Sorry, I meant to say think. It just seems like something anybody who plays with dice a lot would have figured out intuitively.
At that point is key here. At that point, you are beginning to calculate odds. Not before. Most of the 36 possible double-die roll combinations have been eliminated. They do not count. We have no numerator or denominator yet. We are reduced to “what are the chances of rolling a single die and getting N?” Now the numerator is 1 and the denominator is 6.
If that wasn’t your intention, you need to seriously revise the premise.
I disagree. I believe that the question is fundamentally ambiguous; the fact that it can easily be interpreted as having so much “irrelevant” information is a consequence of this.
You’re going through the whole sequence saying “You’re trying to do sleight of hand, but you can’t fool me, you can’t get me to look away from this die that you’re calling the ‘other die’, and the odds of that die rolling a 6 are 1 in 6 and that can’t change…”
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.
I rolled dice for about an hour last night and did the problem as worded. It does seem to be 1/11.
(Breaking a post up to address each point as well as I can.)
I don’t agree with this. You can begin calculating odds at the beginning, with rolling both dice, and then modify them as new information comes in, “at each point.”
True; 25 of them.
We can have a denominator; it’s 11, the number of cases which haven’t been eliminated. Depending on exactly what we’re looking at we could have several different numerators.
No, I don’t think we are.
If it helps, let’s leave aside the question about “the other die” to start with. I’ve rolled a green die and a red die. Skammer tells us that at least one of those dice is a 6. What are the chances that the green die shows a 4?
You could say that the chances are 1 in 6, but I don’t agree. We have some information which might relate to the green die; that either it or the red die (or both) is a 6. I say that we’ve got 11 equal possibilities remaining, and in only 1 of them does the green die have a 4.
Would you agree with that?
What if it were phrased without asking about “the other die”? I’m curious what you believe the answer is if it were phrased like this?:
Then the omniscient viewer says, “Look, you rolled at least one six.” At that point, what is the chance both dice are 6s?
You roll three dice, and remove the highest two. What is the probability that the other die is a 6?
Sigh. As I predicted back in post #24, this thread is generating “endless fruitless debate” because people are disagreeing on the interpretation of the question.
For those of you who think the answer is unambiguosly 1/11, consider this situation:
I roll a pair of dice. I note the highest number showing (call it X), and say to another person, “At least one of the dice is showing an X. What is the probability that the other is showing an X?” Clearly the answer is 1/6. I repeat this multiple times. Each time the answer is 1/6, although X may be different each time.
Now a third person walks in the room just as I’ve rolled a pair with a 6 showing, and hears me say “At least one of the dice is showing a 6.” This is the same as the description in the OP, but the answer here is 1/6 not 1/11. As stated several times in this thread, the answer depends on the randomization procedure, which was not stated in the OP. If you assume you know what the randomization procedure was (eg. “ignore throws that don’t show a 6”) you’ll get a specific answer, but it’s not the only answer because THE RANDOMIZATION PROCEDURE WAS NOT DESCRIBED and the answer changes if the randomization procedure changes.
–Mark
I’m a perfectionist and don’t like half-correct posts to stand.
The actual probability is of course a rational number. I get that it is
52949 / 285384
which has a repeating decimal expansion of period 506:
.185
53597959240882460123903232136349620160906007344490
23070669694166456423625711322288565581812575337089
67566506881955540604939309842177557256188153505452
30286210859753875480054943514702996664143750175202
53412945364841757071174277464749250133153925938384
77279735374092450873209430101196983713172427325988
84310262663639166876909707622011044767751520757996
24365766826451377792728394023491155776077145179827
88103047122473579457853278389818630336669189583158
13079920387968491576262159055868584083200179407394948560
53597959240882460123903232136349620160906007344490
23070669694166456423625711322288565581812575337089
67566506881955540604939309842177557256188153505452
30286210859753875480054943514702996664143750175202
53412945364841757071174277464749250133153925938384
77279735374092450873209430101196983713172427325988
84310262663639166876909707622011044767751520757996
24365766826451377792728394023491155776077145179827
88103047122473579457853278389818630336669189583158
13079920387968491576262159055868584083200179407394948560
…
Assuming ten was meant to be then, the only ambiguity I see is the phrase “you are told that ‘at least one of the dice is a 6’”. That could mean whoever is telling you is a liar. So in that case it’s like saying two fair dice are rolled, one of them is removed, what is the probability of it being a 6?
So the answer is that it is either 1/6 or 1/11 depending on whether or not the die removed is actually a 6. Given that the choices are " 1/18, 1/36, 2/11 and 1/6" we have to assumed we were lied to and the answer is 1/6, and who ever posed the problem is just being annoying.
That’s my conclusion, and I’m sticking to it, because it really doesn’t matter.
ETA: Arggh! Even then there’s no ambiguity, it states a 6 is removed, so there’s no lie. The answer is 1/11 and the answer is none of the above. Unless we assume it’s all a lie it’s not ambiguous.
I’m not sure that’s completely clear, especially if I know you’re picking the highest number to tell me about. If X=1, then my answer is 1. For X=2, I’d say 1/3, for 3, 1/5, for 4, 1/7, for 5, 1/9 and 1/11 for 6.
Now, since you’re more likely tell me a higher number in proportion, the weighted average for my answer comes back to 1 out of 6.
Please clarify. What do you mean by “the other”? Is “the other is showing an X” equivalent to “both are showing an X,” since we are already given that at least one is? If not, what does it mean?
No, this is not “clearly” at all. Can you show your work?
ETA after seeing chrisk’s post: Does the other person know that X is the highest number showing?
As I interpret it, a question of the form “A is true. What is the probability that B is true?” is a conditional probability question. It means, “Given that A is true, what is the probability that B is true?” It’s asking for P(B|A), not P(B).
They wont stop, no matter how many times this is pointed out. :rolleyes:
I can’t see that there is any ambiguity in the stated question at all.
Ironically the idea of removing a 6 is an attempt to eliminate any ambiguity caused by saying “if one die is 6.”
So it becomes like a casino game. You bet $100 and roll two dice. No 6 showing and the croupier pushes both dice back to you and you are coming out again. Any sixes showing the croupier takes one off the table. If the remaining die isn’t a 6 you lose your $100. If it is a 6 you win the payout which should be $1100.
For those people who believe the correct odds are one in six I will happily set up this game and pay them $700 every time the remaining die on the table is a 6.
You’re right, I shouldn’t have said “highest number”. I should have said “I pick one of the dice at random and announce its value”. The rest of my argument holds.
–Mark
Now you’ve moved the goalposts and changed the formula and odds. If you remove the highest two, the chances of the remaining one being a high number are reduced. This is not the original premise at all. Do you see the difference?
What a cool problem!
I think the problem is unambiguous (excepting the typo for “then”) and my first answer was 1/11. I was wrong. The list of combinations showing boxcars twice are, in a sense, correct. There are two ways that a six can be removed when double six is rolled and that case (“a six is removed”) needs to be counted twice. The right answer, as others have pointed out, is 1/6. But, boy, it’s a subtle difference.
Also, this thread is bound for 10 pages plus, easy…
If I remove one because it’s a 6, then the chances of the remaining one being a six are reduced. It’s the same principle!
I fail to see the difference between saying “look, you rolled at least one six” and “we now remove the six.” The remaining “other” die can be any value from 1…6 and is unaffected by the status or even presence of the first die.
How about this version? …the omniscient viewer says, “Look, you rolled at least one six.” At that point, what is the chance the other die is 2? Answer: 1/6
I dare you to try it, either with real dice or a computer simulation, assuming you program it correctly, of course.
What is your definition of programming it correctly? That is, to what specifications should we set up the simulation?