And we’ve always known that’s what you thought it meant, but that doesn’t change the fact that the OP was stated in an ambiguous manner.
So that’s one for “No”.
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.
Curt’s version doesn’t tell us the dice are six-sided, or that they’re equally likely to produce each outcome. But you’re happy to make those assumptions, so welcome, finally, to the club.
Yeah - I figured it’s a long shot but at least one die was a…I mean at least one poster asked this…
…and CurtC came up with a response that answers that very well.
When the wording for an acceptable problem was theoretical it was easy for them to insist their interpretation was the only reasonable one. But now there is an actual acceptable re-wording for them to compare to the OP.
And so I was wondering if any of them would recognize how the different wording has different implications.
LOL - I had considered editing my response to his post to add something about bonus points for not mentioning fair dice, number of sides, or anything that was never an issue.
How unfortunate if you actually think those non-issues are in any way comparable to the ambiguity that was being discussed.
Ah so only the ambiguity you decide is important is actually important. That’s a bit tautological.
There is only one way to roll 6,6, not two. Of the 36 ways to roll 2 dice, there is one way to roll a total of 2(1-1), 2 ways to roll a total of 3(1-2, 2-1), 3 ways to roll a total of 4 (1-3, 2-2, 3-1) and so on, until we see there is only one way to roll a total of 12, (6-6). Your chart double-counts the 6-6 combination, so there are only 11 combinations that contain at least one 6.
Sure, but there are two ways to roll [6 6] and then choose one of the two dice. There are also two ways to roll [1 6] and then choose one of the two dice, but only one of them is consistent with what happened in the original puzzle. Therefore, under these assumptions there are twelve equally likely possibilities, not eleven, of which two are [6 6], not one.
This puzzle had me confused. At first I thought the answer was 1/6, then I was sure
it was 1/11, and now I’m doing some more thinking.
Let’s consider the situation just before the roller removes one die. We don’t care
which non-6 die face might be showing on either die, so we can call it “X”. So the
two dice must be in one of three configurations, as we know there must be at least
one 6:
- Green die 6 and Red die 6.
- Green die 6 and Red die X.
- Green die X and Red die 6.
Note that the fact that there are 11 possible combinations containing a 6 is
irrelevant at this point. It doesn’t matter how we got here.
Hello, finally, Mr. Proper Sample Space.
Now, the original problem didn’t mention how the roller removed the die. He may have looked at the dice while doing so, or he might have chosen randomly. It may make a difference. In the infamous Monty Hall problem, Monty ALWAYS opened a door showing a goat because He always knew where the Grand Prize was. However our original problem here states the roller shows a 6 so I won’t go into this issue further and simply accept the fact it will always be a 6.
I assert the color of the removed die is irrelevant.
The roller may be showing us any of the 4 possible die-face 6’s (1. Green die, 1. Red
die, 2. Green die, or 3. Red die). 2 out of 4 of these die faces are both in the case
1 configuration, so the answer to the original problem is the probability that the
other die is showing a 6 is 1/2.
Bringing back the color issue, if the roller shows a Green 6 it might be either the
case 1 Green where the Red is also 6, or the case 2 Green where the Red is X. Here the probability is 1/2 that the Red is 6. Likewise when a Red 6 is shown.
<no words
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.
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.
The version I’m referring to is the one that CurtC posted, just a few posts before I said that.
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.
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?
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.
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 .
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:
- It does not state that the roller will reveal only 6s.
- 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?