Confidence problem with dice: outcomes in the range [2..12]

BeeCuz

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Rolling 2d6, there are 36 possible outcomes in the range [2..12].

1). Let's say I choose the range [2..10] as success, what then is my confidence? There is only one way to get a 12 and two ways to get an 11, for 3/36 of the outcomes, so my confidence in success is 33/36.

2). Let's say I choose 30/36 as my confidence, what then is the matching range? There are three ways to get a 10 + 2 elevens + 1 twelve = 6/36, so my range in order to achieve that level of confidence is [2..9].

How can I more formally determine a range given a confidence or a confidence given a range?
 
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tkhunny

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"Confidence" is not well-defined as you have stated it. Do you mean "chance of success"? Do you mean "odds"? Do you mean "Probability of rolling 1..10"? Just the word "confidence" may mean something ethereal - just how you feel about it.
 

BeeCuz

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Dr.Peterson

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From Wikipedia at https://simple.wikipedia.org/wiki/Confidence_interval

"In statistics a confidence interval is a special form of estimating a certain parameter. With this method, a whole interval of acceptable values for the parameter is given instead of a single value, together with a likelihood that the real (unknown) value of the parameter will be in the interval."
But that definition does not fit the way you are using the word "confidence". You seem to be using it just to mean "probability", which is a number, not "confidence interval" which is a set like (32/36, 34/36). You can't say "so my confidence interval in success is 33/36".

Am I right that you mean, "my probability of success is 33/36"?
 

BeeCuz

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You seem to be using it just to mean "probability", which is a number, not "confidence interval" which is a set like (32/36, 34/36). You can't say "so my confidence interval in success is 33/36". Am I right that you mean, "my probability of success is 33/36"?
Please start from the premise that no errors have been committed, thank you. At the very least it makes you look good.

The problem states ranges of dice rolls (confidence intervals) and the associated confidence level (probability) of the outcome being a member of the range. What I'm interested in is how to calculate the confidence interval for the associated probability.

For example, given the associated probability 10/36 30/36 ​[correction], what is the range of dice rolls that meet that specification? I know the answer beforehand ( the range of values [2..8] ), but that doesn't resolve how to calculate it.
 
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Dr.Peterson

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Please start from the premise that no errors have been committed, thank you.

The problem states ranges of dice rolls (confidence intervals) and the associated confidence level (probability) of the outcome being a member of the range. What I'm interested in is how to calculate the confidence interval for the associated probability.

For example, given the associated probability 10/36, what is the range of dice rolls that meet that specification? I know the answer beforehand, but that doesn't resolve how to calculate it.

The answer by the way is the range of values [2..9]
Let's forget the terminology, which you are still using inaccurately. That doesn't matter. What I think you want is this:
Given a pair of 6-sided dice and an interval [m, n], that is, 2 <= m <= x <= n <= 12, we can determine the probability that the sum of the dice, x, is in that interval, P(m <= x <= n). Is it possible to find an interval for which the probability will be any given number? That is, given p, we want to find numbers m and n such that P(m <= x <= n) = p.

The question still needs further clarification, though. There are infinitely many possible values you could choose for p (any real number between 0 and 1); but any probability we produce will be a fraction with denominator 36. Are we assuming that p = k/36 for some integer k? Without that, it would be impossible for most values of p.

Secondly, even with this restriction, there may be no interval that gives exactly p; or there might be more than one. Are you willing to accept that as the answer for some p, or do you want an interval that yields the closest possible probability to the given p? Your phrasing, "the confidence interval for the associated probability", assumes that there is always one and only one answer, which is not going to be true.

By the way, for your example, I think you meant 30/36, as in the original question, not 10/36 as you said here. If not, then I'm doing something wrong.
 

BeeCuz

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Let's forget the terminology, which you are still using inaccurately. That doesn't matter.
Either provide a reference to a definition that establishes an objective difference in usage so that I may understand and use that definition, or accept this as the proper usage of terminology.

What I think you want is this:
Given a pair of 6-sided dice and an interval [m, n], that is, 2 <= m <= x <= n <= 12, we can determine the probability that the sum of the dice, x, is in that interval, P(m <= x <= n). Is it possible to find an interval for which the probability will be any given number? That is, given p, we want to find numbers m and n such that P(m <= x <= n) = p.
This is an accurate restatement of the problem in this discrete dice model.

There are infinitely many possible values you could choose for p (any real number between 0 and 1); but any probability we produce will be a fraction with denominator 36. Are we assuming that p = k/36 for some integer k?
There's probably a more succinct way to express the limitations of k in terms of number of sides of the dice, such that k = aX where X is the set {1,2,3,4,5,6} and aX<=36. But that specificity narrows the problem space down to just the discrete dice model.

I would rather pursue an avenue that leads to a continuous model of probabilities where p is any real number in the range [0..1], and the confidence interval is a range across a normalized set of possible outcomes [0..x], where 0 < x <=1.

By the way, for your example, I think you meant 30/36, as in the original question, not 10/36 as you said here. If not, then I'm doing something wrong.
Error corrected. apparently the forum doesn't support strikethrough, so I used faded font color to represent the error text.
 

Dr.Peterson

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Either provide a reference to a definition that establishes an objective difference in usage so that I may understand and use that definition, or accept this as the proper usage of terminology.
You quoted a definition, and it didn't reflect what you are asking for. A confidence interval is a set of values of a population parameter, such that that there is a certain probability (confidence level) that the true parameter is in that interval, based on observation of a sample. You are not talking about estimating a parameter of your dice based on observation of throws. So this is not the right term.

But as I said, I'm not interested in arguing about the meaning of words; I just want to figure out what you actually want, and see if I can help.

There's probably a more succinct way to express the limitations of k in terms of number of sides of the dice, such that k = aX where X is the set {1,2,3,4,5,6} and aX<=36.
I'm not sure what you mean by aX. How are you multiplying a set by a number to get a number?

But that specificity narrows the problem space down to just the discrete dice model.

I would rather pursue an avenue that leads to a continuous model of probabilities where p is any real number in the range [0..1], and the confidence interval is a range across a normalized set of possible outcomes [0..x], where 0 < x <=1.
Your question WAS about dice. If you're taking that as just an example of what you want, state what the real goal is, rather than wasting people's time with other problems that may not need the same technique. There is a huge difference between the dice question we've been talking about and a continuous distribution.

You haven't responded to the most important thing I said. The question as I stated it, which you say is accurate, does not have a solution! For some values of p there will be no intervals (unless you allow some latitude), and for others there will be two very different intervals.

Now, the continuous situation you are talking about appears to have infinitely many solutions, unless you can clarify it. As I picture it, suppose you have a normal distribution. You want to find an interval such that the probability in this interval is a given number p. You can pick just about any value of x for the left end of the interval and find a right end that will give the desired p. But I'm probably completely misinterpreting what you said.
 

BeeCuz

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But as I said, I'm not interested in arguing about the meaning of words; I just want to figure out what you actually want, and see if I can help.
The opportunity has already come and gone; you cannot help. Calculus is about shifting from discrete to continuous.

For anyone else reading this thread:
"In statistics a confidence interval is a special form of estimating a certain parameter. With this method, a whole interval of acceptable values for the parameter is given instead of a single value, together with a likelihood that the real (unknown) value of the parameter will be in the interval."
1). a whole interval of acceptable values: [2..7]
2).
a likelihood that the real (unknown) value of the parameter will be in the interval: 21/36

If we adjust either parameter, how then do we calculate the other?
 
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tkhunny

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Here’s the problem… “Confidence”and “Confidence Interval” are not the same thing.

“Confidence Interval” is well-defined.
“Confidence” is not. “\(\displaystyle 1-\alpha\) Level of Confidence” is better.

Calculus is about continuous things, rather than just discrete things. The difficulty, here, is that rolling dice is discrete, not continuous. You cannot define any confidence interval you want. You have to stick to some subset of [2:12]. You cannot define any \(\displaystyle \alpha\) you want. There are only a few choices. If you really want all the intervals and all the levels of confidence, just build a catalogue. There are not that many entries in this one – 66 in this case. If you can do one, you should have no trouble doing the rest. Let’s see what you get for a couple of examples, maybe [2:2] and [3:7].


Attitude is everything. If you want to control the discussion and berate people who want to help, you should consider paying for the engagement. Abusing our volunteers is not a good idea.
 
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stapel

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Here’s the problem… “Confidence”and “Confidence Interval” are not the same thing.

“Confidence Interval” is well-defined.
“Confidence” is not. “\(\displaystyle 1-\alpha\) Level of Confidence” is better.

Calculus is about continuous things, rather than just discrete things. The difficulty, here, is that rolling dice is discrete, not continuous. You cannot define any confidence interval you want. You have to stick to some subset of [2:12]. You cannot define any \(\displaystyle \alpha\) you want. There are only a few choices. If you really want all the intervals and all the levels of confidence, just build a catalogue. There are not that many entries in this one – 66 in this case. If you can do one, you should have no trouble doing the rest. Let’s see what you get for a couple of examples, maybe [2:2] and [3:7].

Attitude is everything. If you want to control the discussion and berate people who want to help, you should consider paying for the engagement. Abusing our volunteers is not a good idea.
Thank you! ;)
 

BeeCuz

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Here’s the problem… “Confidence”and “Confidence Interval” are not the same thing.
“Confidence Interval” is well-defined. “Confidence” is not. “\(\displaystyle 1-\alpha\) Level of Confidence” is better.
I continue to claim that confidence is the thing being measured, not some other thing also named confidence. You're welcome to disagree, but you'll have to do much better than the last guy in presenting a convincing argument for defining what is being measured, if you wish to claim a ruling on the subject.

Calculus is about continuous things, rather than just discrete things.
One of the ways calculus works is by examining a discrete distribution of steps and increasing the resolution down to a continuous set of points. This seems to me the most fruitful of avenues available for converting the dice model into a area under a probability curve and an attendant probability.

What I need help with is the conversion process: How would one step up to infinitely small increments? My first guess at approaching this problem is to increase the number of sides on a dice, but that's not enough. Rolling 2d10 instead of 2d6 will still result in a steep approximation of a bell curve. As the number of dice goes up the curve becomes more...curvy. Rolling 10d10 would result in a refined version, heading toward continuous.

But here we have two variables, the number of sides of dice and the number of dice.
 

Dr.Peterson

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It will help a lot if you can state the problem you actually want to solve. What continuous distribution do you have (at least an example), and what do you want to do with it? A mere analogy to dice does not define the problem.
 

BeeCuz

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It will help a lot if you can state the problem you actually want to solve. What continuous distribution do you have (at least an example), and what do you want to do with it? A mere analogy to dice does not define the problem.
The final destination if you will, or what I want to do with it, is to re-arrange the problem which is taught as f(x) = y so that I have f(y) = x. More specifically, what I've worked out thus far without assistance, demonstrated in this thread using dice:

1). I can specify a sub-set of outcomes as a confidence interval and then calculate the probability of the outcome being a member of the set.
2). I can specify a probability of outcome and then calculate the sub-set of outcomes as a confidence interval paired with that probability.

I would like the continuous version of both of those calculations. If you have such knowledge then please share an understanding of it.
 
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