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#451




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




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#453




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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




The version I'm referring to is the one that CurtC posted, just a few posts before I said that.

#456




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




Here's a thing:
Mr Dice rolls two (fair, sixsided) 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 doublesix? 
#458




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#459




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 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 non6 Green non6 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




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Of course fair and 6sided 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: Quote:
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




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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 errorfilled 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




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




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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




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#465




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; 10122016 at 12:27 PM. 
#466




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 rereading 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




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1,6 2,6 3,6 4,6 5,6 6,6 6,5 6,4 6,3 6,2 6,1 Quote:
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




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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




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Follow the steps listed by xray vision and see if it makes it more clear. 


#470




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Here's where the difference arises: Quote:
(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




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




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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:
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#473




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




The Final Answer!
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 66 combination can only appear once, as G6,R6, as opposed to, for example, 12, where 12 can appear as either G1,R2 or as G2,R1. Saying there are 12 combinations is saying there are two 66 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 mighthave 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 (modexmodex) 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 modexmodex 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 BoyGirl 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; 10122016 at 06:22 PM. Reason: adding title and bolding 


#475




WTF?

#476




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"...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 6sided die roll being a 6 is 1/6
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




That depends: How did you decide which 399 to remove? Did you, for example, deliberately remove all the ones that showed 5s?

#478




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If I nonrandomly 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




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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




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#481




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




Cliffy, I made a similar point as Haldurson did. Can you answer the question I specifically asked you in post 325?

#483




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Again, if you doubt it then write a small computer simulation following the instructions
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




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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 ultrapedantic questions. 


#485




Musicat, jtur88, you are both interpreting the question to be same as the following:
"You roll two fair sixsided 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 sixsided 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:
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:
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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




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.
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6 6 first 6 6 second 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:
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