Two Dice Puzzle: Part Deux
Because "math guys" are the last people you'd want to rely upon to understand a probability question? Perhaps the answer would have more credibility coming from a medieval historian?Quote: IbeatyouracesLet them believe whatever they want. Alan said over there that he's "given up on WoV" because they are "math guys."
Quote: MathExtremistBecause "math guys" are the last people you'd want to rely upon to understand a probability question? Perhaps the answer would have more credibility coming from a medieval historian?
Remember, we're talking about a guy who believes in DI and win/loss goals. He's also the poster boy for Caesar's Total Rewards.
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.
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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?
But back to the dice question. I've agreed that your take on it is the wording of the original question, not the actual math involved. You read it one way while we read it another. Sort of like the Second Amendment debate.
Quote: Ibeatyouraces
I've agreed that your take on it is the wording of the original question, not the actual math involved. You read it one way while we read it another. Sort of like the Second Amendment debate.
Yes, and on my forum I even explained when the answer 1/11 is the correct answer. And 1/11 is the answer to a different question. But the question under discussion (original question) doesn't call for looking at 11 different dice combinations UNLESS I and the others are reading it wrong. So EXPLAIN your reasoning for using the 11 different dice combinations please.
Quote: AlanMendelsonYes, and on my forum I even explained when the answer 1/11 is the correct answer. And 1/11 is the answer to a different question. But the question under discussion (original question) doesn't call for looking at 11 different dice combinations UNLESS I and the others are reading it wrong. So EXPLAIN your reasoning for using the 11 different dice combinations please.
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." :-)
Quote: IbeatyouracesI 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." :-)
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: AlanMendelsonHere'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.".
And that is where you are wrong. That's not how math works.
Quote: AlanMendelsonHere'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.
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.
Quote: AlanMendelsonBut the question under discussion (original question) doesn't call for looking at 11 different dice combinations UNLESS I and the others are reading it wrong. So EXPLAIN your reasoning for using the 11 different dice combinations please.
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.
Quote: IbeatyouracesYou can do this exact same thing with 12 cards. Ace through 6 of two different suits.
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.
Quote: AlanMendelsonYes, and on my forum I even explained when the answer 1/11 is the correct answer. And 1/11 is the answer to a different question. But the question under discussion (original question) doesn't call for looking at 11 different dice combinations UNLESS I and the others are reading it wrong. So EXPLAIN your reasoning for using the 11 different dice combinations please.
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.
Quote: AlanMendelson... So EXPLAIN your reasoning for using the 11 different dice combinations please.
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.
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.
Quote: thecesspitQuote:
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.
Quote: OnceDearQuote: thecesspitQuote:
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.
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".
To recap;
First die is a 2 (1/6 the time), second is a spinner. You have a 1/6 chance of rolling a hard 4 and a 5/6 chance of rolling a non-hard-4. This happens 1/6 of the time. From this game state, we get a probability of 1/36 of a hard 4 and a 5/36 chance of an non-hard-4.
The other game state is where the first is a non deuce (5/6 the time). And the second die (spinning) is a deuce (1/6 the time). That gives us 5/36 chance for a non-hard-4.
The third state is 5/6 times where the first die is a non-deuce, but the spinner also lands on a non-deuce (5/6 the time). That gives us a 25/36 chance of a "neither die a deuce" roll.
Add them all up and what do you get?
Of the (11) rolls where at least one die was a deuce, there were 10 non-hard-4 rolls and 1 hard-4 roll. Or in other words, 1/11 of the rolls where at least one die was a deuce, both dice were a deuce.
Quote: RSOf the (11) rolls where at least one die was a deuce, there were 10 non-hard-4 rolls and 1 hard-4 roll. Or in other words, 1/11 of the rolls where at least one die was a deuce, both dice were a deuce.
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.
Quote: AlanMendelsonHere'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: RSAnd 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: IbeatyouracesQuit 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: MathExtremistYou'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: MathExtremistI 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: MaxPenIt 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: indignant99Because 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: thecesspitQuote:
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: OnceDearQuote: thecesspitQuote:
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: thecesspitQuote: OnceDearQuote: thecesspitQuote:
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: RSAlan 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: OnceDearIndeedy
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.
Quote: IbeatyouracesI'm not getting involved anymore. I've got better things to do.
Of course you're not going to be involved anymore. NOT ONE of you will do a video with two dice and with at least one die showing a two, to JUSTIFY your answer of 1/11.
Faulty logic would lead you to 1 in 10, since there are only 10 digits for you to choose from.
What are the odds that he picked 22, assuming he randomly picked a number from 10 to 99 with at least one two in it?
Is asking if he picked 22 the same thing as asking if the other digit was a two?
What if you also know that he didn't pick any two digit numbers containing a 7, 8, 9, or 0?
Quote: AlanMendelsonOf course you're not going to be involved anymore. NOT ONE of you will do a video with two dice and with at least one die showing a two, to JUSTIFY your answer of 1/11.
I don't have video service capabilities. I'm not creating an account either, nor do I need video proof. I know the answer. Now I've got a table game to get back to beating. Have fun!
Quote: AlanMendelsonOf course you're not going to be involved anymore. NOT ONE of you will do a video with two dice and with at least one die showing a two, to JUSTIFY your answer of 1/11.
Not one proof in the history of mathematics has required "video justification."
I've already posted an actual proof of the solution using Bayes' theorem. You completely ignored it, but it is unquestionably correct. Here's the link:
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/26/#post454379
I also posted an experimental design that you could run to easily disabuse yourself of your false notions, but you continue to decline to actually roll your own dice and count.
Instead, you make up some new and unrelated problem, which basically involves you rolling a pair of dice until a 2 shows up on exactly one die, and then rerolling the other one and asking "what are the odds this die is a 2?"
Somehow you think that's the same scenario as the problem statement of "what is the probability that a pair of dice each show 2 given that at least one die shows 2?" I don't know how and neither does anyone else. But when you're in a room full of experienced mathematicians, some of whom actually get paid to do probabilities, and you continue to reject all their suggestions and explanations in favor of your own logic, perhaps you should examine that logic rather than assume that everyone else is wrong.
None of you is responding to the conditions set up in the question.
And all of your responses about cards, and which of two dice is showing a 2, and picking numbers, and looking at the 11 combinations of two dice, MISS what is presented in the original question. But if any one of you attempted to do the video with the information presented in the video, and try explain your own 1/11 answer, you will see how the 1/11 answer does not fit the question.
Who will step up to the plate and show your stuff?
Wizard did you try?
SOMEONE shoot a video with two dice, and with at least one of them showing a 2, and explain to me with what is shown on the video how you can come up with the answer of 1/11. Do it.
You Tube accounts are free. I posted my video in less than three minutes. Do it.
Quote: AlanMendelsonQuote: indignant99Because 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?
Okay, Alan, you do another video. Show us exactly which combinations CAN show up under-the-cup, and which ones CANNOT. Use two dice of different colors.
Quote: AlanMendelsonI need each and every one of you to do this -- with two dice -- so you can see for yourself the conditions of the question. By doing the video, with the information provided in the question, you can see for yourself what the question is asking.
None of you is responding to the conditions set up in the question.
And all of your responses about cards, and which of two dice is showing a 2, and picking numbers, and looking at the 11 combinations of two dice, MISS what is presented in the original question. But if any one of you attempted to do the video with the information presented in the video, and try explain your own 1/11 answer, you will see how the 1/11 answer does not fit the question.
Who will step up to the plate and show your stuff?
Wizard did you try?
SOMEONE shoot a video with two dice, and with at least one of them showing a 2, and explain to me with what is shown on the video how you can come up with the answer of 1/11. Do it.
You Tube accounts are free. I posted my video in less than three minutes. Do it.
If no one else has made a video, I'll do one Monday. PM Tuesday if I forget.
Quote: AlanMendelson
I dare you to show this picture to your Las Vegas Forum, and to your Facebook rabble.
I want you to show me your roll of two dice, with one die showing a 2 and THEN you tell me how you can justify your answer of 1/11.
Once again, indignant YOU are changing the question. You are altering the problem. You are making changes to facilitate your answer of 1/11.
One of you man-up and put up your video.
Quote: AlanMendelsonI want you to show me your roll of two dice, with one die showing a 2 and THEN you tell me how you can justify your answer of 1/11.
This statement makes me think you don't actually understand what "probability" means. You don't need to roll any physical dice at all in order to know the probability and answer the question. You only need to understand the basic facts about how dice behave: two dice, each having six faces numbered 1..6, each of six faces is equally likely at 1/6. With just that information you can establish the following facts for a roll of the two dice:
A) The probability of no 2s is 25/36
B) The probability of exactly one 2 is 10/36
C) The probability of two 2s is 1/36
D) The probability of at least one 2 is 11/36 (the sum of B and C)
E) The ratio of C to D is 1/11, which is what the question was asking: when there is at least one 2 on the dice, what is the probability that both dice are 2? Do you dispute that this is what the question is asking?
You can experimentally validate the answer in E by
a) rolling the pair of dice until there is at least one 2 showing (that is, one or two 2s),
b) add one to a first tally, and
c) add one to a second tally if both dice show 2.
After a few hundred rolls, the ratio of the second tally to the first tally will approach 1/11.
I'd estimate the conditional probability that you will perform this experimental validation, given your other responses in this thread, is zero. Instead, based on those responses, I expect you to keep ignoring all of this and insist that the problem is really about just one die. It's not -- it's about the conditional probability for both dice given the information that at least one of the dice is 2 (which happens 11/36 of the time).
I am not here to discuss probability theory or anything else.
The original question is specific. If you think the answer is 1/11, show me how you set up the question and show me your explanation on a video so I can see it and we all can see it.
Frankly, I don't think you can do it.
I don't think you can take the original question, present and justify your 1/11 answer.
If you think I am wrong, show me your video and prove to me I am wrong.
PRESENT THE QUESTION AND ANSWER IT.
Quote: AlanMendelsonI am not here to discuss probability theory or anything else.
This is a probability theory problem. What are you here to discuss instead?
Quote: AlanMendelsonThe original question is specific. If you think the answer is 1/11, show me how you set up the question and show me your explanation on a video so I can see it and we all can see it.
Here's the proof via Bayes' theorem again. I can write it out here as well as I can write it out on whiteboard in front of a camera:
Quote: MathExtremistp(B|A) = p(A|B)*p(B)/p(A)
For this problem:
Event A = There is at least one 2 in a throw of 2 dice
Event B = There are two 2s in a throw of 2 dice.
The probability of at least one 2 in a throw of 2 dice:
p(A) = 11/36
The probability of two 2s in a throw of 2 dice:
p(B) = 1/36
The probability of at least one 2 in a throw of 2 dice, given that there are two 2s in the throw:
p(A|B) = 1
Therefore, the probability of two 2s in a throw of 2 dice, given that there is at least one 2 in the throw:
p(B|A) = 1*(1/36)/(11/36)
= 1/11
If you believe there is an error in this proof, kindly point it out.
It's show and tell time.
Make your video so that when I see it, I will say, "damn they are right!!" When there are two dice rolled and at least one die is showing a 2, the chance that both dice are showing 2-2 is in fact 1/11.
Do that for me. Because the original question asked if at least one of the dice is a 2, what is the probability that both dice are showing a 2?
Show me: With two real dice and one showing a 2 that the answer is 1/11. I am all excited about seeing this.
Edited to add: white board doesn't work. Do it with real dice.
Quote: Alan @ top of p.38But 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.
Are you trying to tell me math isn't real? I am dumbfounded.
If I do a video I'm not going to use a cup. Unless you think that changes anything?
Quote: RSWhat do I get if I do it? I at least want a lunch.
If I do a video I'm not going to use a cup. Unless you think that changes anything?
Absolutely no reason to use a cup. The only thing you need to show is that with two dice, and at least one die showing a 2, the chance of having 2-2 is 1/11.
I'd be happy to buy you lunch if you can show that.
Quote: mipletIf no one else has made a video, I'll do one Monday. PM Tuesday if I forget.
Its 1 in 91 for the three dice problem. 75 will have one 2, and 15 will have two 2s, and of coarse 1 way for three 2s
Save your breath on that video.
If it's simplified down to the atomic level of pre-1st grade fundamentals of numeracy then it will necessarily be long and Alan will say that it does not answer the question or that it answers some other question ( Which it would need to first, to dispel the ignorance fog). If it assumes the viewer has a modicum of intelligence and is following your reasoning, then it might be shorter, but unless it shows an 11 sided die, then the unenlightened viewer will dismiss it.
Indeed if it was short and sweet and still absolutely explicitly showed that Alan is abjectly wrong, then he will do as he always does and ignore it totally or selectively quote some few words from it.
Just as he now ignores and re-ignores the questions. . .
I still wonder what proportion of these events Alan would expect to see a pair of deuces.
and
Why won't he roll a pair of dice a few hundred times and MEASURE an estimate of the chances of both dice showing a two. I already have my 169meg of dice roll video, timestamped. But there's one person ineligible to download it. He would only say ( "They are not real dice", or "So what, this isn't about probability", or maybe "That was a lucky streak of several hundred rolls", or some such complete and utter load of bo*****s M********s
Thanks, you are right about the 1/91, of course. My working out had a flaw.
Quote: AlanThanks 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.
The 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: 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.
The question is if at least one die is showing a 2, what are the chances that both dice are showing a 2? I said the answer is 1/6. You say the answer is 1/11. Make a video showing me how the answer can be 1/11. Show me how, when one die is showing a 2, that there is a 1/11 chance that the other die will show a 2. Just as I did here: https://www.youtube.com/watch?v=uCT-5Rye9bk
There are no other questions. Just this one.
It's the same thing you asked me on my forum.
Take two dice, show a 2 on one of them, then show me how the other die has a 1/11 chance of showing a 2 to have 2-2.
The question states 'at least one' but doesn't specify which one.
I won't make a video mostly because I only have a 256kbps upload rate, but if I did, I would show you this:
Die on the left of the picture starts on a 2, die on the right, a 1:
2-1
I would then rotate the right die, showing you
2-3 2-4 2-5 2-6 2-2
I would then rotate the left die:
1-2 3-2 4-2 5-2 6-2
Each of those cases is unique,
those are all of the possible cases where "at least one die is a two",
each case occurs with equal probability,
and each demonstrates "the other die is a two"
and there are 11 cases.
In only one of those cases is the other die a 2
The other die is a 2 only one time in 11
| possibility | probability |
|---|---|
| 1-2 | 2/11 |
| 2-2 | 1/11 |
| 3-2 | 2/11 |
| 4-2 | 2/11 |
| 5-2 | 2/11 |
| 6-2 | 2/11 |
Quote: Dalex64
Each of those cases is unique,
those are all of the possible cases where "at least one die is a two",
each case occurs with equal probability,
and each demonstrates "the other die is a two"
and there are 11 cases.
In only one of those cases is the other die a 2
The other die is a 2 only one time in 11
Pah!! you are showing the number of possible and equally likely combinations of "At least one of them is a two". What have equally likely combinations got to do with probability? $:o)
It's not as if some idiot would give odds of say 8:1 on such probabilities.
Nooooo. that would be just the kind of stupidity that the myriad maths guys on WOV would try to come up with.
Meanwhile, back to the minority team of 1/6thers....
When are you going to answer the question that I keep banging on about, sort of related to the original question... What proportion of the times that 'at least one of the dice is a two' would we have the outcome of 'a pair of twos'? No need to explain the answer, just give it.
Quote: Dalex64You can also collapse the 11 equally probable events into 6 possible events which do not occur with equal probability:
possibility probability 1-2 2/11 2-2 1/11 3-2 2/11 4-2 2/11 5-2 2/11 6-2 2/11
Kewl. So it's kinda like 1-2 will happen twice as often as 2-2... Wow, I bet that would require some sort of... what's the word for it.... Oh. that's it.... Pair of dice
Quote: in my dreamsLOL. I got you guys good! Did you really think I was totally ignorant all along? I was only yanking your chain and you all fell for it hook line and sinker. Having you all frantically recording videos for me to ignore was the icing on the cake.
But he won't say that because there are now two possibilities: He really is the former, or he is doing the latter. Anyone taking odds.
Quote: OnceDearQuote: Dalex64You can also collapse the 11 equally probable events into 6 possible events which do not occur with equal probability:
possibility probability 1-2 2/11 2-2 1/11 3-2 2/11 4-2 2/11 5-2 2/11 6-2 2/11
Kewl. So it's kinda like 1-2 will happen twice as often as 2-2... Wow, I bet that would require some sort of... what's the word for it.... Oh. that's it.... Pair of diceSorry Dalex, It's not you I'm being facetious for
And yet another way to say it:
From this chart, there are 6 possible values for the other die when one of the dice is a two,
But those 6 possibilities do not have an equal probability of occurring.
Yeah... You could even pick the other die up and point to it's 6 sides in a video! See. 6 sides, not 11. Examine closely, because it must be a dodgy die, weighted or shaved or something. Oh. Hang on, This die is an imposter. This isn't the other die, this is the other other die. I think I put the wrong die in my pocket... Drat, must be my other trousers.Quote: Dalex64Quote: OnceDearQuote: Dalex64You can also collapse the 11 equally probable events into 6 possible events which do not occur with equal probability:
possibility probability 1-2 2/11 2-2 1/11 3-2 2/11 4-2 2/11 5-2 2/11 6-2 2/11
And yet another way to say it:
From this chart, there are 6 possible values for the other die when one of the dice is a two,
But those 6 possibilities do not have an equal probability of occurring.
