Life In 19x19

RobertJasiek wrote:

Bill Spight wrote:In English "to choose at random" means to choose so that each option has the same probability of being chosen.

I'd say that "to choose at random uniformly / with a uniform distribution" would mean this. At least in (precise) German, one needs to say "gleichverteilt zufällig wählen".

Deutlich ist Deutschlich.

Bill Spight wrote:

RobertJasiek wrote:Since it may be any random distribution with any probability, there are an infinite number of such probabilities but only a finite number of available answer probabilities. Hence the answer is: "almost 0". With a model of an increasing number of distributions and finally infinitely many, the answer is: "converges to 0 from its positive side".

In English "to choose at random" means to choose so that each option has the same probability of being chosen.

So you believe the statement, "He played at random", indicates that the 1-1 points were as likely to be chosen as Tengen? I believe that English in the hands, hmmm... in the mouths that is, of us normal folks has nothing like the precision that can be ascribed to mathematical jargon.

What is the correct answer to this question?
(A) Answer B is correct.
(B) Answer A is correct.

mitsun wrote:What is the correct answer to this question?
(A) Answer B is correct.
(B) Answer A is correct.

They are both correct. If A is the correct answer then it says that B is the correct answer, which is fact true. And vice versa.

Alternatively, neither are correct. If A is wrong then it says B is not the correct answer, so A is wrong. And vice versa.

Take your pick.

(A) Statement B is true.
(B) Statement A is false.

This statement is false.

ez4u wrote:

Bill Spight wrote:

RobertJasiek wrote:Since it may be any random distribution with any probability, there are an infinite number of such probabilities but only a finite number of available answer probabilities. Hence the answer is: "almost 0". With a model of an increasing number of distributions and finally infinitely many, the answer is: "converges to 0 from its positive side".

In English "to choose at random" means to choose so that each option has the same probability of being chosen.

So you believe the statement, "He played at random", indicates that the 1-1 points were as likely to be chosen as Tengen? I believe that English in the hands, hmmm... in the mouths that is, of us normal folks has nothing like the precision that can be ascribed to mathematical jargon.

From Merriam-Webster's Dictionary:

One sense: "being or relating to a set or to an element of a set each of whose elements has equal probability of occurrence <a random sample>; also : characterized by procedures designed to obtain such sets or elements <random sampling>"

Another sense: "lacking a definite plan, purpose, or pattern"

The first sense is the one for choosing among given alternatives at random. The second is the one for playing at random.

Bill Spight wrote:

ez4u wrote:blah blah

From Merriam-Webster's Dictionary:

One sense: "being or relating to a set or to an element of a set each of whose elements has equal probability of occurrence <a random sample>; also : characterized by procedures designed to obtain such sets or elements <random sampling>"

Another sense: "lacking a definite plan, purpose, or pattern"

The first sense is the one for choosing among given alternatives at random. The second is the one for playing at random.

Exactly! The first sense is what professionals in the field do. The second sense encapsulates my entire life in seven words!

DrStraw wrote:

entropi wrote:If you choose an answer to this question at random, what is the chance you will be correct?
(A) 25% (B) 50% (C) 60% (D) 25%

Zero. Without further information we can only assume that all answers are possible. Therefore the choices form a finite subset of an infinite number of possible answers. The density is zero.

If, on the other hand, all possible correct answers are listed and each is considered equally likely then the probability is one third.

The fact that this is formulated as a multiple choice question implies that "choose an answer" intends to mean "choose one of (A),(B),(C), or (D)".

How do you come up with one third?

entropi wrote:

DrStraw wrote:

entropi wrote:If you choose an answer to this question at random, what is the chance you will be correct?
(A) 25% (B) 50% (C) 60% (D) 25%

Zero. Without further information we can only assume that all answers are possible. Therefore the choices form a finite subset of an infinite number of possible answers. The density is zero.

If, on the other hand, all possible correct answers are listed and each is considered equally likely then the probability is one third.

The fact that this is formulated as a multiple choice question implies that "choose an answer" intends to mean "choose one of (A),(B),(C), or (D)".

How do you come up with one third?

There are three possible values to choose. If it is random that means one in three.