Two Dice Puzzle: Part Deux
Quote: AlanMendelsonThere are two ways to get from point A to point B. The shortest route is a direct line. And then there is the "1/11 route."
I give up.
But I am still interested in finding out how the Wizard decided the answer is 1/11 ??
There are two ways. Count them out, which has been demonstrated several times in this thread, or maybe he used Bayes' theorem. That goes like this:
p(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
Quote: indignant99Not at all. You're separating the dice into two separate, distinct observations.
IT DON'T WORK THAT WAY! You get to view both simultaneously, and either one - either GD one - can obligate the announcement of "deuce."
This is ridiculous. More changing of the question. In fact, each of you seems to be changing the question your own way to fit your own argument.
I am not changing the question at all, or the conditions of the question.
In the original question you are not told what both dice are. You are told "at least one" is a deuce. You do not know if the judge/partner/friend/witness has seen both dice. And it still doesn't matter. As long as one of two dice is a deuce, whether or not you have 2-2 all rests with the other die -- and that's 1/6.
I am sorry but unless one of you can clearly explain to me why I shouldn't be taking the short, direct route between A and B, I am sticking with the short, direct route between A and B. I see no reason to make the issue any more complex than looking at one die and counting its faces. I see no reason to look at eleven different two dice combinations when the question is 1/6 faces on one die. I see no reason to flip coins or come up with elementary school puzzles that switch the denomination of two coins to trick you.
Quote: AlanMendelsonExcuse me, but where did you get the idea that I am not acknowledging the information that the partner/judge said at least one of the dice is a 2? I've been saying it all along -- that we know one die is a two. And it is that information which makes it clear that the question now revolves on the other die -- which has six sides.
I have been arguing this information all along -- that once the judge/partner truthfully says at least one die is a two the problem shifts to the remaining die (there are only two) and that is the 1/6 answer.
But you have not taken into account this information at all! You are effectively saying the other dice is 2 at a rate of 1 in 6, which is what we'd get if we didn't know anything about the pair of dice. The judge's information gives us info about the set of dice, and we can infer from this. The two events are now not independent and we can summarize the chances of a pair of 2's from 1 in 36 to 1 in 11.
But you can argue this by inference as much as you like. I've done it plenty of times with the EXACT rules stated in the question. It's 1 in 11.
I've already shown if it is 1 in 6, you end up with a deduction that a pair of two's is more than a 1 in 36 chance.
Quote:On the other hand, the 1/11 "faction" claims we need to know which of the dice is showing a two. How that can matter when we start with only two dice is what I can't understand. Perhaps the Wizard would like to explain? I asked him for his video to explain the process of coming up with the 1/11 answer, just as I posted a video explaining the 1/6 answer.
It'd matter if there was 300 dice, right? It'll matter if there's two, as the dice with the two on it is UNKNOWN, so it's not equivalent to the spinner example.
It's not a faction. It's the correct answer. If you'd spend more time actually trying it, and less time arguing from ignorance, you'd discover this for yourself. Spend an hour away from the key board and try it.
Quote: AlanMendelson
There are two ways to get from point A to point B. The shortest route is a direct line. And then there is the "1/11 route."
I give up.
But I am still interested in finding out how the Wizard decided the answer is 1/11 ??
Some things are impossible to figure out and better left alone.
Quote: AlanMendelsonYou are told "at least one" is a deuce. You do not know if the judge/partner/friend/witness has seen both dice. And it still doesn't matter.
Oh it matters very much. If the judge only sees one die and states "it's a two" then 1/6 is indeed the correct answer. But if the judge sees both dice, which he would if he "peeks under the cup" as the problem actually states, then 1/6 is not correct.
As before, the reason it matters is because the probability of observing "at least one 2" is very different depending on how many dice you look at. If you only look at one die, it's 1/6. If you look at two dice, it's 11/36. If you look at 100 dice, it's nearly 100%.
Quote: AlanMendelsonQuote: indignant99Not at all. You're separating the dice into two separate, distinct observations.
IT DON'T WORK THAT WAY! You get to view both simultaneously, and either one - either GD one - can obligate the announcement of "deuce."
This is ridiculous. More changing of the question. In fact, each of you seems to be changing the question your own way to fit your own argument.
I am not changing the question at all, or the conditions of the question.
In the original question you are not told what both dice are. You are told "at least one" is a deuce. You do not know if the judge/partner/friend/witness has seen both dice. And it still doesn't matter. As long as one of two dice is a deuce, whether or not you have 2-2 all rests with the other die -- and that's 1/6.
Quote:
"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?"
Whose changing the question now? The judge sees both dice, so we have to use information about the pair of dice to resolve the answer. It says it right there in the question We can not simply discard one dice and say 1 in 6. We have information about a group of dice, and statement that can only be made 11 times in 36. So we have to look at the possibilities of those 11 times.
But why do I care? I'm wasting my time trying to convince a marketing guy how to do probability, which is clearly not appreciated.
Quote: MathExtremistOh it matters very much. If the judge only sees one die and states "it's a two" then 1/6 is indeed the correct answer. But if the judge sees both dice, which he would if he "peeks under the cup" as the problem actually states, then 1/6 is not correct.
As before, the reason it matters is because the probability of observing "at least one 2" is very different depending on how many dice you look at. If you only look at one die, it's 1/6. If you look at two dice, it's 11/36. If you look at 100 dice, it's nearly 100%.
Okay, now we have something concrete to discuss and you will have to do a better job of educating me on this one.
If you look at two dice and tell me that at least one of the two dice is a six, then I think the other die has a 1/6 chance of being a 6. And I don't care which of two dice you are talking about either. Because there are only two dice and if at least one of them is a 6 then that leaves only one that can be 1/6. Just as I said in my video.
And again, with only one six-sided die to consider there is no way I can come up with 11/36.
Quote: AlanMendelson... with only one six-sided die to consider there is no way I can come up with 11/36.
There's not just one to consider. (You think there is, because you artificially discard a Deuce, and then think you only have one left over to see.)
The announcement of "Deuce Ahoy, Mateys!" came as a result of seeing both.
These are the graphically explicit ways the dice could have rolled, and obligated an announcement of "Deuce Ahoy."
Now you can be my guest. Take a deuce away from all of those pairings.
You will be seeing { 1, 1, 2, 3, 3, 4, 4, 5, 5, 6, 6 }. Eleven little goblins to haunt you.
Quote: AlanMendelsonOkay, now we have something concrete to discuss and you will have to do a better job of educating me on this one.
If you look at two dice and tell me that at least one of the two dice is a six, then I think the other die has a 1/6 chance of being a 6. And I don't care which of two dice you are talking about either. Because there are only two dice and if at least one of them is a 6 then that leaves only one that can be 1/6. Just as I said in my video.
And again, with only one six-sided die to consider there is no way I can come up with 11/36.
First things first. You throw two dice. What are the chances that you observe at least one 2?
You know how to do this, yes? The shortcut is to realize that the opposite of "at least one" is "none" and then, since the chances of each die being "not 2" is 5/6, then the chances of seeing no 2s on both dice is (5/6)*(5/6) = 25/36. Do you agree?
If so, then the rest of the time you see at least one 2 when you throw the two dice. That's 1 - (25/36) = 11/36.
So far so good? Or not?
Quote: MathExtremistFirst things first. You throw two dice. What are the chances that you observe at least one 2?
Stop right there. That was not the question.
The question is specific.
Quote: indignant99There's not just one to consider. (You think there is, because you artificially discard a Deuce, and then think you only have one left over to see.)
The announcement of "Deuce Ahoy, Mateys!" came as a result of seeing both.
These are the graphically explicit ways the dice could have rolled, and obligated an announcement of "Deuce Ahoy."
Now you can be my guest. Take a deuce away from all of those pairings.
You will be seeing { 1, 1, 2, 3, 3, 4, 4, 5, 5, 6, 6 }. Eleven little goblins to haunt you.
Nice graphic, but if a 2 shows up on one die, the chance that it shows up on the other die is still 1/6.
Quote: AlanMendelsonQuote: indignant99There's not just one to consider. (You think there is, because you artificially discard a Deuce, and then think you only have one left over to see.)
The announcement of "Deuce Ahoy, Mateys!" came as a result of seeing both.
These are the graphically explicit ways the dice could have rolled, and obligated an announcement of "Deuce Ahoy."
Now you can be my guest. Take a deuce away from all of those pairings.
You will be seeing { 1, 1, 2, 3, 3, 4, 4, 5, 5, 6, 6 }. Eleven little goblins to haunt you.
Nice graphic, but if a 2 shows up on one die, the chance that it shows up on the other die is still 1/6.
Looking at the graph, it only shows up 1/11.
Ask yourself this question. Why do you think AP's are salivating to make bets against the 1 in 6ers? Surely you can't think they're stupid.
Quote: Ibeatyouraces
Looking at the graph, it only shows up 1/11.
Ask yourself this question. Why do you think AP's are salivating to make bets against the 1 in 6ers? Surely you can't think they're stupid.
I guess we are looking at different things. You are looking at two-dice combinations. God bless you.
I am looking at one die and one die only.
The original question tells me that a 2 is showing on one die. Therefore, the chances of a 2 showing on the other die (whichever die that is in a two dice problem, I don't care) are 1 out of 6.
Now I am going to bet the Wizard a $50 lunch because he is going to pay me more "points" than 6:1 for each time a 2 shows up on that "other die." And the contest is triggered when at least one 2 shows up when two dice are thrown.
The Wizard has done his simulations and believes he will win. Great. I will still take the bet with the Wizard. If I lose, I lose. But I think it's a good bet. But that is a side issue, because when rolling dice anything can happen. It's all random.
Getting back to the original question: the situation is specific. One die is known to be a 2. The question is what are the odds for a 2 showing on the other die. I still think that's 1/6 and not 1/11. Unless you have an 11-sided die, I am going to stick with 1/6.
Quote: AlanMendelsonStop right there. That was not the question.
The question is specific.
The question *is* specific yet you're ignoring half of it. The question is "In a throw of two dice, what is the probability that both dice show 2 given that at least one of them does?" There's an order to things. You can't answer the actual question without understanding the problem first.
The first thing that happens is you throw two dice but you don't see either of them. Then the peeker says "at least one of the dice is a 2". How often "at least one 2" happens is directly related to the answer for the final question. The final question, remember, is "What is the probability that both dice are showing a 2?" That's a direct quote from you.
If you think what the peeker says is not related at all to the answer for "what is the probability that both dice are showing a 2?", then that explains your error. Here are some simple examples to demonstrate why what the peeker says matters:
a) The peeker says "at least one of the dice is a 3." What is the probability that both dice are showing a 2?
b) The peeker says "both of the dice are showing 2." What is the probability that both dice are showing a 2?
c) The peeker says "none of the dice are odd." What is the probability that both dice are showing a 2?
d) The peeker says "at least one of the dice is even." What is the probability that both dice are showing a 2?
e) The peeker says "none of the dice are showing 2." What is the probability that both dice are showing a 2?
Do you think the answer is 1/6 in all cases?
Quote: AlanMendelsonI guess we are looking at different things. You are looking at two-dice combinations. God bless you.
I am looking at one die and one die only.
The original question tells me that a 2 is showing on one die. Therefore, the chances of a 2 showing on the other die (whichever die that is in a two dice problem, I don't care) are 1 out of 6.
Now I am going to bet the Wizard a $50 lunch because he is going to pay me more "points" than 6:1 for each time a 2 shows up on that "other die." And the contest is triggered when at least one 2 shows up when two dice are thrown.
The Wizard has done his simulations and believes he will win. Great. I will still take the bet with the Wizard. If I lose, I lose. But I think it's a good bet. But that is a side issue, because when rolling dice anything can happen. It's all random.
Getting back to the original question: the situation is specific. One die is known to be a 2. The question is what are the odds for a 2 showing on the other die. I still think that's 1/6 and not 1/11. Unless you have an 11-sided die, I am going to stick with 1/6.
Because you're thinking in terms of a specific die and not both.
I am still interested in hearing from the Wizard (seeing his video would be great) about why he thinks the question does not relate to one die?
Quote: IbeatyouracesBecause you're thinking in terms of a specific die and not both.
Exactly.
Now I'm going back to "work" :-)
Quote: AlanMendelsonME I am sorry, but I am not going to respond to you when you ask questions like the one above. Clearly we have a difference of opinion. I think the question relates to only one die. You say it doesn't. That's the end.
I am still interested in hearing from the Wizard (seeing his video would be great) about why he thinks the question does not relate to one die?
Fair enough, if you're done responding, I'm done trying to explain things. After you lose your lunch bet with the Wizard, if you decide you're interested in learning why you lost, you can find the answers in this thread.
Quote: IbeatyouracesLook at the graph again. How many "different" times is the peeker going to say, "there's at least one two.." Now how many of those are 2-2? If you answer 6 then 1 respectively, you're blind!
Now I'm going back to "work" :-)
Roll two dice. If one shows a 2, what are the chances of a two on the other die? Is it still 1/11 ??
Do you understand that 1/11 is an answer to a different question? I guess not.
Quote: MathExtremistFair enough, if you're done responding, I'm done trying to explain things. After you lose your lunch bet with the Wizard, if you decide you're interested in learning why you lost, you can find the answers in this thread.
I am going to ask you a different question: Why do you interpret the original question to consider more than one die in your answer? It's the same question I have for the Wizard and for everyone else who says the answer is 1/11.
Quote: AlanMendelsonRoll two dice. If one shows a 2, what are the chances of a two on the other die? Is it still 1/11 ??
Do you understand that 1/11 is an answer to a different question? I guess not.
To answer the first question, an emphatical yes! If I roll them SEPARATELY and the first ends up a two, then yes it's now 1/6.
To answer the second, no it's not.
Quote: IbeatyouracesTo answer the first question, an emphatical yes! If I roll them SEPARATELY and the first ends up a two, then yes it's now 1/6.
To answer the second, no it's not.
Very good.
But if you ask me, when you roll two dice together the chance of 2-2 showing up is 1/36 and not 1/11.
Quote: AlanMendelsonI am going to ask you a different question: Why do you interpret the original question to consider more than one die in your answer? It's the same question I have for the Wizard and for everyone else who says the answer is 1/11.
Here's your original question, emphasis for clarity.
Quote: AlanMendelson"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?"
I honestly don't understand how you're misreading that as being related to only one die. Every sentence in that problem statement contains the word "dice."
Quote: MathExtremist
I honestly don't understand how you're misreading that as being related to only one die. Every sentence in that problem statement contains the word "dice."
To answer your question: when you have a two dice problem and at least one die is known to be a 2 the question involves the other die. Do you disagree with that?
Or does the word "dice" make you think everything must be a two dice combination?
Quote: AlanMendelsonVery good.
But if you ask me, when you roll two dice together the chance of 2-2 showing up is 1/36 and not 1/11.
Correct, but only if you count all possible combinations. Now if we don't count the 1-3's, 5-4's, or any other combo without a two, now we improve, but ONLY to 1/11, NOT 1/6.
Quote: IbeatyouracesCorrect, but only if you count all possible combinations. Now if we don't count the 1-3's, 5-4's, or any other combo without a two, now we improve, but ONLY to 1/11, NOT 1/6.
Do me a favor and try this:
Roll two dice until at least one of the two dice shows a two. Then pick up the non-2 die and tell me what the odds are that the non-2 die could also show a 2? Do you get 1/6 or 1/11?
If, when you roll two dice, both dice show 2s, pick up one of the dice and look at it. The chance that the die you picked up is a 2 was what? 1/6 or 1/11?
That's how I read the original question.
Quote: AlanMendelsonDo me a favor and try this:
Roll two dice until at least one of the two dice shows a two. Then pick up the non-2 die and tell me what the odds are that the non-2 die could also show a 2? Do you get 1/6 or 1/11?
If, when you roll two dice, both dice show 2s, pick up one of the dice and look at it. The chance that the die you picked up is a 2 was what? 1/6 or 1/11?
That's how I read the original question.
I'm going to say this loud and clear....
THEY ARE ROLLED SIMULTANEOUSLY!! NOT SEPARATELY!!
You keep wanting to reroll the non two'd die after one two has been established. Quit doing that.
No wonder you can't beat casinos. So I quit trying to help.
Quote: IbeatyouracesI'm going to say this loud and clear....
THEY ARE ROLLED SIMULTANEOUSLY!! NOT SEPARATELY!!
Yes, and the two dice in my exercise are also rolled together and not separately. Now try it.
Quote: AlanMendelsonYes, and the two dice in my exercise are also rolled together and not separately. Now try it.
I did and it came out 1/10.8. Close enough to 1/11.
Quote: AlanMendelsonIf, when you roll two dice, both dice show 2s, pick up one of the dice and look at it. The chance that the die you picked up is a 2 was what? 1/6 or 1/11?
That's how I read the original question.
It's neither. The probability of picking up a 2, if both dice show 2s, is 1. Or 1/1 if you prefer.
Additional information informs your evaluation of the odds. Without any information, the probability of two 2s is 1/36. If you know that both dice are already 2s, it's 1. If you know that one of the dice is a 3, it's 0. If one of the dice is 2 and the other is still spinning, it's 1/6. And if you know that at least one of the dice is a 2 and both have already stopped, it's 1/11. Yes, there is a difference between the last two examples.
Quote: AlanMendelsonME your lessons are all very informative but answer this question please: why do you insist on looking at two dice instead of one die?
Because that's what the problem says. Your interpretation, that the probability of two 2s is really about only a single die, is wrong.
The probability of a single die showing 2 is different than the probability of at least one of two dice showing 2. The latter is how the problem starts, but you don't comprehend the distinction (yet). But the answer to the final question, "what is the probability of both dice showing 2" depends on that. I pointed this out a few pages ago but you ignored it. If you've changed your mind about trying to understand it, re-read the past half-dozen pages. It's all there.
Quote: MathExtremistBecause that's what the problem says. Your interpretation, that the probability of two 2s is really about only a single die, is wrong.
So that's it. We disagree.
Quote: AlanMendelsonSo that's it. We disagree.
We do, but this isn't a matter of opinion. This disagreement is equivalent to me (and everyone else) saying "2+3=5" while you believe "2+3=3 because the first 2 doesn't matter."
That's just not the way the math works. Again, if you're willing to re-read and comprehend the past dozen pages or so, it's all in there.
The dice don't care how many of them you're looking at. The odds are what they are whether you interpret the question properly or otherwise.Quote: AlanMendelsonThanks but no. I read the question. You can interpret the question anyway you like, but the way it was written tells me to look at one die.
Good luck at your lunch. You'll need an awful lot of it, because the bet you're making behaves exactly the same way as the hard 8 bet: 1 way to win at 9 to 1, 10 ways to lose, 25 ways for no-resolution. That's a 9.1% house edge. The only difference between hard 8 and the "hard 4 when at least one 2" bet is which dice combinations are winners, losers, and no action.
Quote: MathExtremist
Good luck at your lunch. You'll need an awful lot of it, because
You realize this makes no sense? Is that how you read and interpreted the original question?
I realize it makes no sense to you. I'm at a loss for how to explain it another way than the several ways I've already posted. The only other thing I can suggest is to get a pair of dice and play out the bet on your kitchen table. Keep track of how often wins/losses/no-actions happen and see for yourself whether the ratio of wins to (wins+losses) is 1/6 or 1/11. That suggestion has already been made by others, so I really don't have anything else to add.Quote: AlanMendelsonYou realize this makes no sense? Is that how you read and interpreted the original question?
Good luck.
Even McDonald's has some people believing their food is not crap.
This thread needs to die. The problem has been sliced and diced to death.
One can either see the truth or live in a fantasy land.
Quote: MathExtremistI realize it makes no sense to you. I'm at a loss for how to explain it another way than the several ways I've already posted. The only other thing I can suggest is to get a pair of dice and play out the bet on your kitchen table. Keep track of how often wins/losses/no-actions happen and see for yourself whether the ratio of wins to (wins+losses) is 1/6 or 1/11. That suggestion has already been made by others, so I really don't have anything else to add.
Good luck.
Please... go to your kitchen table and roll two dice. When one of the dice is a 2, look at the other die and ask yourself will another 2 show up 1/6 of the time or 1/11 of the time?
Quote: AlanMendelsonPlease... go to your kitchen table and roll two dice. When one of the dice is a 2, look at the other die and ask yourself will another 2 show up 1/6 of the time or 1/11 of the time?
Please go to your kitchen table and roll the dice 300 times, and tell us.
Quote: AlanMendelsonPlease... go to your kitchen table and roll two dice. When one of the dice is a 2, look at the other die and ask yourself will another 2 show up 1/6 of the time or 1/11 of the time?
That's the wrong tense. You've already rolled the dice. The question is what they are, not what they will be if you re-roll them. Just record how many times you see "two 2s" and "at least one 2". That's all you have to do. The ratio will be nowhere close to 1/6.
Quote: AlanMendelsonPlease... go to your kitchen table and roll two dice. When one of the dice is a 2, look at the other die and ask yourself will another 2 show up 1/6 of the time or 1/11 of the time?
What do you do when both are a two?
Also, will anybody in the 1/6 camp be attending the WoV Spring Fling. If so, please bring gambling money and I'll bring the dice.
Quote: WizardWhat do you do when both are a two?
Also, will anybody in the 1/6 camp be attending the WoV Spring Fling. If so, please bring gambling money and I'll bring the dice.
They'll think you'll bring biased dice.
Quote: WizardWhat do you do when both are a two?
I addressed this before. It has no impact on the question or the answer.
Again, the question asks about a situation with at least one of the dice showing a two. As long as one die shows a 2 the question now applies to the other die which has six sides.
It doesn't matter if die A is a 2 or if die B is a 2. Having a 2 on either die or both dice only triggers the question.
Quote: MathExtremistQuote: AlanMendelsonPlease... go to your kitchen table and roll two dice. When one of the dice is a 2, look at the other die and ask yourself will another 2 show up 1/6 of the time or 1/11 of the time?
That's the wrong tense. You've already rolled the dice. The question is what they are, not what they will be if you re-roll them. Just record how many times you see "two 2s" and "at least one 2". That's all you have to do. The ratio will be nowhere close to 1/6.
You see this is not the question either. It's not even the bet for lunch that I have with the Wizard because the Wizard is giving me odds that the 2 will show up more than 1/6 times.
But let's return to the question in the problem. The question again is exactly this:
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?
When you read THAT QUESTION what do you consider? Do you think of two dice combinations or do you think of one die? I think of one die.
I really want to know what makes you consider two-dice combinations?
Quote: AlanMendelsonI really want to know what makes you consider two-dice combinations?
You look at both, not one.
