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#451
10-12-2016, 06:21 AM
 Cliffy Biro Guest Join Date: Oct 2016 Posts: 46
When choosing a sample space, you must also consider the relative probabilities of the outcomes in it. It can make things simpler if you choose the sample space so that each outcome in it is equally likely. Otherwise you have to get into weighting things by their probabilities.
You can't just say "there are four possible outcomes in the sample space, so the probability that either one of two particular outcomes will occur is 2/4." That's true if the four outcomes are equally likely, but not necessarily true otherwise.
#452
10-12-2016, 07:40 AM
 misterdls Guest Join Date: Apr 2006 Location: Wisconsin Posts: 29
Quote:
 Originally Posted by Cliffy Biro When choosing a sample space, you must also consider the relative probabilities of the outcomes in it. It can make things simpler if you choose the sample space so that each outcome in it is equally likely. Otherwise you have to get into weighting things by their probabilities. You can't just say "there are four possible outcomes in the sample space, so the probability that either one of two particular outcomes will occur is 2/4." That's true if the four outcomes are equally likely, but not necessarily true otherwise.
Each of the 3 possible outcomes in my last post are equally likely.
#453
10-12-2016, 08:05 AM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 16,293
Quote:
 Originally Posted by Chronos I would expect that anyone in the 1/11 crowd would claim that the only reasonable interpretation of the original problem is the one that's equivalent to CurtC's version.
The "bit rot" of this thread is just getting too high to deal with.

What version are you referring too??????

There are over 450 posts in this thread!

As to the OP saying that 1/11 isn't one of the listed answers: Yeah, this happens.

I've seen it on multiple standardized tests.
#454
10-12-2016, 08:54 AM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 75,283
The version I'm referring to is the one that CurtC posted, just a few posts before I said that.
#455
10-12-2016, 09:42 AM
 The Great Unwashed Guest Join Date: Mar 2002 Location: France Posts: 2,081
Quote:
 Originally Posted by ftg The "bit rot" of this thread is just getting too high to deal with. ... There are over 450 posts in this thread!
That might explain why you've overlooked this and this.
#456
10-12-2016, 09:43 AM
 Telemark Charter Member Join Date: Apr 2000 Location: Again, Titletown Posts: 20,569
Quote:
 Originally Posted by ftg As to the OP saying that 1/11 isn't one of the listed answers: Yeah, this happens. I've seen it on multiple standardized tests.
As explained upthread, it is because the version he quoted was changed from the originally published version and the people who changed it didn't realize that they'd also changed the outcome. The original version (where the all knowing observer sees a 4 and asks what is the probability that the other die is a 6) does yield the answer 2/11. By using the groundrules of that version of the question it's likely that the 1/11 version of the question was intended by the folks who changed it, but they screwed it up.
#457
10-12-2016, 09:53 AM
 The Great Unwashed Guest Join Date: Mar 2002 Location: France Posts: 2,081
Here's a thing:

Mr Dice rolls two (fair, six-sided) dice. He says, "There's at least one six!"

Mr Dice is a bit of a troll though and he always says that.

Turns out, this one time, he is right.

What is the probability that he has rolled a double-six?
#458
10-12-2016, 09:56 AM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,057
Quote:
 Originally Posted by Thudlow Boink Yes, 1/11 was not one of the available answers. There are three possible reasons for this, all of which I find plausible enough not to rule out: 1. 1/11 was not the intended answer. 2. 1/11 was the intended answer, and they made a mistake in not including it. 3. 1/11 was the intended answer, and they deliberately left it off the list to try to get people to click on the link to see why the answer they came up with wasn't one of the choices.
After reading here that the original source question was "what are the chances that the other die is a 4?", I'm convinced that the original, intended answer was 2/11 which would be correct given the intended assumptions. When the 4 was changed to a 6, they failed to change the 'correct' answer to 1/11. Because there are two ways to roll a 6 and a 4 but only one way to roll a 6 and a 6.
#459
10-12-2016, 09:58 AM
 Cliffy Biro Guest Join Date: Oct 2016 Posts: 46
Quote:
 Originally Posted by misterdls Each of the 3 possible outcomes in my last post are equally likely.
OK, so you're saying:
- a green and a red dice are rolled
- someone truthfully announces, for unspecified reasons, that at least one die is a six
- it follows that the following three possibilities are equally likely:
Green 6 and Red 6
Green 6 and Red non-6
Green non-6 Red die 6

I mean, I can think of a somewhat peculiar condition that would make this true, but there's no mention of it in your post or in the OP.

The second half of your scenario seems to introduce a novel twist, though-- not only do we not know what prompted someone to announce that there was at least one six, we also don't know if they then knowingly removed a six, or removed a die at random and it just happened to be a six. A whole extra layer of ambiguity, nice! I think you can get a result of 2/7 if you assume the "1/6" scenario plus "randomly removed dice just happens to be a six". Makes a nice change from 1/11 and 1/6 .
#460
10-12-2016, 11:02 AM
 sich_hinaufwinden Guest Join Date: Jan 2003 Posts: 515
Quote:
 Originally Posted by octopus Ah so only the ambiguity you decide is important is actually important. That's a bit tautological.
That would be tautological, however it's not what I wrote, not what I meant, and not what I implied. But thanks for trying to put words in my mouth.

Of course fair and 6-sided dice are important to the question. And if they're not specified you would have to assume those parameters. But here's the thing - if we treat this as not a trick question, then not only are those reasonable assumptions, but more importantly we all agree to them.

That's why I was careful to word my post as such:

Quote:
 Originally Posted by sich_hinaufwinden ...or anything that was never an issue...
The issues with the OP are that:

1. It does not state that the roller will reveal only 6s.
2. It does not state that the roller will always reveal a 6 when he can.

While it is reasonable to assume those things, those are not the only reasonable assumptions given the wording in the OP.

And you specifically are the one who asked:

Quote:
 Originally Posted by octopus What details are sufficient to satisfy everyone who may claim the question is ambiguous? Can you write it in such a way that no one could claim ambiguity?
Now that you have your answer you refuse to acknowledge that the wording proposed by CurtC is superior to the OP or give a reason why it isn't.

And if you are incapable of recognizing the difference between his wording and the OP and think they are equivalent, you should still consider his wording superior since it satisfies both the "1/11 only" and the "not enough info" camps.

Why will you not do that?
#461
10-12-2016, 11:07 AM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 16,293
Quote:
 Originally Posted by The Great Unwashed That might explain why you've overlooked this and this.
I did see them. I'm just flat out tired of clearing up people's mistaken notions and faulty comparisons.

Like I said, this thread is rotting now. The ratio of faulty stuff to factual stuff is getting too high to deal with. (As well as unhelpful partial references to earlier posts.)

I wonder why these types of threads just go on and on with people constantly throwing in more error-filled stuff. What is it about Math threads? (Please don't answer. This is rhetorical.)

Eventually I bail when I realize some people are just completely resistant to simple Math and Logic. I'm around this point now.
#462
10-12-2016, 11:46 AM
 LSLGuy Charter Member Join Date: Sep 2003 Location: Southeast Florida USA Posts: 21,037
I honestly can't say whether I'm regarded as part of the problem or part of the solution here. A bit of each is probably a fair assessment.

I have to agree with ftg that by the standards of GQ this is painful. And won't be improved by continuing. I'm out.
#463
10-12-2016, 12:02 PM
 The Great Unwashed Guest Join Date: Mar 2002 Location: France Posts: 2,081
Quote:
 Originally Posted by ftg I did see them. I'm just flat out tired of clearing up people's mistaken notions and faulty comparisons.
I was simply hoping that you would clear up your own mistaken notions and faulty comparisons.

For instance you seem to have said that the sequential rolling of two dice (one after the other) is an essentially different experiment to the "simultaneous" rolling of two dice. It isn't, I'm wondering why you think it is.

You also seemed to have said (it was hard to judge) that the expected number of rolls to have seen at least 2 sixes is less if you roll dice in pairs rather than one at a time. That isn't right, I'm wondering why you think it is.
#464
10-12-2016, 12:22 PM
 octopus Guest Join Date: Apr 2015 Posts: 6,055
Quote:
 Originally Posted by sich_hinaufwinden That would be tautological, however it's not what I wrote, not what I meant, and not what I implied. But thanks for trying to put words in my mouth. Of course fair and 6-sided dice are important to the question. And if they're not specified you would have to assume those parameters. But here's the thing - if we treat this as not a trick question, then not only are those reasonable assumptions, but more importantly we all agree to them. That's why I was careful to word my post as such: The issues with the OP are that: 1. It does not state that the roller will reveal only 6s. 2. It does not state that the roller will always reveal a 6 when he can. While it is reasonable to assume those things, those are not the only reasonable assumptions given the wording in the OP. And you specifically are the one who asked: Now that you have your answer you refuse to acknowledge that the wording proposed by CurtC is superior to the OP or give a reason why it isn't. And if you are incapable of recognizing the difference between his wording and the OP and think they are equivalent, you should still consider his wording superior since it satisfies both the "1/11 only" and the "not enough info" camps. Why will you not do that?
As I've learned interpretations are subject to a greater deal of subjectivity then I care to attempt to fix. Because at some point all language is either circularly defined or established by the assumption that shared experiences are largely universal. In other words it's too much effort for no reward to deal with linguistic issues when even if people were 100% clear those reading still couldn't do the correct calculation.
#465
10-12-2016, 12:26 PM
 pulykamell Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 43,165
Can we at least agree that if two dice are thrown together until at least one 6 shows up, then the other one being a 6 has a 1/11 probability? And if one die is thrown (or only one die is seen) until a 6 shows, that the chance of the other die being a 6 is 1/6? And that everything else is just a matter of how we interpret the OP?

Last edited by pulykamell; 10-12-2016 at 12:27 PM.
#466
10-12-2016, 12:31 PM
 MonkeyMensch Charter Member Join Date: Jul 2001 Location: Encinitas Posts: 2,520
I was going to do a head count of which answers were favored but the number of hypothetical situations and side topics discussed makes that unappealing. I planned, some five or six pages ago, to post a nice explanation of why the probability being discussed in the OP is 1/6. But after a very little re-reading I saw LSLGuy had made that exact explanation in post #5! Reading matters!

If at this point some other poster has made this explanation I apologize in advance for the repetition. The OP only states:

""Two fair dice are rolled together, and you are told that 'at least one of the dice is a 6.' A 6 is removed, and you are t(h)en shown the other die. What is the probability of the other die showing a six?"

When the experimenter (I'm calling her that) says truthfully, "At least one of the dice is a 6" she is giving you information about the possible outcomes. When she removes a 6 she effectively takes away that very information. At that point the only information you have is that there is a fair die left behind, randomly rolled, and the probability of a 6 (or of any number, really) is 1/6.

And my prediction of running to at least ten pages has come true. Steve Holt!!
#467
10-12-2016, 12:38 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,427
Quote:
 Originally Posted by MonkeyMensch When the experimenter (I'm calling her that) says truthfully, "At least one of the dice is a 6" she is giving you information about the possible outcomes.
Here they are. I assume you agree:

1,6
2,6
3,6
4,6
5,6
6,6
6,5
6,4
6,3
6,2
6,1

Quote:
 When she removes a 6 she effectively takes away that very information. At that point the only information you have is that there is a fair die left behind, randomly rolled, and the probability of a 6 (or of any number, really) is 1/6.
When she removes a 6, we are left with the following possibilities as being the die left:

1
2
3
4
5
6
5
4
3
2
1

1 out of 11 times, the die left over is a 6. If you think I did something wrong above, tell me at what point you think it was.
#468
10-12-2016, 12:46 PM
 sich_hinaufwinden Guest Join Date: Jan 2003 Posts: 515
Quote:
 Originally Posted by octopus As I've learned interpretations are subject to a greater deal of subjectivity then I care to attempt to fix. Because at some point all language is either circularly defined or established by the assumption that shared experiences are largely universal. In other words it's too much effort for no reward to deal with linguistic issues when even if people were 100% clear those reading still couldn't do the correct calculation.
Since I understand the calculations for each of the scenarios discussed you must not be referring to me with that remark.

And you admit now that you will not address the answer to a question that you brought up.

I've only asked you that directly one time, so lest I be accused of badgering, I won't waste either of our time by asking you again.
#469
10-12-2016, 01:05 PM
 Telemark Charter Member Join Date: Apr 2000 Location: Again, Titletown Posts: 20,569
Quote:
 Originally Posted by MonkeyMensch When the experimenter (I'm calling her that) says truthfully, "At least one of the dice is a 6" she is giving you information about the possible outcomes. When she removes a 6 she effectively takes away that very information. At that point the only information you have is that there is a fair die left behind, randomly rolled, and the probability of a 6 (or of any number, really) is 1/6.
No, this is where you go wrong. The set of possible die rolls is not a random set with an equal distribution of 1/6 for each number. What is left is a set of rolls with a probability of 2/11 for numbers 1 through 5, and 1/11 for number 6. This is critical to understand the problem. That information given (one of the die is a 6) doesn't disappear then the die is removed. That information is built into the probability of the remaining die.

Follow the steps listed by x-ray vision and see if it makes it more clear.
#470
10-12-2016, 04:14 PM
 MonkeyMensch Charter Member Join Date: Jul 2001 Location: Encinitas Posts: 2,520
Quote:
 Originally Posted by x-ray vision 1 out of 11 times, the die left over is a 6. If you think I did something wrong above, tell me at what point you think it was.
I'd be happy to. Those are ordered pairs you list and depict permutations, not combinations. This is clear because you list 1,6 and 6,1 both. I agree that after the first roll this represents the set of outcomes in the probability space, all equally likely.
Quote:
 Originally Posted by x-ray vision Here they are. I assume you agree: 1,6 2,6 3,6 4,6 5,6 6,6 6,5 6,4 6,3 6,2 6,1

Here's where the difference arises:

Quote:
 Originally Posted by x-ray vision When she removes a 6, we are left with the following possibilities as being the die left: 1 2 3 4 5 6 5 4 3 2 1
You list 11 possibilities left and that's incorrect. I'll use REM to denote the die that was removed. Say the second die was removed. Here's the possible permutations:

(1, REM)
(2, REM)
(3, REM)
(4, REM)
(5, REM)
(6, REM)

Or if the first die was removed:

(REM, 1)
(REM, 2)
(REM, 3)
(REM, 4)
(REM, 5)
(REM, 6)

Those are now two disjoint probability spaces since either the first die or the second die was removed. Only one of those spaces is the case since you can't remove the first die and/or the second die simultaneously. It's one or the other. And it's clear that all values remaining have equal probability. Your error arises because you are conjoining two disjoint probability spaces. To wit, you equate the pairs (6, REM) and (REM, 6) when they're actually two distinct ordered pairs.
#471
10-12-2016, 04:17 PM
 MonkeyMensch Charter Member Join Date: Jul 2001 Location: Encinitas Posts: 2,520
I refer Telemark to the solution previously given.

Joking aside my response is just, in essence, using notation to elaborate what others have probably said.
#472
10-12-2016, 04:34 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,427
Quote:
 Originally Posted by MonkeyMensch Say the second die was removed. Here's the possible permutations: (1, REM) (2, REM) (3, REM) (4, REM) (5, REM) (6, REM) Or if the first die was removed: (REM, 1) (REM, 2) (REM, 3) (REM, 4) (REM, 5) (REM, 6)
Why are we going to say something that we don't know happened? We don't know which die was removed, so we don't remove possibilities before calculating.

I'm going to post two interpretations of the problem that Chronos posted earlier. Are you interpreting the OP as the first, second, or are you saying the answer is 1/6 either way? Or are you interpreting it another way?

Quote:
 1: I roll a pair of dice. I then choose one die at random, look at the number on the die, and announce "At least one die is a ___", where the blank is the number I looked at. On this particular iteration of the experiment, the die I looked at was a 6, and I announce as such. What is the probability that the other die is a 6?

Quote:
 2: I roll a pair of dice. I then examine both dice, and if it is the case that at least one of them shows a 6, I announce "At least one die is a 6", and otherwise I say nothing. On this particular iteration of the experiment, I saw at least one 6, and so I announce as such. What is the probability that both dice are 6?
#473
10-12-2016, 05:48 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 75,283
Sure, you can call the two 6 cases different cases, but the probability for both of them combined is equal to the probability for any one of the others.
#474
10-12-2016, 06:19 PM
 misterdls Guest Join Date: Apr 2006 Location: Wisconsin Posts: 29

Rolling a Green die and a Red die gives the following sample space:

G1,R1 G2,R1 G3,R1 G4,R1 G5,R1 G6,R1
G1,R2 G2,R2 G3,R2 G4,R2 G5,R2 G6,R2
G1,R3 G2,R3 G3,R3 G4,R3 G5,R3 G6,R3
G1,R4 G2,R4 G3,R4 G4,R4 G5,R4 G6,R4
G1,R5 G2,R5 G3,R5 G4,R5 G5,R5 G6,R5
G1,R6 G2,R6 G3,R6 G4,R6 G5,R6 G6,R6
36 possibilities, equally likely, obviously.

The roller truthfully says that a 6 appears on one or both of the dice. This reduces
the sample space to only the 11 possible combinations bolded above and listed below:

G6,R1
G6,R2
G6,R3
G6,R4
G6,R5
G6,R6
G1,R6
G2,R6
G3,R6
G4,R6
G5,R6

Again, obvious.

Note that:
1. There are 11, not 12, equally possible combinations which contain one or two 6's.

2. The 6-6 combination can only appear once, as G6,R6, as opposed to, for example,
1-2, where 1-2 can appear as either G1,R2 or as G2,R1. Saying there are 12
combinations is saying there are two 6-6 combinations. There are not. This is
fundamental. There are no "other circumstances" that can give anything different
than 11 combinations.

3. There is nothing "peculiar" about any of this; it all flows from the original
problem. There's no reason to consider other factors, such as the roller's motive,
or any might-have been, or things we are not told. We have a simple,
straightforward problem to solve. And we have before us all the facts we need.

We can put each die face of the 11 listed combinations into one of two "modes". Either
the die face shows 6 or it doesn't. If it shows 6 we'll say it's a "mode6" face.
If it isn't a 6 then we will call it a "modex" face. The actual number shown on a
modex die in the box isn't important in our deliberations here.

Now consider the initial dice throw. After the throw and BEFORE seeing if either die
is a 6 (and therefore mode6), the content of each of the 36 combinations will ALWAYS
consist of one of the following mode combinations:

Green Red
mode6 mode6
mode6 modex
modex mode6
modex modex.

4 combinations, each equally likely. If one of the 4 combinations are removed
from this list, the remaining 3 combinations remain equally likely.

When the roller excludes the possibility of the fourth combination (modex-modex) by
stating, truthfully, that at least one of the dice is showing a 6, AS HE ALWAYS DOES, the sample space we are given when
we are now facing the following 3 possibilities comprising our sample space:

Green Red
mode6 mode6
mode6 modex
modex mode6

All equally likely.

In plain words, either 1. the Green shows a 6 and the Red also shows a 6, or
2. the Green shows a 6 and the Red does not show a 6, or 3. the Green does not show
a 6 and the Red shows a 6.

These are all and the ONLY possible configurations of the two dice in the box at this
point.

Note that there are 4 mode6 faces and 2 modex faces. This is the direct result from
the roller's exclusion of the modex-modex combination, and is normal and expected, and
will be the critical factor in determining the final answer.

Let's label each combination with a case letter, to more easily distinguish them:

Green Red
mode6 mode6 Case A
mode6 modex Case B
modex mode6 Case C

The roller removes one die, and it's a 6, a mode6 face. So it could be any of
these four faces:

Green mode6 from Case A
Red mode6 from Case A
Green mode6 from Case B
Red mode6 from Case C

Bingo.

Counting the Cases, we see there are 2 Case A's, 1 Case B, and 1 Case c. Two of the
four mode6 faces come from Case A where both faces are 6's.

The answer to the question "What is the probability that the die remaining in the box
is a 6" is 2/4 or 1/2.

QED.

This problem is very similar to the Boy-Girl problem where the probability of a
certain child's sex is a boy (or girl, as the case may be) equals 1/2.

I really do like working on (simple) probability problems. Kind of a hobby. I've
done this work simply because I enjoy it. I had no intent to embarrass or anger
anyone.

Best wishes to EVERYONE.

Last edited by misterdls; 10-12-2016 at 06:22 PM. Reason: adding title and bolding
#475
10-12-2016, 07:00 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,427
Quote:
 Originally Posted by misterdls Green Red mode6 mode6 mode6 modex modex mode6 modex modex. 4 combinations, each equally likely.
WTF?
#476
10-12-2016, 09:28 PM
 Grestarian Member Join Date: Aug 2011 Location: Garage & Lab Posts: 1,261
Quote:
 Originally Posted by Skammer A math teacher friend of mine shared this problem on Facebook and we are disagreeing about the solution. Here is the problem verbatim: "Two fair dice are rolled together, and you are told that 'at least one of the dice is a 6.' A 6 is removed, and you are ten shown the other die. What is the probability of the other die showing a six?" My friend thinks the answer is 1/6, because the dice rolls are independent events and regardless of what one shows, the other has a 1/6 chance of being a six. I'm quite positive this is incorrect.
Quote:
 Originally Posted by Turble Charma got it. The roll of each die is an independent process. You can completely ignore the fact that one of them shows a 6. The chance of any one 6-sided showing a 6 is always 1/6 no matter how many dice you roll and no matter what the other dice show.
For Ogg's sakes! It's not a math, statistics, or probability challenge. It's a test of reading ability!

"...fair dice are rolled .....what is the probability....die showing a six?"
[We are assuming these are standard cubes, not D&D/wargaming dice]

Haldurson said it best: The red herring doesn't change the fact that the probability of any single 6-sided die roll being a 6 is 1/6
• You could just as easily say the probability of any 6-sided die-roll being a 4 is 1/6
• You could just as easily say the probability of any 10-sided die-roll being a 2 is 1/10
• You could just as easily say the probability of any ten billion-sided die-roll being a 2 is (okay, once the darned thing stopped rolling)....

I roll 400 octahedrons, each with numbers 1 through 8 on each face. I remove 399 of those dice. What's the probability that the one I left behind shows a 5*?

There is no Lady Luck. There is no force or entity with any vested interest in making dice rolls sequential or logical or pretty. Why make things more complex than they already are?

---G!
* Now the real challenge!
What's the probability that the one left behind shows a 9?
#477
10-12-2016, 10:56 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,534
Quote:
 Originally Posted by Grestarian I roll 400 octahedrons, each with numbers 1 through 8 on each face. I remove 399 of those dice. What's the probability that the one I left behind shows a 5*?
That depends: How did you decide which 399 to remove? Did you, for example, deliberately remove all the ones that showed 5s?
#478
10-13-2016, 07:28 AM
 Haldurson Guest Join Date: Oct 2013 Posts: 144
Quote:
 Originally Posted by Thudlow Boink That depends: How did you decide which 399 to remove? Did you, for example, deliberately remove all the ones that showed 5s?
If I said instead that I use some pre-specified algorithm to remove dice, and one die remains, you still have to assume that the odds are 1 out of 8 because you don't know what the criteria is. Information about that remaining die doesn't exist, therefore, it's irrelevant to the question.

If I non-randomly selected a face on a die, and I asked you what the odds are that I chose an 8 on that die, without knowing my criteria for selecting that die, you'd be correct if you said that it's 1 out of 8. On the other hand, if you knew the criteria, then the odds might be different.

Once you add knowledge to the mix, you can change those odds. But without such knowledge, you have to assume that all possible results are equally likely. It really doesn't matter if the dice are picked deliberately or not, unless you have access to the actual criteria used.
#479
10-13-2016, 08:14 AM
 Cliffy Biro Guest Join Date: Oct 2016 Posts: 46
Quote:
 Originally Posted by Haldurson If I non-randomly selected a face on a die, and I asked you what the odds are that I chose an 8 on that die, without knowing my criteria for selecting that die, you'd be correct if you said that it's 1 out of 8. [...] without such knowledge, you have to assume that all possible results are equally likely.
I'm not sure that there's a rational basis for assuming that the probability is 1/8, or any particular value.

Say we have a million bags each containing 100 balls that are either black or white, in proportions that may differ in each bag. We don't know how many black balls are in each bag and have no reason to expect any particular number. For all we know, every single ball in every bag could be black, or they could all be white, or whatever. Anything's possible.

A goes to take a ball out of the first bag, saying "we have no information, so it's a 1/2 chance that the ball will be black." Meanwhile B says "I reckon that it is a 1/3 chance." Afterwards, they open up the bag and see that there were 70 black balls in there, so for this bag it was actually a 7/10 chance. Neither of them was right this time.

They do the same with all the bags, with A doggedly sticking to 1/2 while B persists with 1/3. After they've done the million bags, given what we know, we can only expect that A will have been correct about the same number of times as B. So A's guess of the probability was no better than B's.
#480
10-13-2016, 08:23 AM
 Haldurson Guest Join Date: Oct 2013 Posts: 144
Quote:
 Originally Posted by Cliffy Biro Say we have a million bags each containing 100 balls that are either black or white, in proportions that may differ in each bag. We don't know how many black balls are in each bag and have no reason to expect any particular number. For all we know, every single ball in every bag could be black, or they could all be white, or whatever. Anything's possible.
I roll a die, and only I know what is on that die. For me, it's either 0% or 100% that the number showing is a 6. As far as you are concerned, thoiugh, it's 1 out of 6. Knowledge changes everything. In your example, you are cheating by looking ahead in time. If all you know is that each ball may be either black or white, then yes, as you say, anything is possible. if I'm a betting man, and without foreknowledge, If I bet on I pick from a random ball from a random bag on what color it's going to be, and someone else without foreknowledge of the results were betting against me, if you wanted to make the bet fair it would be 50-50. If I were cheating, obviously, the odds would be different. What you are suggesting is that someone is cheating.
#481
10-13-2016, 08:31 AM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,534
Quote:
 Originally Posted by Haldurson Knowledge changes everything.
Yes; and the reason that people in this thread (at least the ones who understand probability) are getting different answers is because of different assumptions/interpretations about what knowledge you have in the situation described in the OP.
#482
10-13-2016, 11:58 AM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,427
Cliffy, I made a similar point as Haldurson did. Can you answer the question I specifically asked you in post 325?
#483
10-13-2016, 01:41 PM
 Telemark Charter Member Join Date: Apr 2000 Location: Again, Titletown Posts: 20,569
Quote:
 Originally Posted by Grestarian Haldurson said it best: The red herring doesn't change the fact that the probability of any single 6-sided die roll being a 6 is 1/6
This is true, but it's not the question at hand. Some of the possible rolls have been removed from the choice set, and what we're left with has a 1/11 chance of being a six. The manner in which the die that we're asking about is chosen isn't the same as if you threw a single die and asked if it was a 6.

Again, if you doubt it then write a small computer simulation following the instructions
1. Roll two dice
2. If neither is a 6 then ignore roll and go back to step 1
3. If AT LEAST one die is a 6 then increment your throw counter, select a die that shows six as your primary die and call the other the remaining die
4. If the remaining die is a 6, increment your success counter
5. Go back to step 1

After running this simulation 10,000 times do you think this test will produce a success/throw ratio of 1/6 or 1/11?

If you remove step 2 and the first part of step 3 then your answer will be 1/6, but that's not the setup.
#484
10-13-2016, 04:13 PM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 16,293
Quote:
 Originally Posted by The Great Unwashed For instance you seem to have said that the sequential rolling of two dice (one after the other) is an essentially different experiment to the "simultaneous" rolling of two dice. It isn't, I'm wondering why you think it is.
You left out something very key about what I said: Rolling one die until you get a 6 and then rolling the second die is what I was objecting to. This conditional rolling of the second die changes things quite a bit.

You can roll one die on Monday and the second die on Friday. As long as the second roll happens regardless of what the first roll was. It suffices to not look at the first roll until the second roll is finished. Note that not saying anything to the other person until the 2nd roll is finished is implied by all this.

OTOH, rolling the first die. Looking at it and if it's a 6 telling the other person before rolling the 2nd die is a separate problem.

So any situation with the events in this order:

1. Roll the dice. One at a time, together, doesn't matter.
2. Look at both dice.
3. Tell the other person "At least one is a 6.", if that is indeed the case.

satisfies the conditions for the 1/11 answer.

Some variations of the above also give 1/11. But if you push it a bit too far, you get something else.

The OP's statement is clear enough that the 1/11 answer is the only one that fits unless you get into: What if the dice are 23 sided? What if the roller is lying? and other ultra-pedantic questions.
#485
01-31-2017, 08:35 PM
 Irishman Guest Join Date: Dec 1999 Location: Houston, TX, USA Posts: 12,253
Musicat, jtur88, you are both interpreting the question to be same as the following:

"You roll two fair six-sided dice. You then ignore one of them. What are the odds that the remaining die is a six?"

Going back the the original problem as stated and given that the list of answers excludes "1/11" but includes "1/6", I am inclined to agree that the test writer meant the statement "a six is removed" to be a meaningful severing of the combined information you were given - that at least one die was a six.

This appears to be a trick question trying to get the statisticians to assume it is a different problem than it is. You are given irrelevant information - at at least one is a 6. It is irrelevant because that 6 is then removed from consideration, and you are told to look only at the remaining die.

The problem it resembles would be equivalent to:

"You roll two fair six-sided dice. If one of them is a six, what are the odds they are both sixes?"

For those attempting to solve that problem, then:
Quote:
 Originally Posted by chrisk But the real sleight of hand, as it were, is that I'm possibly not deciding which die I'm calling "the other die" until after I've rolled both dice, and THAT changes the odds.
Agreed. The selection of a red and green die is irrelevant, because this version of the question does not specify the order. There are 11 equally likely possible outcomes. Of those, only one outcome has double sixes.

You can't conflate the action of selecting the red 6 versus the green 6 as a separate condition. That is a separate step in the sorting process, and is another term in the probability calculation.

In this problem, (6,6) has the same likilihood as (1,6) or (6,1). If you pick red 6 half the time and green 6 half the time, then you have to split the probability of (6,6) in two for show Red (6,6) and show Green (6,6).

Alternately, you have to lump (1,6) and (6,1) as one option.

Think of it this way, if the dice are Red/Green (6,6), and half the time the selector picks Red 6 and half the time Green 6, you have to compare against the condition that if the dice are Red/Green (1,6), then half the time the selector will pick Green 6 and the other half the time Green 6. That is the level of equivalent likelihood options.

Analogy: think of an estate of \$100,000 to be divided among 4 siblings equally. Everyone should get \$25,000. But what if one of the siblings died, but left two children of his/her own? They get 1 share equal to 1/4 the whole pot, and they have to split that share between the two of them.

Or tell the equivalent with the lottery. 6 winning lottery tickets, but 5 tickets are owned by 1 person apiece while the sixth ticket has two owners sharing the ticket. Each ticket gets an equal share of the winnings, and the two people sharing a ticket have to split their single share. They don't get a full share apiece and force the pot to be divided seven ways.

Quote:
 Originally Posted by octopus Sorry for the ugly formatting but padding spaces seem to be removed.
Try using [code][/code] tags.

Code:
```Double 6	At least one 6	Odds	Inverse

1        	9	0.111111111	9
2	      13	0.153846154	6.5
3	      41	0.073170732	13.66666667
4	      48	0.083333333	12
5	      66	0.075757576	13.2
6	      80	0.075	        13.33333333
7	      85	0.082352941	12.14285714
8	      86	0.093023256	10.75
9	      95	0.094736842	10.55555556
10	      97	0.103092784	9.7
11	     106	0.103773585	9.636363636
12	     135	0.088888889	11.25
13	     149	0.087248322	11.46153846
14	     150	0.093333333	10.71428571
15	     154	0.097402597	10.26666667
16	     181	0.08839779	11.3125
17	     182	0.093406593	10.70588235
18	     201	0.089552239	11.16666667
19	     207	0.09178744	10.89473684
20	     227	0.088105727	11.35```
#486
02-02-2017, 07:28 PM
 Irishman Guest Join Date: Dec 1999 Location: Houston, TX, USA Posts: 12,253
I see now that I'm late to the party. When I posted that, I had eliminated the other option: that the test writer did not know what they were testing, so wrote the problem incorrectly. I now see that is actually the case.

Quote:
 Originally Posted by LSLGuy @CHronos: As I said a bit ago, I wonder what the original puzzle maker was thinking when he/she asserted 2/11 was the right answer. To be sure there are conceivable rule sets that produce that result. What IMO there aren't, is simple & plausible rule sets that do so. Any ideas?
When throwing two fair six-sided dice, and specifying a 4 and a 6, but not specifying which die is which number, you have two dice to pick from and 11 possible pairs. This differs from the specifying 6 case by virtue that (6,6) is only one option, but (4,6) is different than (6,4).

Quote:
 Originally Posted by Thudlow Boink This is the interpretation I'd lean towards if the original wording had said "one of the dice is a 6" instead of "at least one of the dice is a 6." If it had simply said "One of the dice is a 6," it would be natural and reasonable (IMHO) to interpret this as "I'll tell you what one of the dice was: it happened to be a six. Now if I remove that one, what's the probability that the other one is a 6?" And then your reasoning would be correct, and the answer would be 1/6. But (again IMHO) the words "at least" don't fit that interpretation. To me, "at least one of the dice is a 6" means "Out of all the possible ways the pair of dice could come up, the result was one of the ones in which at least one of the two dice shows a 6." There are 11 such possibilities, and in only one of them would the remaining die be a 6 if a 6 were removed.
Ah, but someone else could just as easily say "you told me one of the dice is a 6, so the other dice is not a 6, QED." Thus the verbage "at least one is a 6", allowing that the other might be a 6 as well.

Language is a bitch inherently ambiguous.

Quote:
 Originally Posted by septimus Wrong. I have two children and I tell you at least one of them is a girl. What is the chance both are girls? Do the people getting OP's problem wrong also get the two-children problem wrong? They are really the exact same problem. The correct answer is 1/(2N-1) not 1/N, where N=6 and N=2 for the respective problems.
Once you remove a specific die (the 6), you have narrowed the field just like specifying the older child.

Quote:
 Originally Posted by Cliffy Biro Does this help illustrate the 1/6 (2/12) argument? (Not that I agree with it.) The bold lines are the equally likely possibilities that are consistent with the event described in the problem statement. There are twelve of them, of which two are [6 6]. Code: ```Dice Die you learn about ==== =================== 1 6 first 1 6 second 2 6 first 2 6 second 3 6 first 3 6 second 4 6 first 4 6 second 5 6 first 5 6 second 6 1 first 6 1 second 6 2 first 6 2 second 6 3 first 6 3 second 6 4 first 6 4 second 6 5 first 6 5 second 6 6 first 6 6 second```
I think this post describes the flaw.

Quote:
 Originally Posted by Cliffy Biro When choosing a sample space, you must also consider the relative probabilities of the outcomes in it. It can make things simpler if you choose the sample space so that each outcome in it is equally likely. Otherwise you have to get into weighting things by their probabilities. You can't just say "there are four possible outcomes in the sample space, so the probability that either one of two particular outcomes will occur is 2/4." That's true if the four outcomes are equally likely, but not necessarily true otherwise.

Code:
```6 6	first
6 6	second```
[/QUOTE]

are not equally likely options to the other options. There are two steps in that selection, the die roll with it's inherent probabilities, and then the selection of which to reveal. Note that for the other options, all iterations with that outcome provide the same choice of which to reveal, but for (6,6), half the time you pick one and half the time you pick the other.

Quote:
 Originally Posted by BigT This is adding information not stated in the problem. At no point does it say that they picked one die at random and read the number. The number 6 is given in the problem without any means for it to be different, and thus is fixed. Any simulation must keep the 6, or else it is modeling a different problem. And once you do require there to be a 6, the 1/11 interpretation is the only one that works. I do not understand why such an explanation is still being promoted, when one of the basic rules of word problems is that all required information is given.

Quote:
 Originally Posted by misterdls We can put each die face of the 11 listed combinations into one of two "modes". Either the die face shows 6 or it doesn't. If it shows 6 we'll say it's a "mode6" face. If it isn't a 6 then we will call it a "modex" face. The actual number shown on a modex die in the box isn't important in our deliberations here. Now consider the initial dice throw. After the throw and BEFORE seeing if either die is a 6 (and therefore mode6), the content of each of the 36 combinations will ALWAYS consist of one of the following mode combinations: Green Red mode6 mode6 mode6 modex modex mode6 modex modex. 4 combinations, each equally likely. If one of the 4 combinations are removed from this list, the remaining 3 combinations remain equally likely.
False. That's just like saying "Either my lottery ticket is a winner or a loser, so I have a 50% chance of winning."

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