I was concerned that septimus was in the 1/11 camp because he is so good at stuff like this. I have rethought things. As a result of post #218 by cerberus I withdraw all my previous statements and formally switch to the 1/11 camp.
:smack:
Again, nobody but you and **Xema **is suggesting the game host ever lies. Nobody. That’s a gigantic red herring.
The point isn’t even that the game host might say different albeit truthful things on different rolls just for the lulz.
The point is that the text of the problem does not say: “The game host will always announce either A) ‘There is at least one 6’ when there is such, or announce B) ‘This is a do-over; ignore this roll’ otherwise”.
The game host’s rule for his announcement *could *be much more complex. And still be applied honestly and consistently to every roll. For some rule sets the answer is 1/6. For others it’s 1/11. For others it’s yet different odds.
We’re all left guessing what the rule set really is. Because the OP’s problem statement is silent on this point. Once we’ve guessed that, the determination of the resulting odds is trivial. The whole point is that we’re stuck guessing at step 1. So whatever calculation we make at step 2 is a product of guesswork. And hence utterly flawed as a definitive conclusion.
I think you may have misunderstood Xema’s point (although I’m not certain). It’s not necessarily that the observer might choose to lie. It’s that there is of course more information that can be garnered that can alter probability and it’s not relevant, although Cliffy Biro seems to think it is.
Do you see what Cliffy is stating there, especially given the last quote from him above? He’s not only stating that the boy/girl problem is either 1/2 or 1/3 depending on how it is phrased, he’s saying that since there is other possible information we could gather, the answer could be anything.
Also, he is even saying something close to stating there is a possibility the observer may lie by stating there may be a rule in place that he has to:
“say that at least one of them is black only if both are black”
Saying that there could be variables like this that will affect the answer makes almost any probability question impossible to answer.
You pick someone in a crowd and ask him to either say “country” or “rock” out loud. The person complies. What are the chances he/she said “country”?
According to Cliffy, “there are infinitely many possible answers, depending on which assumptions you use” and “the answer depends on how the observer decided to” pick a name.
Maybe this person picks first choices 4 out of 5 times consistently. Maybe there’s been a study done proving that when given a choice between saying two words, people pick two syllable words over one syllable words 2 to 1. Or maybe there’s a rule in place that the questioner looks for people that are Country music fans. The additional information will change the probability. But we don’t have additional information. So the probability is based on the information we do have alone. The probability the person in the crowd said “country” is 1/2.
I admit I haven’t read all the posts, but I’m perplexed on why the correct answer (1/11) isn’t obvious.
I have two die. Here are the possible combinations:
(1, 1)
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(3, 5)
(3, 6)
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
I now want to play the “game” described in the OP. Example:
I roll 2 and 4. Nope, roll again.
I roll 3 and 5. Nope, roll again.
I roll 4 and 4. Nope, roll again.
I roll 1 and 6. Stop!
I keep rolling until one or both die is a 6. And then I announce, “One or both die is a 6. Let’s continue…”
So in other words, all combinations that do not contain a 6 can be completely ignored; you need to pretend they do not exist:
[del](1, 1)[/del]
[del](1, 2)[/del]
[del](1, 3)[/del]
[del](1, 4)[/del]
[del](1, 5)[/del]
(1, 6)
[del](2, 1)[/del]
[del](2, 2)[/del]
[del](2, 3)[/del]
[del](2, 4)[/del]
[del](2, 5)[/del]
(2, 6)
[del](3, 1)[/del]
[del](3, 2)[/del]
[del](3, 3)[/del]
[del](3, 4)[/del]
[del](3, 5)[/del]
(3, 6)
[del](4, 1)[/del]
[del](4, 2)[/del]
[del](4, 3)[/del]
[del](4, 4)[/del]
[del](4, 5)[/del]
(4, 6)
[del](5, 1)[/del]
[del](5, 2)[/del]
[del](5, 3)[/del]
[del](5, 4)[/del]
[del](5, 5)[/del]
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
Another way of thinking of the problem: two “very special” die are rolled, and the only - ONLY!! - combinations they can produce are:
(1, 6)
(2, 6)
(3, 6)
(4, 6)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
The odds of both die being 6 are 1/11.
How is that “close to a lie”? If there are two black cats, and the observer says “there is at least one black cat”, that is absolutely true and not misleading at all.
In the other cases he will say, under this rule, “there is at least one white cat”, again completely truthfully. And yet the statement “there is at least one black cat” would, under this rule, give a probability of 1, that is 100%, that the other cat is black, not 1/3 or 1/2. Assuming you know the rule.
In the OP’s question, of course, you don’t know the rule, so all such calculation is futile.
The probability cannot be determined. 1/2 is merely your best estimate of the probability. It is not the actual probability.
Why roll again? There’s nothing in the OP rules that says you would have to. You’ve made assumptions not specified in the OP.
You could in this case say “At least one of the dice is a 4” and remove that die then ask “What is the chance the other die is a 4?”
Many people are assuming that the game began by specifying that 6 was a magic number and rolls not including a 6 would be ignored.
Other assumptions lead to different answers, such as the one I described above - that no number is magic and that on every roll a die is removed and the question asked what the probability is that the other die is the same. That is consistent with the OP which happens to describe a particular roll in which there was a 6 and it was the one removed.
I’ll retract my statement if you think how it’s labeled is relevant. But the label doesn’t affect the points of my post in total.
If you say so. You are taking away from the spirit of discussing probability, IMHO, but proceed as you will.
Yes. I can think of two possible reasons for people making that assumption:
- the wording of the question, "“at least one is a 6”, perhaps suggests that 6s are particularly noteworthy?
- in many games involving dice, 6 *is *noteworthy, what with it being the biggest number.
But there are other dice games in which 1, for example, is important. Why can’t we assume that the observer would be keen to note any 1 with the statement “there’s at least one 1”, but otherwise just describe one die at random? That leads to the probability of a throw of 6-6 being 1/5, by my calculations.
So you admit it is an assumption?
And that some assumptions are necessary to calculate an answer?
And that there are other assumptions consistent with the OP that lead to different answers?
And that in the end, the problem as described is ill-defined, since assumptions have to be made to answer it?
And that if you have to choose an answer from those given in the OP, that 1/11 is not one of the answers, but 1/6 is?
So maybe assumptions that lead to 1/11 are not consistent with the game actually being played in the OP?
I was agreeing with you, but you seem to be arguing?
Anyway, you are right of course that 1/11 was not one of the answers given in the OP. But all of the answers in the OP that are greater than or equal to 1/11 are possible, under appropriate assumptions. So it could be 2/11. Don’t ask me to come up with the rule that the observer would have to have been using, though!
The OP states,
There are many, many ways this could occur:
I roll 2 and 4. Nope, roll again.
I roll 3 and 5. Nope, roll again.
I roll 4 and 4. Nope, roll again.
I roll 1 and 6. Good! Let’s continue…
I roll 5 and 4. Nope, roll again.
I roll 2 and 2. Nope, roll again.
I roll 6 and 1. Good! Let’s continue…
I roll 4 and 6. Good! Let’s continue…
I roll 5 and 1. Nope, roll again.
I roll 2 and 1. Nope, roll again.
I roll 2 and 4. Nope, roll again.
I roll 1 and 1. Nope, roll again.
I roll 6 and 6. Good! Let’s continue…
As I stated in my previous post, all combinations that do not contain a 6 can be completely ignored; you need to pretend they do not exist.
Here is another way of thinking of the problem:
You have a 2-digit, digital display. When you press a button, one of the following will be displayed:
(1, 6)
(2, 6)
(3, 6)
(4, 6)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
There is an equal chance for each of the above to be displayed:
The odds of (1, 6) being displayed is 1/11.
The odds of (2, 6) being displayed is 1/11.
The odds of (3, 6) being displayed is 1/11.
The odds of (4, 6) being displayed is 1/11.
The odds of (5, 6) being displayed is 1/11.
The odds of (6, 1) being displayed is 1/11.
The odds of (6, 2) being displayed is 1/11.
The odds of (6, 3) being displayed is 1/11.
The odds of (6, 4) being displayed is 1/11.
The odds of (6, 5) being displayed is 1/11.
The odds of (6, 6) being displayed is 1/11.
And that’s the question in the problem… what are the odds that (6, 6) is displayed.
Actually, I think the rule is quite simple:
- if there is a six and a non-six, say “six” with probability 0.45, or the other number with probability 0.55 .
By my calculations, under that rule, if the observer says “there is at least one six”, the probability that they are both six is 100/550, that is 2/11 .
Yes - I see what you had written in previous posts where you acknowledge at least two different sets of assumptions that lead to answers including, but not limited to, 1/11 and 1/6.
But in response to my post, you appear to be playing devil’s advocate and making the case for why people could justify certain assumptions to themselves.
And that doesn’t help my point of trying (along with you and others) to get some posters here to realize that assumptions do in fact have to be made to calculate any answer - there’s simply not enough information given in the OP to justify a single answer.
Some posters seem to think that there is only one way to analyze the problem and so I was replying more for their sake than directly to you.
In any event, I’m just discussing - not arguing, if that would imply my response as being combative on any level.
Got it, sorry for the misunderstanding.
I understand how to calculate the probability under assumptions that lead to 1/11.
Do you acknowledge that 1/11 is not an answer in the OP?
Or do you further assume there is an error in the choices given for answers in the OP?
As long as were assuming things, I guess why not continue assuming beyond an ill-defined problem that there is an error in the possible answers.
It’s all good - hope you’re having a great weekend!
But Cliffy posted this:
I don’t think this is the clearest statement of the problem. I think the OP made the problem reasonably clear:
But clearest of all would be to state it as “Two fair dice are rolled together; at least one of them is a 6.”
IOW, there’s no need to introduce an ‘observer’ at all, nor to consider which of many possible recondite strategies he may be using.
Just think about a roll of two fair dice, whose result includes at least one 6.
That is rather like talking about past events (because you have excluded the possibility of no six being rolled, which cannot apply to fair dice that have yet to be rolled.) Probability does not apply to events that have already happened. The probability of last week’s lottery numbers is not one in fourteen billion or whatever it is, because those numbers did actually come up.
If you say “no, I am talking about future rolls, but only those ones with at least one six”, that is reintroducing an assumption that is not in the OP, i.e. that it will always be noted when at least one die is a six.
Are you familiar with the concept of conditional probability? It’s a very common type of question to ask “What is the probability that B is true, given that A is true” (symbolized P(B|A)). This is how I would interpret a statement of the form “A is true. What is the probability that B is true?” And it seems to me that your “rule” would wreck discussions of conditional probability:
Question: A black card is drawn from a deck. What is the probability that it is a spade?
Person used to thinking in terms of conditional probability: If it’s a black card, there are 26 possibilities for what it could be. In that case, the probability that it’s one of the 13 spades is 13/26, or 1/2.
You: If a black card has already been drawn, it is what it is, and probability does not apply.
That is a complete listing of the possibilities as to what the pair of dice showed after the roll. Now what is the list of possibilities as to what die remains? Well, for each of the above possibilities, we remove a 6:
1
2
3
4
5
6
5
4
3
2
1
And now we ask, what is the probability that the remaining die is a 6?
1/11.