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:
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:
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?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.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?
<<Originally Posted by arcimede$....chuckle....
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.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.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?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.
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