have at least one of them showing a 2 and explain to me in a video while using the dice as props, how you came up with a 1/11 answer?? Since it is an issue for you that you don't know which die is showing the 2 start with the die on your left and complete your explanation, and then repeat your explanation with the die on your right (switching the position of the 2). Oh-- if you want to set both dice to show a 2 that's okay too. Just pick up one of the two dice and tell me how the chance of a 2 showing on the other is 1/11. Okay? And remember, the original problem told you at least one of the dice is showing a 2.
Moving on:
Quote: MathExtremist
You're reading it wrong. That's why it's an interesting question. This wouldn't have generated so many posts if the answer were trivially-obvious to everyone. That's what makes it a "puzzle."
But the correct solution and reasoning behind it has been explained dozens of times in this thread, in at least three different ways. It's also been suggested that you conduct your own experiment. You have declined to read the explanations and declined to conduct the experiments. You should try what I suggest in the following post.
Actually, I did conduct my own experiment and I put the video on YouTube and it was posted here as well. I shook two dice in a cup and slammed the cup on the table and when at least one dice was showing a two I looked and asked what were the chances that there could be 2-2. And knowing that there was a 2 on one die the answer was clear that with only six sides on the other die the answer was 1/6.
But I will move on to your suggested experiment:
Quote: MathExtremist
I didn't read through the whole thread to see if anyone has previously suggested this:
Write down each possible dice combination on 36 cards, then play the game using this deck of cards. Shuffle every time, then deal a single card. Keep track of (a) how often at least one 2 shows up and (b) of those, how often the card shows 2-2. The ratio will be 1/11.
Unfortunately this is not the original question, nor does it demonstrate the original question -- which is typical of all of the responses from those who say the answer is 1/11. You guys keep changing the question to match your answer. So MathExtremist I am going to ask you to do the same thing: take two dice and when at least one die is showing a two (which is what the original question tells us) what are the odds/chances that both dice are showing 2-2 ?? The key to your answer is that you are told at least one of the two dice is already a 2. Still think it's 1/11 ??
Moving on:
Quote: MaxPen
It is IMPLIED.
It is tacitly understood.
I understand that you are a marketer. You must be a damn good one at that because your persistence is relentless. However, you will never convince a mathematician that you are correct. Why? Keep looking at the trees. You're missing the forest.
I am also going to ask you to demonstrate what you understand about the original question. Take two dice and shake 'em up and when at least one of the two dice is showing a two, with your cell phone camera rolling, explain in a show-and-tell manner why you say it's still 1/11. Please, I really want you to do this so I can actually see your explanation. Yes, have at least one die a two and using the other die -- why is the answer 1/11??
Moving on:
Quote: indignant99
Because any/all of those 11 combinations can show up under-the-cup. And all ELEVEN combinations qualify for the announcement "at least one of the dice is a two." And one of those combinations wins for you. Ten lose.
Really? If one of two dice under a cup is showing a 2 you are telling me that all 11 combinations can show under the cup? With only two dice to begin with? Aren't you forgetting that at least one of the two dice is already on 2 and that leaves only the six sides on the other die?
Moving on:
Quote: thecesspit
Quote:
A question about odds and two dice:
You have two 6-sided dice in a cup (standard dice). You shake the dice, and slam the cup down onto the table, hiding the result. An independent judge peeks under the cup, and announces truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
Just to repost the question one more time... Alan keeps answering a very different question, despite the question above being what he swears is the question.
I am going to ask you to do the same thing: Use your cell phone and demonstrate the question and your 1/11 answer using two dice with at least one of the two dice showing a 2. Show me, using show-and-tell, how the answer is 1/11. Remember the question: "at least one of the dice is a 2. What is the probability that both dice are showing a 2?"
Moving on:
Quote: OnceDear
Quote: thecesspit
Quote:
A question about odds and two dice:
You have two 6-sided dice in a cup (standard dice). You shake the dice, and slam the cup down onto the table, hiding the result. An independent judge peeks under the cup, and announces truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
Just to repost the question one more time... Alan keeps answering a very different question, despite the question above being what he swears is the question.
And NOWHERE does it ask anything about the probability of the other die being a two.
Well, let me suggest this: if there are only two dice in the problem, and at least one of them is showing a 2, there aren't many remaining options about which die remains. Why don't you shoot a video as well demonstrating your response and how the answer is 1/11 when at least one of two dice already shows a 2.
Moving on:
Quote: thecesspit
Quote: OnceDear
Quote: thecesspit
Quote:
A question about odds and two dice:
You have two 6-sided dice in a cup (standard dice). You shake the dice, and slam the cup down onto the table, hiding the result. An independent judge peeks under the cup, and announces truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
Just to repost the question one more time... Alan keeps answering a very different question, despite the question above being what he swears is the question.
And NOWHERE does it ask anything about the probability of the other die being a two.
Exactly. Neither is it 'throw two dice, find one is a 2 and one is a spinner, whats the chance of having a hard 4'? That's a different question, but one he keeps answering instead of the above. The question is about a group of dice.
I love this. Now we're being told that instead of a two-dice question the question is about "a group of dice."
Moving on:
Quote: RS
Alan earlier brought up the point like "at the craps table, if the first die lands on a 2 and the other is spinning, what's the change of rolling a 2-2 (hard 4)." Of course the answer to that question is 1/6.
What you didn't take into account is the situation where the first die lands on a non-deuce (let's say it lands on "1"). The chance of rolling a hard 4 (2-2) is 0% in this situation. But you still run the risk of the second die landing on a deuce....and that is one of the cases to be brought into comparison when you say " at least one die is a 2".
Thanks RS but the illustration of the "spinner" was to mimic the question. If we used your example there the first die lands on a non-deuce there is no question to be answered -- BECAUSE THE ORIGINAL QUESTION SPECIFIES THAT AT LEAST ONE OF TWO DICE IS A 2. Now if you rolled two dice and at least one of the dice is a 2 then the answer still is 1/6 because you have a 2 on one die. I am going to ask you, again, to shoot a video and post it on YouTube, using two dice with one of them showing a 2 -- and tell me, while knowing at least one of two dice is a 2, how the answer can be 1/11. Please, do it.
Moving on:
Quote: OnceDear
Indeedy
And just in case Alan ever gets a clue as to what probability is and what it means...
In the original question, WE NEVER GET TO SEE WHAT IS UNDER THE CUP
There are exactly 11 possible ways that the dice landed under that cup. They were equally likely. Of those 11 ways there is one way that they show a pair of deuces. That's how probabilities are worked out. Alan is not allowed to f*** with the dice. He never gets to see or touch them.
Nobody gives a flying fig what 'one of' those dice shows and nobody gives a flying fig what 'the other die' shows and as per the original question, nobody gives a flying fig which die is the two and which die is 'the other die'.
I still wonder what proportion of these events Alan would expect to see a pair of deuces.
So, OnceDear, we never get to see what is under the cup? WHY DO WE HAVE TO SEE WHAT IS UNDER THE CUP? We are told that at least one of two dice is showing a 2. Again, take two dice and your cell phone camera, and show me how WITH AT LEAST ONE DIE SHOWING A TWO how you can come up with 1/11. Please, do it.
I look forward to your demonstrations.
I've already explained it to you (as have others). You just don't want to accept the answer. You've seen the 11 combinations that contain at least one 2. Those are the possible results of the experiment. Only one of those is a pair of 2s. Hence 1-11 is the correct answer. How simple can it be?Quote:
Originally Posted by Alan Mendelson
Arc do the video. Put two dice on a table and with at least one of them showing a two, explain in your video how you came up with the answer of 1/11. Do it.
By the way, no one on the WOV forum will rise to the challenge. Like Arc they are all falling back on their previous explanations. But they won't present the problem with a video demonstration that clearly says at least one of the dice is showing a 2.
Would you agree to an answer of "one time out of eleven" if the question was rephrased to "How often will both dice show a two?" instead of the original phrasing?
Both dice show a 2 1/36 times.
Answer the original question with your video. Enough arguing. The problem is specific. Shoot your video and explanation according to the problem and then try to justify 1/11.
I'm tired of this crap. I've been nice for too long.
No, both dice would show a two 1/36 times if you didn't have your peeker limiting the action to only rolls that yield at least one two. If you maintain the setup, could you add the "how often" version to your list of questions that could be answered with "one out of eleven"?Quote:
Originally Posted by Alan Mendelson
Whoa synergistic. Don't start changing questions.
You wrote this:
And I responded to this: "How often will both dice show a two?" And that answer is 1/36.Quote:
Originally Posted by synergistic
Now, I am telling you right now that I have been pushed to the wall. There is a specific question and I want you and everyone else who says the answer is 1/11 to justify their answer by first recreating the question with a video and then explaining their answer. You can't do it -- unless you lie through your teeth and make up a barrel of hogwash. Because when you present the question with props, and do a hands-on show and tell the ONLY answer you are going to come up with is 1/6.
Enough of this gibberish about creating scenarios. As I said from the beginning: the question is specific. Don't change anything to facilitate your 1/11 answer. Show me how you create the problem on a video and then justify your 1/11 answer if you can.
Frankly, I wonder why the Wizard didn't post the video he said he recorded? I wonder if when he did the video he discovered how 1/11 was impossible. I want to see HIS video. And I asked him on his forum.
You know, initially, I thought you were trying to sucker someone you didn't particularly care for into a bad bet. I really have no idea what your motivation is at this point.
My motivation has nothing to do with the bet. That's a side issue.
This is all about the original question. If you think it's 1/11 show your stuff.
Everyone on the WOV forum has been trying to justify their 1/11 answers by changing the question. I want someone to show me the answer to the original question is 1/11... if you can.
Alan, would you accept the results of the following: If I were to roll the dice many many times, record the times where at least one 2 was present, then show the results (ie: frequency of "other" die being a 2).
If the correct answer is 1/6, then we should see the other die show up as a deuce 1/6 of the time at least one die is present, yes?
Edit. Alan, we've showed (shown?) you many times why the answer is 1/11 but you are not willing to accept it. You can lead a horse to water but you can't make him drink.
PS: Did you ever wonder, why just about everyone on the WOV forum was willing to bank the bet I presented a while ago? Why would a bunch of APs be willing to risk their money (thousands of $$$) for somethijg where they don't expect to win?
No I will not accept any number of rolls. And it's because that was not the question that was asked. YOU and many others have said the answer is 1/11 to the original question. Show me how you arrived at that answer given the information in the original problem. And again that information is: there are two dice and at least one is showing a 2. No additional rolling needed. You said the chance of both dice showing 2-2 is 1/11, so go ahead and demonstrate that to me just as the original question lays out the problem.Quote:
Originally Posted by RS__
I look forward to it.
Why won't you accept many rolls of the dice? If what you say is true (1/6 answer), then you should expect 1/6 of the "at least one die is showing a deuce" for the other die to also be showing a deuce....correct?
I mean, you can't honestly believe something has a 1/6 chance of occurring....but not accept a simulation?
What's next, you're going to say there's a 33% chance of flipping a coin on heads...and when a "math guy" tries to prove you wrong by flipping a coin many times, you say "No! That won't do! Only flip the coin once, not many times!!" ?
there you go again... changing the question to fit your answer. ANSWER THE ORIGINAL QUESTION. SHOW ME.Quote:
Originally Posted by RS__
I do believe the original question was something like "what is the chance the other die is a 2" (or maybe it was "what is the chance both dice are a 2").
How is this any different? No math involved here [you should get this right at least..].
I posted a video using two dice showing that the answer to the question was 1/6. You and others maintain the answer is 1/11. The question involves two dice, with at least one showing a 2. Let's see your video. Very simple. You have an answer -- show it, just as I showed it.
https://www.youtube.com/watch?v=uCT-5Rye9bk
Why in the world would you need a video? This is as simple as it gets. All the possible throws have been identified. The difference in the two scenarios has been explained to you. I did it again above. If you specify one particular die is a 2 then the answer is 1-6, if either die can be a 2 then it is 1-11. Why do you persist in arguing?Quote:
Originally Posted by Alan Mendelson
Your video showed nothing.Quote:
Originally Posted by Alan Mendelson
The video will show how the question is interpreted. In reality when you have two dice with a 2 showing -- can the 2 jump from one die to another? This reality destroys the 1/11 answer.
Edited to add: I just responded to another post on the WOV forum. It seems that no one wants to do a video but they are all continuing to argue that the answer is 1/11 because it isn't specified which of the two dice is a 2. The WOVer's question is in bold, and my response is in italics:
Quote: Dalex64
You can't just show a two on one of them - you have to show a two on either one of them.
The question states 'at least one' but doesn't specify which one.
I see that there is still some discussion about "at least one of the dice shows a 2." Let's deal in the real world, please, because in the real world the "2" cannot jump from one die to another. So you have the following options when creating your video (unless you are using Disney animation):
You can set Die A to show a 2 and Die B to be any of the six faces;
You can set Die B to show a 2 and Die A to be any of the six faces;
Or you can set both Die A and Die B to show a 2.
Any of those three will fulfill the information in the original question that "at least one of the dice shows a 2."
Now please, quiet on the set, and begin your video production.
Thanks.
Arc, would you agree that in the real, physical world, that when two dice are rolled and you are told that at least one of the dice is showing a 2, that the answer is 1/6??Quote:
Originally Posted by arcimede$
Again, this is the real, physical world -- not a world where the "2 face showing" can jump from one die to another.
There must be a glitch in the WoV video equipment. Or Schrodinger's Dice are on interlibrary loan.
Of course not, the answer is 1-11. The key phrase is "at least". That means either one could be the 2. To get 1-6 you have to specify one particular die, then the odds the other die is a 2 is pretty obvious.Quote:
Originally Posted by Alan Mendelson
Let's say you have a red die and a blue one. If the red die is a 2 then the odds the blue one is a 2 is clearly 1-6. However, no one is claiming that one particular die is a 2. That is what "at least" means. Either die could be a 2 and hence you have consider BOTH situations which leads to the 1-11 conclusion. Think about it. How can the answer be the same when either die could be a 2 vs. when one particular die is a 2.
Arc let me clue you in on the English language. The key phrase "at least" means this:
There is a 2 on one die, or there is a 2 on both dice.
That is what "at least" means. It does not mean the #2 can jump from one die to another.
True but not the complete picture. There is a 2 on one die (the red one), or there is a 2 on the other die (the blue one) or there is a 2 on both of them.Quote:
Originally Posted by Alan Mendelson
No one said it did. Back to the red/blue dice. The 2 can be on the red die or the blue die. That is what "at least one die" means. If the 2 is on the red die then the blue die can be 1-6, that gives you 6 possibilities. If the 2 is on the blue die then the red die can have 1-6, that also gives you 6 possibilities. Since the 2-2 is a duplicate we eliminate one and you are left with 11 possible results.Quote:
Originally Posted by Alan Mendelson
This has all been explained to you multiple times.
Arc try to think this through: as long as one die shows a 2 the solution rests on the second die in a two dice problem.
OMG Arc you can't use the same die twice to solve the problem.Quote:
Originally Posted by arcimede$
RS___ made the same mistake on the WOV forum that Arc made here. The phrase "at least" does not mean the 2 can change which cube it is on. "At least" means that in a two dice problem one or both shows a 2. It doesn't mean the 2 on the first die can move the second die for convenience.
Really guys. Speak English and improve your reading comprehension and stop "over thinking" the problem.
I can't decide if it would be good or bad to play craps with those dice that Arci uses. What do you think Alan?
I'm afraid that with his dice the faces can change to increase 7-outs.Quote:
Originally Posted by regnis
At this point, it might be helpful to re-work the same exercise but with different parameters.Quote:
Originally Posted by regnis
If at least two of three dies show a 2, then what is the probability all three dies show a 2? If at least one of three dies shows a 2, then what is the probability all three dies show a 2? (Does the latter question revert to the original question?)
Would rolling one die consecutively three times change the results?
One hit now you are sounding like the hacks on the WOV forum. There is no need to change the question. They either misinterpreted the question or can't read or understand English or -- as they have shown -- they can't use two real, physical dice to understand that the "2" can't flip from one cube to another to fulfill their 1/11 answer.
Still on the WOV forum they won't look at two real, physical dice. The Wizard has revealed that his video was not an examination of the problem using real dice but was showing him roll dice hundreds of times tracking results.
They are dodging and weaving and just won't come to grips with the condition of the question. And OneHit I'm not going to let you dodge and weave around the question either.
I'll say it again: the answer 1/11 does not fit this question as asked. 1/11 is valid math but for a different question.
In the closed thread, http://vegascasinotalk.com/forum/showth...-vs-1-11/page7 , I asked only about extending yours and the Wizard's thinking to three dies, to gain a better appreciation for both sides of the argument. I was and still am in the camp of the one-sixth chance.
However, I have lost faith in the forum process here. I doubt that the Wizard as a forum host would meet such an honestly innocuous attempt at further enlightenment with, "I'm not going to let you dodge and weave around the question either," much less therefore promptly close the thread.
As far as I am aware, the Wizard's thread is still very much open to you after your numerous replies there.
Sorry that the thread was accidentally closed. It's reopened. Both Dan and I have closed threads accidentally when we've posted. We'll have to find out why. OneHitWonder's post from another thread has been moved here -- just above -- which is where it belongs.
OneHit I am sorry the thread was closed.
It appeared to me that when you suggested considering a three dice problem it was a way to dodge the question, because WOV forum members have suggested changes to the original question both here on this forum and on the WOV forum.
If you do in fact support the 1/6 it is better if you clearly state why you do. Thanks.
I received an email that on WOV they are saying I closed this thread. My explanation about what happened is above. Still they have not created a video explaining their 1/11 answer but continue to create side issues. Of course they won't create a video with two dice explaining their 1/11 answer because with two real dice the answer 1/11 is impossible.
Not a vid, because I'm not daft enough to waste the time and energy, Besides, I can't find an actor capable of portraying Alan in all his belligerent glory. http://wizardofvegas.com/forum/quest...46/#post456034
Kudus for the followup table layout from Indi in a later post there.
I read your dramatic script. That's all it was.
Instead describe your proof using descriptions of two dice, OnceDear.
And you can tell "indignant" that his layout and bet do not justify the 1/11 answer either.
When will one of you apply the question and answer to two real, physical dice that cannot magically shift the 2 from one cube to another and where one die can only be included in the answer one time? Ya know... like in the real world?
The reason some have presented a three-dice question or a 2-coin problem or the 2-child problem is not to change the question -- but to show you how your logic/reasoning isn't correct.
The 2-coin and 2-child problem are EXACTLY the same (except one uses heads/tails and the other uses boy/girl). These two problems are almost the exact same thing as the 2-dice problem, except instead of having something with a probability of 50% (heads/tails or boy/girl), the dice problem has a probability where the outcomes are 16.66%. But just because the chance of any individual coin being a head or tail (at 50%) or a child being a boy or girl (also 50%)....that doesn't mean the other coin or the other child has a 50% chance [of being a heads/tails or boy/girl]. And, in the same way, the other die does not have a 16.66% chance of being a 2.
Now, instead of ignoring the 2-coin puzzle, you can try to learn from it. You are told a mother has at least one boy, and are asked what is the chance the second is a boy. When you realize the probability/chance of the other child being a boy is not 50% but is 33.33% -- at this point, you can (hopefully) relate the two problems and learn that 1/6 is not the correct answer to the question......just like 50% isn't the answer to the 2-coin or 2-child question. Baby steps, Alan. Baby steps.
The two coin or child question is not the same as the dice question. What makes you think it is? In the coin question the coins can be flipped this way:
HH, TT, HT, TH
but in the dice question you use each die only once. Once a die is identified as a 2 that die does not get used again in figuring the solution. That 2 face is frozen.
The dice can be rolled three ways with X = a number other than 2 and they are:
22, 2X, X2. But once "at least one of the dice" is known to be a 2 that die is removed from further use. And that leaves ONE die with SIX faces.
If you would simply roll two dice you would see that once at least one die shows a 2 it cannot be used again to come up with the 1/11 answer.
Roll the dice. Show me the video.
Edited to add: with the dice the rolls of XX aren't used as defined by the problem. When there is no 2 there is nothing to consider.
I am going to break down the dice rolls in detail to show you again why the answer is 1/6.
Roll 2,2. Use one of the dice and the remaining die has six faces to be 2,2.
Roll 2, X and you have a 2 that is frozen leaving six faces on one die.
Roll X,2 and the 2 is frozen leaving six faces on one die.
In all cases the answer can only be 1/6 in the real, physical world.
RS___ I have a question for you. Are these two questions the same?Quote:
Originally Posted by RS__
1. Maria has two kids what are the chances both are boys?
2. Maria has two kids. One is a boy. What are the chances the other is a boy?
Please sir, no sir... Not a chance. you are not even quoting a puzzle correctly.
The answer to q1 is simple and determined. 1/4 (in a gender neutral world)
The answer to q2 depended on what population sample Maria was chosen from and so is indeterminate.
One perfectly valid answer to q2 is 1/2
One other perfectly valid answer to q2 is 1/3
No perfectly valid answer is 1/4
http://www.curiouser.co.uk/puzzles/kids3.htm
The last paragraph gives the answer to the second question.
Did you get it right Alan, did you? All on your own? Do you now understand probability a bit better?
Or is it back to the drawing board...
...Or is everyone else in the world just plain wrong?
Now.... In my video script..... what proportion of bets on 'pair of deuces' would you expect to win? This isn't maths or charts or excel. A simple question. Do you expect to win 1 in 6 or 1 in 11.
Really no need for your explanation at this point.
Maybe also, what proportion of your friends' bets would you expect to win... the bets on other outcomes?
OnceDear I am glad you read the Maria questions. Yes the answer to #2 is 1/2. There is no reason to consider the first child so it's not 1/3. This is the same problem you are having with the dice question. You aren't reading properly.
Once again Alan, you show your ignorance.Quote:
Originally Posted by Alan Mendelson
Read the very, very clear explanation on that web site. Until you can grasp where you are wrong on this you will never qualify to answer questions about dice.
By the way, speaking of ignorance. Why are you ignoring my questionQuote:
Now.... In my video script..... what proportion of bets on 'pair of deuces' would you expect to win? This isn't maths or charts or excel. A simple question. Do you expect to win 1 in 6 or 1 in 11.
I expect to win a pair of deuces 1/36 times. When one deuce is showing on one die the chance of a deuce showing on the second die is 1/6.
OnceDear regarding the children question until you understand what you are reading you will continue to make errors as you have with the dice question. You offered two answers to #2. Make up your mind BASED ON THE WORDING because there can't be two answers.
I didn't want to, but I returned to the WOV forum to ask the Wizard a question since it seems I am now being painted as ignorant. Since they are now talking about boy children and girl children, and heads or tails on coins, I asked this:
Wizard: If I am wrong, the same way you can visualize the coin problem by flipping heads and tails, and the children problem with stick figures of boys and girls, please use two dice to help me visualize the dice problem.
Thanks.
I am ready to be shown how when at least one die is a 2 how it is a 1/11 chance that both will show 2? As I've said and others have said, with one die a 2 (and in the case of both dice showing a 2 you just choose one) the "other die" only has 6 faces to choose among -- and not 11.
Again I am being criticized for referring to "the other die."
So, a visual presentation will help me understand what they're talking about.
Just look at the bloody table layout that indi kindly drew for you. place your money on pair of deuces or any other bet you like and then figure out how you can make a profit. It's as real world as any scenario that you can conceive of.Quote:
Originally Posted by Alan Mendelson
http://wizardofvegas.com/forum/quest...46/#post456141
Stop thinking about bloody dice, and start thinking of odds.
Unfortunately, OnceDear I have to think about two dice in a two dice question. It should be very easy for you to show, using two dice, why the chance of 2-2 with at least one die showing a 2, is 1-11. After all, it only involves two dice.
The coin flip problem is simple to illustrate.
So is the boy-girl problem -- simple to illustrate.
Certainly illustrating a problem with two dice must also be simple.
I am waiting for you and the Wizard (and the Wizard says he has video capability) to take two dice and show why -- when at least one die is a 2 -- the chance of showing 2-2 is 1/11.
This is simple to do, right? You've been showing me enough graphs and layouts of bets. Show me how the dice work to show 1/11 chances. Do it already.
OnceDear and RS___ in case you missed this on the WOV forum I explained MY PROBLEM in understanding how the answer is 1/11 when there are just two dice with at least one die showing a two. I responded to Math Extremists' post. His statement is in bold and mine follows.
MathExtremist
Please answer:
a) How many combinations of two dice have "at least one die showing a 2"?
b) How many combinations of two dice have "both dice showing a 2"?
First I want to say that I am sure you are a very nice and intelligent and worldly man and I appreciate your help in making me understand.
The problem I AM HAVING is that I can't consider 11 different combinations with at least one die showing a 2 in a two-dice problem. And let me explain why.
With each die, when a 2 is showing, there can no longer be 11 different combinations. Let me show you:
2....1
2....2
2....3
2....4
2....5
2....6
So, when you tell me that at least ONE DIE is showing a 2, I look at my combinations above and see that there are only 6 -- not 11.
If we look at the mirror image of the combinations above I still get only 6 combinations -- not 11.
1....2
2....2
3....2
4....2
5....2
6....2
Now, it is also possible that when we say "at least one die" it can also mean that BOTH dice are showing a 2. Well, when I look at two dice, both showing a 2, I treat either of the dice as being the "variable die." And by that I mean, this:
2....2 (but it could have been a 1)
2....2
2....2 (but it could have been a 3)
2....2 (but it could have been a 4)
2....2 (but it could have been a 5)
2....2 (but it could have been a 6)
And what this indicates to me, is that even if both dice were showing a 2, you still have only a 1/6 of getting that 2-2 combination.
So this is why I ask: using two dice, show me how the answer is 1/11 ??
YOU MUST treat either of the dice as being the "variable die." You don't victimise and nail one of the little 6u99ars down.Quote:
Originally Posted by Alan Mendelson
You simply can't nail one down.
It's like having two naughty children and a broken window. At least one of them broke it, but you cannot just take one of them and give him a slap.
2....2 (but it could have been a 2...1 AND it could have been a 1...2)
2....2
2....2 (but it could have been a 2...3 AND it could even have been a 3...2)
2....2 (but it could have been a 2...4 AND it could easily have been a 4...2)
2....2 (but it could have been a 2...5 AND it could quite possibly have been a 5...2)
2....2 (but it could have been a 2...6 AND it could unless someone had nailed one specific die to the table have been a 6...2)
And what this indicates to me, is that even if both dice were showing a 2, you still have only a 1/11 of getting that 2-2 combination.
Below is a photo of a photo of 11 combinations of 2 dice with at least 1 die a 2. This is supposed to prove to me that the answer to the original question is 1/11 and my comments above about there being only 6 combinations are wrong. This was posted on the WOV forum.
Well, here's what I wrote on the WOV forum (in bold): Not quite. You're not using TWO DICE. This is a two dice problem.
Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.
OnceDear wrote above: YOU MUST treat either of the dice as being the "variable die." Unfortunately, I can't do that. If at least one die is showing a 2, I have to treat the die showing a 2 as the 2. And if both dice happen to be showing a 2, then I can pick either one and still get 1/6 as my answer.
Again, this goes back to what I said before: you have a two dice problem and you can't shift the value of the cube with a 2 to fit your 1/11 answer.
You guys are nucking futs.
And as I said on that forum. Now you are just talking stupid!!!
Goodbye Alan.
May you either learn to read, or die wondering if you might ever have figured this out, perhaps by listening, looking, or God forbid, thinking.
One would think a cadre of geniuses could make a simple video. C'mon, dudes -- pool your gambling bankrolls for 2015 and get a Super Eight.
redietz they wouldn't dare attempt to make a video using ONLY two dice. They keep on using multiple dice to come up with 1/11. Even worse: when a die is settled showing a 2 they still use the other 5 faces on that same die to explain their answer is 1/11.
Hi Alan,
I don’t want to muddy the waters or anything, but I wanted to take a moment to describe how my own past personal experiences made me believe the 1/11 figure to be the answer to the question.
Although I feel like I have eventually come to understand the math that has been provided explaining the 1/11, it was my real world experience that had the most initial influence on my opinion (versus the math).
I have enjoyed craps for a while, so I thought back on my own time at the tables over the years. I thought about the many different rolls of the dice I have seen and then applied two things to my recollections:
1. I assumed any roll of the dice on the craps table that included one (or more) 2’s would count as satisfying the initial statement of “two dice are thrown and at least one die is a 2”.
2. I asked myself: “For all of those times I saw any roll on the table that had a 2 in it, do I feel like I saw a pair of 2’s one-sixth of the time?”
I didn’t think about cups, peekers, multiple combinations, specific die, etc., etc. I just applied the two items above to my memories of the tables.
It’s not very scientific (or mathematical) and is obviously based upon my own personal “feelings” and memories of playing craps, but I just didn’t feel like I saw a pair of 2’s every sixth time (on average) when there was a 2 on the table.
If my assumption #1 is felt to be correct, I’d be curious to hear what other people’s feelings are when applying the two criteria to their own craps table experiences & memories.
Anyway, just thought I’d share my own thoughts on the question.
Thanks for providing the forum!
blount2000 thanks for joining and for posting.
I'd like you to do me this favor: take two dice -- any two dice will do -- and put them on your table or desk with at least one of the two dice showing a 2. Knowing that one of the two dice is showing a 2 ask yourself: what are the odds that the other die can also show a 2? What is your answer?
Thanks.
BULLETIN NEWS. A video has been shot. Please view and comment:
https://www.youtube.com/watch?v=zT4yAJejRe8
Hey, I had a hard enough time getting my mind wrapped around the original question! I’m not mentally prepared at this point for a follow up! :)
Like I mentioned, I had to apply my own personal experiences at the craps tables to the situation in order to make it understandable to me.
There has already been much back & forth discussion on the wording, interpretation, intent, etc., of the original question. My own interpretation of the original question is reflected in how I described looking back on my past craps play.
I don’t want to “make it all about me”, but in the spirit of keeping it within my own experiences (and hence my understanding) I’d like to know if my assumption #1 written above does (or doesn’t?) satisfy the statement in the original question (i.e. two dice rolled, at least one is a 2).
If my assumption does satisfy the initial question, then I come back to my feeling that I’m just not seeing a pair of twos 1/6 of the time that any two is showing up on the table.
If my assumption doesn't satisfy it, well, then I must have interpreted the question wrong and that will be that. It won’t be the first (or last) time I was wrong!
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.
And here's what I said about the video shot my miplet on the WOV forum: he ignored the original question. He simply rolled two dice until "at least" one 2 was showing and then proceeded to show us that there are 11 combinations of two dice with at least one 2. He didn't answer the question.
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.
the Wizard himself has posted a video: https://www.youtube.com/watch?v=Ej-sZk3_Idk
Again, he shows the question but instead of answering the question with the information provided (at least one die shows a 2) he proceeds to once again refer to the eleven dice combinations including a 2.
Mike threw in the possibility that 2-2 shows on the initial throw and yes, Mike, as I said before -- it doesn't matter. Yes, you take one of the 2s arbitrarily because it does not affect the question. The question simply asks for you to figure the chance that another die in a two dice problem matches the number shown on the other die in the two dice problem.
Anyone else care to comment? I also asked to be suspended from the Wizard's forum. No one there will actually address the actual question asked. Even Mike has sidestepped the question in his video response.
Well, at least I finally think I understand what you're saying now - yes, the odds of rolling a pair is 1/6. I still think you're parsing the question unnaturally at best, and find it hard to believe you don't think our interpretation is at least equally valid.
Yes, the odds of rolling a pair with two dice is 1/6. Thank you.Quote:
Originally Posted by synergistic
synergistic your interpretation does not include that one die has been identified. And that's the big sticking point. The 1/11 answer is the correct answer to the question about dice combinations. But it doesn't answer a specific question when one particular number has been identified.
Again: the original question asks you to solve the problem of a 2 being rolled and the chances of another 2.
The original question does not ask for anyone to discuss all of the combinations showing a 2 and then to say how many of those combinations would show 2-2.
I think it was regnis who said something early-on in the first discussion about taking the simplest, most direct route and answer. And when you know there is a 2 showing you should take the shortest, most direct route -- and not do the mathematical gymnastics.
"Miplet" did the gymnastics and then the "Wizard" did the gymnastics. The gymnastics is "valid" for showing a solution to this kind of problem, but it didn't answer this particular problem.
As I said over on the WOV forum it's amazing how many "non math" people using common sense answered 1/6. Does that make "common sense" stupid? Or does it mean that the math guys overthought the problem? They overthought it and missed it.
Alan: I respectfully submit to you that the die was not identified on purpose! The whole point of the question was a trick designed to lull the casual and unsuspecting into believing the answer is 1/6. It really is not 1/6, and the question was intentionally worded that way to stimulate conversation from those who initially believed it was 1/6.Quote:
Originally Posted by Alan Mendelson
Think of it this way: If the creator of that original question was clever enough to know that many people would be fooled into thinking it was 1/6, then maybe the question was not poorly worded at all? Rather, it was intentionally worded that way to provoke the exact reaction you are providing right here on this thread and on WOV?
Judging from your reaction and the length of this thread along with the one over on WOV, I would say the original question did a beautiful job of stimulating that conversation.
Whoever authored that question should be collecting some royalties by now based on the number of collective cerebral cortexes it managed to churn....
No, my interpretation does NOT account for one die being identified - because the way I parse the sentence, you haven't been told anything about a specific die, you've been told something about the two dice as a unit. Your version is, to me, a less intuitive way to read the question, and I believe it yields a misleading answer.
Count room if you read some of my posts in the early discussion, I said the question was worded to fool people into getting the 1/11 answer because it mentioned the two-coin problem. I wrote that again today on the WOV forum. I think that set up the math guys to answer the question using the wrong methodology.
Here's what I wrote on the WOV:
As I said early-on in the discussion, the question was written to make you consider the 11 combinations because of this phrase that preceded the original question:
"Along the vein of the Two Coin Puzzle,..." It was that phrase that made everyone make the question more complicated than it really is.
It was that phrase "Along the vein of the Two Coin Puzzle,..." that had you all thinking you had to solve the question as you would with the two coin puzzle and looking at all of the combinations of flipping coins. But again, this wasn't a coin flipping question. There are two dice and at least one of those two dice landed on a 2.
synergistic when you say the original question "does not account" for one die being identified I am completely lost. When someone looks at two dice and "truthfully" tells you "at least one" is a 2 that to me is all the identification I need to proceed with the question and coming up with the answer of 1/6. Are you suggesting that the statement "at least one" is a 2 was a lie?
What really frustrated me with the entire WOV bunch was when they started to attack the meaning of English sentences and phrases. They even attacked the meaning of "at least one" and more than one WOVer actually said that made it okay for the 2 to move from one die to another. And then there were those who said "contained a 2" or "showed a 2" on the two dice meant different things.