Yeah it is.
Someone rolls 2 fair dice. Are they six sided dice? There’s an assumption.
If at least one is a 6 and that 6 is shown to another and then removed from the set of 2 dice what is the probsibility that the other die is a 6?
This is all clear English. Can you imagine writing a test that would satisfy this crowd?
Simple assumptions like the person rolling the dice actually looking and selecting a die are what reasonable people would do when reading the question.
Why do we need to see the the dice? Why can’t this be a thought experiment?
(I’m starting to share xrv’s sense of frustration.)
Several of us on the other side are starting to be frustrated at the willful blindness loudly maintaining that there is exactly one possible way to interpret the OP and perform the experiment.
This despite the fact the OP itself outlined the two different interpretations.
There are none so blind as those who have their eyes screwed shut.
Where in the OP are two correct interpretations?
I read the OP. I read a few posts, then skipped to the last page to see where everyone was on the question.
After reading the last page of posts, I’m reminded of the airplane on the treadmill, for some reason… :eek::smack:
Actually it turns out that the tiniest little piece of the plane was missing.
I mean a really small piece.
Infinitely small.
So it’s more correct to say that there was 0.999… of a plane on the treadmill.
A PROPOS THE AMBIGUITY OF THE EXPERIMENT:
An experimenter rolls two dice and observes the outcome. In accordance with some unspecified strategy (S) the experimenter announces that at least one of the two dice is a 6 and they remove a 6 from the pair.
The question is : What is the probability that the remaining die is also a 6?"
It should be apparent that this probability § depends on the particular strategy used. Some examples :
S1: if the experimenter sees exactly one 6 they makes their announcement. Here P = 0
S2: if the experimenter sees exactly two 6s they makes their announcement. Here P = 1
(As has been pointed out, we can specify a strategy to yield whatever value we would like between 0 and 1.)
S3: if the experimenter sees at least one 6 they make their announcement and randomly choose one of the 6s to remove. Here P = 1/11
S4: if the experimenter sees at least one 6 they make their announcement and *choose according to some arbitrary rule *one of the 6s to remove.
An example of the arbitrary rule: suppose the dice are of different colours, say red and green (or otherwise distinguisable).
If the red die is 6 it is removed, otherwise green (which must be 6) is removed
Here if you see a green die removed P = 0
And if you see a red die removed P = 2/11
S5: the experimenter announces the highest value they see, in this case it just happened to be 6. Here P = 1/11 (but note, for example, if they had announced “at least one 5” P(remaining dice is also 5) = 1/9
S6: the experimenter arbitrarily chooses one die and announces its value, in this case it just happened to be 6. Here P = 1/6
Of these several possibilities, only S3 and S6 seem to me to be reasonable readings of the puzzle. But S6 is no puzzle at all because the outcomes of the two dice are independent, knowing that a 6 was shown on an arbitrarily chosen die is no more use than knowing its colour.
Are there other “reasonable” readings that I am missing?
This is why people fail on tests and then argue with the professor until they are blue in the face:
Two dice, one roll. There is nothing in the problem about any rolls that happened earlier, or subsequent rolls. The problem restricts itself to a single roll of two dice.
As it happens on this roll at least one of the dice came up with a six. (unless the experimenter is a lying scumbag in which case this is a psychology experiment, not a probability one)
If only one of the dice was a six it is removed and you are shown the other non-six dice, else both are sixes and one or other is removed. There is nothing in then problem about which of the two was a six, or which is removed when both are a six, so under the problem’s conditions these factors are arbitrary.
If you take the problem at face value it is an easy conditional probability problem-answer: 1/11
If you feel compelled to consider all possibilities which were not explicitly excluded in the problem description you will likely fail the test. No amount of arguing will turn your D- into a B+. Solve the problem you are given, not the set of all problems that are vaguely similar.
In my post immediately before yours I proposed two strategies, S3 and S6, both of which correspond to “face value” readings of the question but lead to two different answers (S3 -> 1/11 , S6 -> 1/6).
I don’t like S6 because to my mind that leaves us with a non-puzzle, but then again 1/6 was one of the possible multi-choice answers and 1/11 wasn’t.
On a test that’s a good idea. It’s also a good idea at work. On a math puzzle designed specifically to generate multiple interpretations it’s failing to get the joke.
In this case, “the problem we are actually given” is, in fact “a set of problems that are all vaguely similar”.
Or how about this one? I ask you a randomly-selected probability question. What are the odds that the answer to my question is 6?
Assumptions, again. You are assuming that that is the rule, but it does not say so in the OP. As has been pointed out, one of the dice could have been chosen at random, and just happened to be a 6. Just as reasonable as your assumption, but it yields a different answer.
Right: we’re not considering the interpretation that’s clearly incorrect* - and so noted by the OP.
*Under the specification that “at least one of the dice is a 6”, the two rolls are clearly not independent, so the 1/6 answer cannot be right.
That’s not “clearly” at all, since there are ways to ensure that at least one die is a 6 while still retaining independence.
In the OP it says “A 6 is removed, and you are ten [sic] shown the other die.” This covers what ticker posted.
Or chosen according to a scheme that prefers small numbers, or red dice on even-numbered days - or by any of a vast number of arcane schemes not mentioned in the OP.
I’d be interested to hear how this is possible while conforming to the OP.
No. If you see a red die removed P = 1/6.
You may have guessed P = 2/11 because the average of the two cases must be 1/11. But it is the weighted average that will be 1/11 and red reveals outnumber green reveals 6 to 5.
It doesn’t say that whoever-it-is must tell you if there’s at least one six, as **ticker **suggests. It just says that on this occasion they did.
You’re right in all respects.
ETA: I jumped rather hopefully to that answer (2/11) because it was one of the multi-choice options in the OP (and I had some meta-point floating around in my head somewhere).
To think that the person rolling the dice would only make the statement “At least one is a 6.” under restricted circumstances such as only when the dice were a 6 and a 3 is perverse.
You may as well go whole hog and consider the situation that the roller is lying and there is no 6 at all! If that is the case, then the poser would have to include an insane number of “axioms” about the problem. “The roller isn’t lying, I’m not lying about the roller, I’m not lying about not lying about the roller.” Etc. Where will it end?
Taking the statements to mean what they say avoids all this nonsense. And that is definitely not ambiguous like the boy/girl problem.
You can always be a nitpicker and get into “What is the mean of the word ‘is’?” weirdness. But then you should avoid all posed problems entirely. You’ll go crazy finding fault with everything.