Two Dice Puzzle: Part Deux
Quote: RSThe die that already settled is a non-deuce. There is absolutely no possible way to get a 2-2. But, there is a possibility to get "at least one die is a deuce" -- the spinning die lands on a 2. And this is perfectly acceptable because at least one die is a 2.
Quote: Someone who bets nickelsWhaaaa. You mean to say you let me bet at 8:1 on this sure fire 1/6 game and just because the first die didn't come up as a two, but the second die did, you are going to hold me to my wager.
Well nobody told me that could happen!
Next you'll tell me that the payout table should somehow be based on possible outcomes and not just the number of faces on a die.... or is it A PAIR OF DICE?
Quote: AlanMendelsonMake your video so that when I see it, I will say, "damn they are right!!"
You're asking for an education, Alan, and you're attempting to start with an advanced topic when you don't have the fundamentals down. That won't ever work. You can't put up the roof of a new house without first building the foundation and the walls.
First, watch the Khan Academy video series on probabilities. This will give you a foundation.
https://www.youtube.com/playlist?list=PL06A16C388F14E6FE
Watch at least the first 25 videos up to "Conditional Probability" because that's what we're discussing here.
Then watch the following two videos that are specifically about conditional probabilities with dice:
https://www.youtube.com/watch?v=c7rwD0Z6V14
https://www.youtube.com/watch?v=OiXguSD_qdM
Once you have given yourself the foundation, and then watched and understood the above two videos, you should be able to figure out the odds for this specific two-dice problem and say "damn they are right!!"
Quote: MathExtremistOnce you have given yourself the foundation, and then watched and understood the above two videos, you should be able to figure out the odds for this specific two-dice problem and say "damn they are right!!"
I loved the title of this one. Bayes' Theorem - Explained Like You're Five !!!
Quote: AlanMendelsonNo this is not a multiple choice, pick your two-combination game.
Those 11 possible combinations, under-the-cup, are exactly why you're going to lose your a$$/lunch in your little Showdown with the Wizard.
You mean like...Quote: OnceDear...Nooooo. that would be just the kind of stupidity that the myriad maths guys on WOV would try to come up with.
So, this is how you guys determine a "good bet"?Quote: OnceDear...No need to explain the answer, just give it.
Good grief. Get the right answer; or get over it.Quote: indignant99Those 11 possible combinations, under-the-cup, are exactly why you're going to lose your a$$/lunch in your little Showdown with the Wizard.
Swiped a quote from a fridge magnet, now Indignant? http://www.amazon.com/mistake-once-thought-wrong-turned/dp/B001CG71RGQuote:Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
Add on: Some funny stuff going on here.
Quote: Dalex64You 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.
And supposedly we're the ones that can't comprehend. *rolls eyes*
Quote: IbeatyouracesSince when does "setting" and "rolling" the dice mean the same thing?
And supposedly we're the ones that can't comprehend. *rolls eyes*
Alan's just being ignorant.
Quote: IbeatyouracesSince when does "setting" and "rolling" the dice mean the same thing?
And supposedly we're the ones that can't comprehend. *rolls eyes*
Oops my bad. I should have said "roll the dice until you have at least one of the two dice showing a 2 and if both dice show a 2 that is okay to start your video."
But for those of you who want to save time for the purposes of shooting your video explaining your 1/11 answer, you can skip the rolling part and just "set" one die on a table showing a "2" and the other die on the table showing whatever face you'd like it to show. But remember -- in the real world that "2" cannot jump from one die to another.
Sorry Ibeatyouraces. This should clarify my error.
I am looking forward to your video.
Quote: RSAlan's just being ignorant.
Really, you are calling me ignorant? I explained my error. I look forward to your video. Remember -- the 2 cannot jump from one die to another.
Quote: RSAlan's just being ignorant.
Maybe. But I truly think he really wants to understand this even though it seems he's being hard headed. Otherwise he wouldn't be debating this.
Ever have a teacher get frustrated at someone because they can't learn something?
Quote: IbeatyouracesMaybe. But I truly think he really wants to understand this even though it seems he's being hard headed. Otherwise he wouldn't be debating this.
Ever have a teacher get frustrated at someone because they can't learn something?
Thank you Ibeatyouraces. I really want to understand how your answer is 1/11 so please, someone -- maybe the Wizard -- do a video for me. I grew up watching TV and it's my business, so I can easily relate to information via video. Just remember -- in the real, physical world the "2 face" cannot jump from one die to another. Thanks.
It'd be like if the Wizard got a probability problem wrong, especially an easy one. He'd likely be embarassed, especially if it was in public where people look up to him. (But I don't think Wizard would be acting this irrationally.)
Same with Alan. Except only one who looks up to him is Rob Singer. And none of us truly care (I don't).
IBYA: Yes. And it usually takes 2 or 3 attempts for the kid to understand. If not, a day or two later in class the kid finally figures it out.
But the kid who sat next to me, he didn't care at all. He understood nothing. (He didn't know which numbers were odd or even, in 5th grade.) But my friend was smart enough not to make a fuss. He would just fail his math homework and all was fine. Had he been Alan, the teacher would still be explaining how remainders work when you divide a number (ie: 17/3 = 5r2).
Quote: IbeatyouracesI don't know if this will help, but here is a reason you cannot set a die on 2 and roll the second separately. If you do that, you eliminate the possibility of the set die to become any if the other 5 numbers.
I'm sorry, I don't follow you. What I am looking for is an explanation of how the chance of two dice, with at least one die showing a 2, is 1/11 for both dice to show 2s. I said the answer is 1/6 and I was told this is the wrong answer since we don't know which of the two dice is showing a 2. Hence, my request for the video. Because in the real, physical world, the 2 face from one die cannot jump to the other die.
So, in translating the original question, it seems to me that one of the dice is showing a 2. Maybe both dice have 2. But at least one of the two dice has a 2 on it and that "2 face" cannot jump from one die to another die in the real physical world.
Knowing the real, physical world conditions, I would like to see a video explanation of the 1/11 answer.
And yes, it seems to me that in the real, physical world, 5 faces would be eliminated from one die.
This is a real, physical world question, isn't it?
I look forward to the videos. Thanks.
Quote: AlanMendelson(1) You can set Die A to show a 2 and Die B to be any of the six faces;
(2) You can set Die B to show a 2 and Die A to be any of the six faces;
(3) 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."
Correct. Now, since you've already watched the videos on probability that I posted earlier:
If you rolled the dice instead of setting them, what is the probability of each of your three situations? I numbered them for convenience.
I'm sure you'll agree that the probability of scenario 3 (both dice are 2) is 1/36. What about the other two?
Can you answer this question? Specifically, when rolling two distinguishable dice (that is, you can tell which one is A and which is B), what are the probabilities of scenario (1) and (2)?
Quote: AlanMendelsonI'm sorry, I don't follow you. What I am looking for is an explanation of how the chance of two dice, with at least one die showing a 2, is 1/11 for both dice to show 2s...
In this case, all you need to do is look at all of the possible 36 total outcomes. 25 have no 2, 11 do, and 1 has both. It's just that simple. The cup and peeker is nothing more than to cause confusion.
Quote: RSIf Alan cared, he'd have watched the videos posted by MathExtremist (I think ME posted them...maybe someone else did). The second to the last one is perfect for this situation.
I do care, and unfortunately this video: https://www.youtube.com/watch?v=c7rwD0Z6V14 gives options for each of the two dice and does not specifically say that a particular die is a fixed number.
In our problem we have a 2 on at least one of the two dice. Our problem does not say "at least one of the dice is showing a 2 or a 3." Our problem says "at least one of the dice shows a 2."
I look forward to your video. Remember -- no Disney animation.
Quote: IbeatyouracesIn this case, all you need to do is look at all of the possible 36 total outcomes. 25 have no 2, 11 do, and 1 has both. It's just that simple. The cup and peeker is nothing more than to cause confusion.
I am sorry. That's not what happens with two dice in the real, physical world. This is why I would like to see your video.
Quote: AlanMendelsonI am sorry. That's not what happens with two dice in the real, physical world.
That's flat-out wrong. Did you not review the videos I linked to?
What are the chances of rolling a single die and not seeing a 2?
What are the chances of rolling two dice and not seeing a 2 on either?
If you can't answer these questions, you need to go back and review the videos again.
EDIT: I found this handy chart. You've seen something similar many times before in this thread. Just look at it and count how many possible combinations involve no 2s and how many involve at least one 2.
Having looked at that chart, do you actually dispute the counts listed by Ibeatyouraces?
Quote: AlanMendelsonI am sorry. That's not what happens with two dice in the real, physical world. This is why I would like to see your video.
So you're saying math doesn't work out in the real & physical world?
Quote: Alan
In our problem we have a 2 on at least one of the two dice. Our problem does not say "at least one of the dice is showing a 2 or a 3." Our problem says "at least one of the dice shows a 2."
You gotta be f*cking kidding me.
OK, so instead of highlighting all the ones with at least a 2 or a 3 in them....just highlight the ones with a 2. That's not so difficult to figure out, is it?
Or the casino?Quote: RSSo you're saying math doesn't work out in the real & physical world?
A woman is chosen at random from all women in the world with two children. She is asked, "Do you have at least one boy?" She answers, "yes." What is the probability she has two boys?
Quote: WizardThis may have been asked before, but let me see if this will help Alan and the 1/6 camp.
A woman is chosen at random from all women in the world with two children. She is asked, "Do you have at least one boy?" She answers, "yes." What is the probability she has two boys?
You should edit to add: Assume the probability of having a boy and a girl are equal at 1/2. That's not actually true in human society but it should be for this problem. I thought this puzzle was done with fair coins anyway (and heads/tails rather than girls/boys)...
Quote: RS
OK, so instead of highlighting all the ones with at least a 2 or a 3 in them....just highlight the ones with a 2. That's not so difficult to figure out, is it?
But we are not highlighting all the dice combinations with a 2 in them. We have at least one die with a 2, and we are asked about the other die. I would like you to shoot your video and show at least one die with a 2, or both dice showing a 2, and then show me how your answer is 1/11.
Very simple. Do it.
You shake the two dice in a cup and and slam it upside down on the table. Someone peeks and truthfully says "at least one of the dice is either A or B." What it's the probability that both A and B are showing?
Quote: WizardThis may have been asked before, but let me see if this will help Alan and the 1/6 camp.
Mike,
You surely noted by now that every time alan is presented with primary school examples of probability, he not only refuses point blank to answer, but he also accuses us of changing the question.
In particular, now he refuses to answer 'What proportion of times wold he expect to see a pair of deuces when at least one die is a deuce'
ME even gave him the MOST deadly simple exercise 'What is the probability of throwing two dice and NOT seeing at least one deuce'
If Alan would spend one minute on that simple question he'd have half of a sound grounding in his education.
He doesn't want to understand. He never will dare.
Quote: WizardThis may have been asked before, but let me see if this will help Alan and the 1/6 camp.
A woman is chosen at random from all women in the world with two children. She is asked, "Do you have at least one boy?" She answers, "yes." What is the probability she has two boys?
You did a video Wizard about the actual problem under discussion. Please post it. Show us how with at least one die showing a 2 the answer is 1/11 for the other die.
I am sure you didn't use Disney animation and the 2 doesn't jump from die to die.
Oh, about the woman with boys? Just a variation of the coin flip problem. But it has nothing to do with rolling two dice, and at least one of the two dice showing a 2.
Quote: AlanMendelsonBut we are not highlighting all the dice combinations with a 2 in them. We have at least one die with a 2, and we are asked about the other die. I would like you to shoot your video and show at least one die with a 2, or both dice showing a 2, and then show me how your answer is 1/11.
Very simple. Do it.
I might be lost. Is Dween's question the one at issue, or is there a new one?
Quote: Dween
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?
Because Alan's request is not equivalent to Dween's question.
Quote: MaxPenThis thread is starting to get beyond stupid now. I would be embarrassed if I still didn't understand it was 1/11 by now.
I wouldn't. I admittedly have a very hard time with all math, certainly including this kind. And I remember one very long conversation with Doc about thermodynamics which, despite his best efforts and using terms a 12yr old could understand, still left me partially unsure of the concept.
I still don't fully understand this. By "fully understand", I mean I can't break it down piece by piece, explain it fully, work it forwards and backwards, or apply it to another example. That's not embarrassing. I just don't have the mind for math.
But when a vast majority claims something, and said majority includes experts who do this kind of stuff for a living, and all the experts agree, and they all break it down as much as possible so I can understand parts and pieces (if not the entire concept), there comes a point when I just accept it. Probably somewhere back around pg11.
Some people just aren't made like that.
Quote: MathExtremist
EDIT: I found this handy chart.
Sadly Chart isn't visible to us.
Quote: LoquaciousMoFWI might be lost. Is Dween's question the one at issue, or is there a new one?
Because Alan's request is not equivalent to Dween's question.
LoquaciousMoFW I think my request is indeed the equivalent:
Show me a video where you have two 6-sided dice in a cup. Shake the dice and slam the cup down onto the table. Have at least one of the two dice showing a 2. Now tell me -- and explain it in your video -- what is the probability that both dice are showing a 2.
This is a real world presentation of the problem. And in the real word, the 2 showing on at least one die cannot jump to the "other die."
If you show a video with both dice showing a 2 (because that's how they ended up after the shake) that's okay. In that case also please explain how the answer is 1/11 which is what most -- if not all -- of the other forum members are saying.
Now, let me cut to the chase and probably end this discussion:
In the real world, the 2 cannot jump from one die to another. If you think the 2 can jump from one die to another, and if you do that in your video with the help of animation, then yes the answer can indeed be 1/11.
But the 2 face doesn't jump. It's stuck on the die it shows on. That leaves the question pertaining to the other die of which there are six faces.
You can use your graphs and excel spreadsheets to come up with the 1/11 answer, or you can take two real physical dice and work out the problem. And in the real, physical world, the answer is 1/6.
I have said this from the beginning: you "math guys" have either "overthought" the problem or "ignored" the real, physical world to come up with your answer.
The "2" can't jump from one cube to another cube to facilitate your 1/11 answer. Sorry.
Quote: LoquaciousMoFWI might be lost. Is Dween's question the one at issue, or is there a new one?
Because Alan's request is not equivalent to Dween's question.
Yes, it's still Dween's question. If you can stomach it, read this.
Quote: Kerkebet...Get the right answer; or get over it...
I do have the right answer. What's your problem/excuse?
Quote: AlanMendelsonBut we are not highlighting all the dice combinations with a 2 in them. We have at least one die with a 2,
How do you understand how likely it is that "we have at least one die with a 2"? You can understand that by "highlighting all the dice combinations with a 2 in them."
I don't understand the confusion on this point. Do you actually think these two points are unrelated?
Quote: AlanMendelsonShow us how with at least one die showing a 2 the answer is 1/11 for the other die.
This sentence does not make sense. When you understand why, you'll understand the actual problem statement and the solution.
Quote: MathExtremistThis sentence does not make sense. When you understand why, you'll understand the actual problem statement and the solution.
the sentence "Show us how with at least one die showing a 2 the answer is 1/11 for the other die" should have ended with a question mark. It was a question. Forgive my typo.
Look guys, you can look at two real, physical dice or you can look at your graphs and spreadsheets to determine the answer.
If the number 2 could jump from one die to another the answer would be 1/11. But with the number 2 fixed on one die the answer is 1/6.
You won't hear from me anymore about this.
Quote: AlanMendelsonBut with the number 2 fixed on one die the answer is 1/6.
If that were the question, you'd have the correct answer. Unfortunately (for you) the 2 is not fixed, hence "at least one die shows a 2".
Quote: AlanMendelsonLook guys, you can look at two real, physical dice or you can look at your graphs and spreadsheets to determine the answer.
Ironically, that's true -- both yield the same answer. But you think they don't so now you're disputing the ability of spreadsheets to model real-world dice games?
Quote: IbeatyouracesLet's try this. Instead of using the 2's on the dice, let's label one 2 "A" and the other "B".
You shake the two dice in a cup and and slam it upside down on the table. Someone peeks and truthfully says "at least one of the dice is either A or B." What it's the probability that both A and B are showing?
Try this^^^
Quote: IbeatyouracesQuote: IbeatyouracesLet's try this. Instead of using the 2's on the dice, let's label one 2 "A" and the other "B".
You shake the two dice in a cup and and slam it upside down on the table. Someone peeks and truthfully says "at least one of the dice is either A or B." What it's the probability that both A and B are showing?
Try this^^^
No, no, no! Whoa, whoa! You'd better label them both "A." For a lot of reasons...
- "A" and "B" both showing up, might not be considered a pair.
- A-plus-anything is NOT equivalent to B-plus-anything (smirk).
- Alan will certainly claim "that's not the original problem, as stated."
- Even substituting Deuce with A won't pass the "original problem" hurdle.
Quote: MathExtremistYou should edit to add: Assume the probability of having a boy and a girl are equal at 1/2. That's not actually true in human society but it should be for this problem. I thought this puzzle was done with fair coins anyway (and heads/tails rather than girls/boys)...
Yes, let's assume every child has as 50/50 chance to be male or female (the actual probability is 50.5% male, and there is a positive correlation between siblings). I think it would help to Alan see the light in dealing with babies as opposed to things that induce math phobia like coins and dice.
Quote: AlanMendelsonYou did a video Wizard about the actual problem under discussion. Please post it. Show us how with at least one die showing a 2 the answer is 1/11 for the other die.
I am sure you didn't use Disney animation and the 2 doesn't jump from die to die.
Oh, about the woman with boys? Just a variation of the coin flip problem. But it has nothing to do with rolling two dice, and at least one of the two dice showing a 2.
Yes, I spent three hours on a video on this but the audio was awful and I didn't want to embarrass myself with it. My wife thought I was nuts making a video of me shaking a dice shaker for some 655 times it took to get at least one two 200 times.
I think it would help to grasp the fundamental principle here to at least answer the boy/girl question.
Again, a woman chosen randomly has two children. She is asked, "Do you have at least one boy?" She says "yes." What is the probability she has two boys? Assume each child as a 50/50 chance to be a boy or girl.
Please humor me and answer the question.
Quote: indignant99I do have the right answer. What's your problem/excuse?
My issue here is that something clearly remains unresolved. And, the onus lays on the side of the mathematicians to clear it up. Certainly, not on television reporter Alan alone.
Why the apparently protracted angst against one or two hold-outs? Off hand, I see nothing to gain from thus trying to convert Alan to the forum's way of thinking here; except, some sort of sensationalism for the site. But, the apparent angst continues.
Lastly, how do you know that you have the only right answer? There are other ways to look at the gist of the problem as written. There almost always are.
Quote: Wizard...A woman is chosen at random from all women in the world with two children. She is asked, "Do you have at least one boy?" She answers, "yes." What is the probability she has two boys?
Just saw this on the news so I looked for an internet article. What are the odds on this, LOL??
http://www.foxnews.com/us/2015/05/07/michigan-couple-with-12-sons-waiting-to-find-out-whether-baby-no-13-keeps-all/
Quote: WizardThis may have been asked before, but let me see if this will help Alan and the 1/6 camp.
A woman is chosen at random from all women in the world with two children. She is asked, "Do you have at least one boy?" She answers, "yes." What is the probability she has two boys?
Someone let me know if this is correct like I believe. It took me a little bit to get the 1/11. I find this very interesting.
She has at least one boy so it's narrowed down to Bb, Bg, Gb (3 possibilities). Large letter representing first child born. So being that there are 3 possible outcomes with only one of those being boy boy, my answer is 1/3.
Quote: MaxPenOk. So the possibles are BB, BG, GG, GB with 2 children.
She has at least one boy so it's narrowed down to Bb, Bg, Gb (3 possibilities). Large letter representing first child born. So being that there are 3 possible outcomes with only one of those being boy boy, my answer is 1/3.
That is correct.
Quote: KerkebetMy issue here is that something clearly remains unresolved...
Only in the minds of unclear thinkers.
