How We Went To War

[Ramius]

I actually did find a link that supports him when I was looking for more info: http://www.ds.unifi.it/VL/VL_EN/expect/expect1.html

 

This site says that the expected value is the mean of the distribution (whether possible or not possible) and thus 3.5.

 

I don't have a friggin' clue if thats the accepted statistical definition though.

 

The problem comes with reading comprehension, and an actual factual understanding of a concept, something HA is incapable of. All he can do is spit out wikipedia quotes that he doesnt even begin to attempt to comprehend.

 

Tom's post a little bit ago explained things fairly well. The problem comes in when you try to treat a die like it a continuous variable, when in fact it is discrete. An expected value would be the most likely value you'd get by rolling a die. In rolling a die, you get a flat line distribution of all possible outcomes. rolling a 1 = a 2 = 3 = 4 = 5 = 6. So how is 3.5 the expected outcome? Its the AVERAGE, not the expected. There is NO expected value for rolling 1 die, because all 6 possibilities have an equal chance of being rolled. (1/6) Case in point: I have a die with sides cat, dog, bird, mouse, horse, and pig. Whats the expected outcome for THAT die? mouse.5?

 

Things DO make sense when rolling multiple die, however. If i roll 2 dice, i get a distribution of possible outcomes, where the most likely (highest percentage) outcome is 7. (6/36) So 7 is the expected outcome for a roll of 2 dice. Also, it IS possible to roll a 7 by rolling 2 die. But you cant just say that since 7 is expected for 2 die, then 3.5 is the expected for 1 die, because 1 die does not have the same distribution as rolling 2 die.

 

Mathematically, if you roll a die thousands of times, and calculate the AVERAGE value of each die roll, you will get 3.5. Thats a mathematical definition of average. By rolling a die multiple times, you've turned a set of discrete variables into a system of continuous variables. But you cannot back extrapolate that onto 1 die roll. 1 roll has equal probabilities of getting a 1,2,3,4,5,or 6. Therefore it has no "expected" value, and you cannot claim that the "expected roll" is 3.5, and that any other roll is "error".

 

This level of reading comprehension and basic understanding of mathematical concepts is what gets HA in trouble with his each and every post.

[DC Tom]

The problem comes with reading comprehension, and an actual factual understanding of a concept, something HA is incapable of. All he can do is spit out wikipedia quotes that he doesnt even begin to attempt to comprehend.

 

Tom's post a little bit ago explained things fairly well. The problem comes in when you try to treat a die like it a continuous variable, when in fact it is discrete. An expected value would be the most likely value you'd get by rolling a die. In rolling a die, you get a flat line distribution of all possible outcomes. rolling a 1 = a 2 = 3 = 4 = 5 = 6. So how is 3.5 the expected outcome? Its the AVERAGE, not the expected. There is NO expected value for rolling 1 die, because all 6 possibilities have an equal chance of being rolled. (1/6) Case in point: I have a die with sides cat, dog, bird, mouse, horse, and pig. Whats the expected outcome for THAT die? mouse.5?

 

Things DO make sense when rolling multiple die, however. If i roll 2 dice, i get a distribution of possible outcomes, where the most likely (highest percentage) outcome is 7. (6/36) So 7 is the expected outcome for a roll of 2 dice. Also, it IS possible to roll a 7 by rolling 2 die. But you cant just say that since 7 is expected for 2 die, then 3.5 is the expected for 1 die, because 1 die does not have the same distribution as rolling 2 die.

 

Mathematically, if you roll a die thousands of times, and calculate the AVERAGE value of each die roll, you will get 3.5. Thats a mathematical definition of average. By rolling a die multiple times, you've turned a set of discrete variables into a system of continuous variables. But you cannot back extrapolate that onto 1 die roll. 1 roll has equal probabilities of getting a 1,2,3,4,5,or 6. Therefore it has no "expected" value, and you cannot claim that the "expected roll" is 3.5, and that any other roll is "error".

 

This level of reading comprehension and basic understanding of mathematical concepts is what gets HA in trouble with his each and every post.

 

I see I've brainwashed you into spreading my propaganda...

[Chilly]

Makes sense Raimus. Looking at the site, their dice example is spread out over 100+ rolls and specifically pertains to distributions.

[Simon]

Do YOU have anything to contribute to the discussion?

Aside from a fruitless effort to discourage you from polluting this fine board with your shrill and repetitive wailing, no not really. How about if I just say that if your Dem heroes cared as much about their country as they do about themselves, they might have done something to stop this idiocy before it started.

[Orton's Arm]

Remember, this IS coming from someone who consistently quotes wikipedia, while at the same time discrediting thousands of published scientists and journals, because they prove him wrong.

You consistently claim that thousands of published scientists and journals support your own views, yet you're consistently unable to provide any links to credible sources to back up that claim. The only one who's been supporting his views with links to credible sources is me.

[Orton's Arm]

Anyone who's taken a sophmore-level college physics class knows that definition is wrong. Basic stat mech or quantum physics: sometimes the ensemble average is the expectation value, sometimes it isn't.

 

Of course, statistics 101 will show the same thing. It's very easy to establish a simple data set (a list of numbers) where the expectation value is less than the average. It's basically the difference between the mean and median values of the data set. Of course, you'll never understand that, as it involves math.

I'm sorry, but the only type of refutation I'll accept is one that involves a credible source. That means a link. You can't just throw around terms like "statistics 101" or "sophomore [sic]-level physics class" and pretend that you speak for all the statistics or physics professors out there. You don't. You speak only for yourself; and you do need to back up your claim with a link if you want to be taken seriously.

[Orton's Arm]

An expected value would be the most likely value you'd get by rolling a die.

That's your idea, and isn't supported by the links that either BlueFire or I found. In BlueFire's link, the expected value of a discrete distribution is found by the probability-weighted summation of all possible outcomes. In other words, the expected value of a die roll is 3.5.

[Bungee Jumper]

I'm sorry, but the only type of refutation I'll accept is one that involves a credible source. That means a link.

 

 

[Orton's Arm]

Since you can't find a link which supports your views, you may as well use an emoticon instead.

[Ramius]

That's your idea, and isn't supported by the links that either BlueFire or I found. In BlueFire's link, the expected value of a discrete distribution is found by the probability-weighted summation of all possible outcomes. In other words, the expected value of a die roll is 3.5.

 

And you still dont get it.

[Orton's Arm]

And you still dont get it.

No, I have a strong grasp on this situation. You decided to mock my explanation of expected value, without first bothering to look up the concept. As usual, you united ignorance with overconfidence to make a fool out of yourself.

[Ramius]

No, I have a strong grasp on this situation. You decided to mock my explanation of expected value, without first bothering to look up the concept. As usual, you united ignorance with overconfidence to make a fool out of yourself.

 

This might be insulting if you had even the slightest clue about what you are trying to discuss. But you dont, and all you've done in make your self like look an ass that cant comprehend basic math and stats, like usual.

 

Lets try this again.

 

The average/expected individual die roll over the course of n rolls (n > 1) will be 3.5

 

There is no average/expected die roll for n = 1 rolls of a die. Its an even distribution.

[Orton's Arm]

This might be insulting if you had even the slightest clue about what you are trying to discuss. But you dont, and all you've done in make your self like look an ass that cant comprehend basic math and stats, like usual.

 

Lets try this again.

 

The average/expected individual die roll over the course of n rolls (n > 1) will be 3.5

 

There is no average/expected die roll for n = 1 rolls of a die. Its an even distribution.

The distinction you're attempting to create does not exist in the article BlueFire found. Here's the relevant quote:

As usual, we start with a random experiment that has a sample space and a probability measure P. Suppose that X is a random variable for the, taking values in a subset S of R.

 

If X has a discrete distribution with density function f then the expected value of X is defined by (probability density formula follows)

Your statements are also directly contradicted by the Wikipedia article

the expected value from the roll of an ordinary six-sided die is 3.5

You claim that the phrase "expected value" has a different meaning depending on whether it's applied to a single trial, or to multiple trials. Care to support that with a link?

[DC Tom]

Since you can't find a link which supports your views, you may as well use an emoticon instead.

 

The simple fact that you think "link" and "reliable source" are synonymous demonstrates more clearly than anything that's come before just how much of a complete and utter fool you are.

[DC Tom]

The distinction you're attempting to create does not exist in the article BlueFire found. Here's the relevant quote:

 

Your statements are also directly contradicted by the Wikipedia article

 

You claim that the phrase "expected value" has a different meaning depending on whether it's applied to a single trial, or to multiple trials. Care to support that with a link?

 

And oh-by-the-way...you never said a die has an expected value of 3.5. You said it has a expected roll of 3.5.

 

So this whole "expected value" argument is irrelevent anyway. Tell me, O Great Mathematician...what's the formal statistical definition of "expected roll"?

[Orton's Arm]

The simple fact that you think "link" and "reliable source" are synonymous demonstrates more clearly than anything that's come before just how much of a complete and utter fool you are.

The problem is that you haven't supplied any link to support your views, not even one from some guy's free homepage. Until you provide a credible link, you're just wasting everyone's time by arguing about definitions you think you remember a lot better than you actually do.

[Orton's Arm]

And oh-by-the-way...you never said a die has an expected value of 3.5. You said it has a expected roll of 3.5.

Care to back that up with a link?

[DC Tom]

The problem is that you haven't supplied any link to support your views, not even one from some guy's free homepage. Until you provide a credible link, you're just wasting everyone's time by arguing about definitions you think you remember a lot better than you actually do.

 

 

Of course, any link - any source, link or not - that disagrees with you is automatically going to be "not credible". So what's the point? Never mind that as a TA in grad school I used to teach this stuff...

[DC Tom]

Care to back that up with a link?

 

No.

 

I mean, I can. I've already verified those are your exact words. As can anyone who can use the friggin' search function of the board. So do it yourself.

[Orton's Arm]

 

Of course, any link - any source, link or not - that disagrees with you is automatically going to be "not credible". So what's the point? Never mind that as a TA in grad school I used to teach this stuff...

It's so much easier to provide excuses than links, isn't it? BTW, you must have made a really lousy TA.