Mixing
What is the probability that a third randomly chosen real number is between two earlier randomly chosen real...
$begingroup$
Question 1: Suppose you pick three random real numbers A, B and C all at the same time.
Are the following outcomes all equally likely or are some more likely than others?
A < B < C
A < C < B
B < A < C
B < C < A
C < A < B
C < B < A
Question 2: Suppose you pick two random real numbers A and B and it happens that A < B. Next, you pick a third random real number C.
What is the probability that A < C < B?
probability
$endgroup$
|
show 8 more comments
$begingroup$
Question 1: Suppose you pick three random real numbers A, B and C all at the same time.
Are the following outcomes all equally likely or are some more likely than others?
A < B < C
A < C < B
B < A < C
B < C < A
C < A < B
C < B < A
Question 2: Suppose you pick two random real numbers A and B and it happens that A < B. Next, you pick a third random real number C.
What is the probability that A < C < B?
probability
$endgroup$
$begingroup$
What are your thoughts on this?
$endgroup$
– Bram28
Jan 28 at 1:23
$begingroup$
It seems the answer to the first question is yes and the answer to the second question is 0 since there is only a finite distance between A and B.
$endgroup$
– user637421
Jan 28 at 1:24
$begingroup$
@michael35 Do you think it matters for the order the 3 numbers will be in whether we pick the 3 numbers at the same time or one after the other? Or whether we pick all 3 at once but reveal them one at a time?
$endgroup$
– Bram28
Jan 28 at 1:27
$begingroup$
That's basically my question.
$endgroup$
– user637421
Jan 28 at 1:28
2
$begingroup$
This question doesn't make sense. You can't choose real numbers uniformly at random.
$endgroup$
– Matt Samuel
Jan 28 at 1:40
|
show 8 more comments
$begingroup$
Question 1: Suppose you pick three random real numbers A, B and C all at the same time.
Are the following outcomes all equally likely or are some more likely than others?
A < B < C
A < C < B
B < A < C
B < C < A
C < A < B
C < B < A
Question 2: Suppose you pick two random real numbers A and B and it happens that A < B. Next, you pick a third random real number C.
What is the probability that A < C < B?
probability
$endgroup$
Question 1: Suppose you pick three random real numbers A, B and C all at the same time.
Are the following outcomes all equally likely or are some more likely than others?
A < B < C
A < C < B
B < A < C
B < C < A
C < A < B
C < B < A
Question 2: Suppose you pick two random real numbers A and B and it happens that A < B. Next, you pick a third random real number C.
What is the probability that A < C < B?
probability
probability
asked Jan 28 at 1:21
user637421
$begingroup$
What are your thoughts on this?
$endgroup$
– Bram28
Jan 28 at 1:23
$begingroup$
It seems the answer to the first question is yes and the answer to the second question is 0 since there is only a finite distance between A and B.
$endgroup$
– user637421
Jan 28 at 1:24
$begingroup$
@michael35 Do you think it matters for the order the 3 numbers will be in whether we pick the 3 numbers at the same time or one after the other? Or whether we pick all 3 at once but reveal them one at a time?
$endgroup$
– Bram28
Jan 28 at 1:27
$begingroup$
That's basically my question.
$endgroup$
– user637421
Jan 28 at 1:28
2
$begingroup$
This question doesn't make sense. You can't choose real numbers uniformly at random.
$endgroup$
– Matt Samuel
Jan 28 at 1:40
|
show 8 more comments
$begingroup$
What are your thoughts on this?
$endgroup$
– Bram28
Jan 28 at 1:23
$begingroup$
It seems the answer to the first question is yes and the answer to the second question is 0 since there is only a finite distance between A and B.
$endgroup$
– user637421
Jan 28 at 1:24
$begingroup$
@michael35 Do you think it matters for the order the 3 numbers will be in whether we pick the 3 numbers at the same time or one after the other? Or whether we pick all 3 at once but reveal them one at a time?
$endgroup$
– Bram28
Jan 28 at 1:27
$begingroup$
That's basically my question.
$endgroup$
– user637421
Jan 28 at 1:28
2
$begingroup$
This question doesn't make sense. You can't choose real numbers uniformly at random.
$endgroup$
– Matt Samuel
Jan 28 at 1:40
$begingroup$
What are your thoughts on this?
$endgroup$
– Bram28
Jan 28 at 1:23
$begingroup$
What are your thoughts on this?
$endgroup$
– Bram28
Jan 28 at 1:23
$begingroup$
It seems the answer to the first question is yes and the answer to the second question is 0 since there is only a finite distance between A and B.
$endgroup$
– user637421
Jan 28 at 1:24
$begingroup$
It seems the answer to the first question is yes and the answer to the second question is 0 since there is only a finite distance between A and B.
$endgroup$
– user637421
Jan 28 at 1:24
$begingroup$
@michael35 Do you think it matters for the order the 3 numbers will be in whether we pick the 3 numbers at the same time or one after the other? Or whether we pick all 3 at once but reveal them one at a time?
$endgroup$
– Bram28
Jan 28 at 1:27
$begingroup$
@michael35 Do you think it matters for the order the 3 numbers will be in whether we pick the 3 numbers at the same time or one after the other? Or whether we pick all 3 at once but reveal them one at a time?
$endgroup$
– Bram28
Jan 28 at 1:27
$begingroup$
That's basically my question.
$endgroup$
– user637421
Jan 28 at 1:28
$begingroup$
That's basically my question.
$endgroup$
– user637421
Jan 28 at 1:28
2
2
$begingroup$
This question doesn't make sense. You can't choose real numbers uniformly at random.
$endgroup$
– Matt Samuel
Jan 28 at 1:40
$begingroup$
This question doesn't make sense. You can't choose real numbers uniformly at random.
$endgroup$
– Matt Samuel
Jan 28 at 1:40
|
show 8 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I see your dilemma. On the one hand we have:
Argument 1
It does not matter whether you pick all three numbers at once or one at a time. For example, suppose we pick three numbers A,B,D all at once, but reveal only two of A and B. And, also suppose that after we have revealed A and B, we pick a new number C.
Now, it makes little sense to think that the probability of D being between A and B would be any different than the probability of C being between A and B: the only difference is that one number was picked before the reveal, and the other after the reveal, and of course the picking of a number is not going to be affected by the reveal.
Also, as the first question makes clear, the probability that D is between A and B is $frac{1}{3}$ .. so therefore the probability that C is between A and B is also $frac{1}{3}$
But on the other hand we have:
Argument 2
Whatever A and B are, their difference is finite, and hence any third random number has a $0$ probability of being between them.
... so ... maybe this is a reductio ad absurdum against the very assumption that we can randomly pick numbers from all real numbers with equal likelihood?
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
$endgroup$
$begingroup$
Yes, randomly choosing a real number is not well defined. A uniform distribution requires a finite width. There do exist probability functions on the reals, so if you specify one of those then your question will be answerable.
$endgroup$
– dbx
Jan 28 at 2:40
$begingroup$
There are infinitely many pairs A and B to be integrated over.
$endgroup$
– WW1
Jan 28 at 3:27
add a comment |
$begingroup$
You can not choose a real number at random with uniform probabilities. However, there are various probability distributions you can choose from. Assuming you choose one of these PDFs, then for question one, all the possible orderings are equally probable because you are choosing them independently or all at the same time. For the second question the probability of A < C < B completely depends on the PDF.
answered Jan 28 at 2:53
SomosSomos
14.7k11337
$endgroup$
add a comment |
$begingroup$
Let A, B and C be uniformly distributed on $(-N,N)$
Then each has pdf = $frac 1{2N}$
The probability $A<C<B$ is given by
$$ begin{eqnarray} P = & &frac 1 {(2N)^3}int_{-N}^Nint_A^Nint_A^B dC ; dB ; dA
\=&&frac 1 {(2N)^3}int_{-N}^N int_A^N (B-A); dB ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N ( frac 12 N^2-AN)-(frac 12 A^2-A^2 ) ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N (frac 12 A^2-AN+ frac 12 N^2 ) ; dA
\= & &frac 1 {(2N)^3} (frac 13 N^3 +N^3) \ = & &frac 16 end{eqnarray} $$
This result is independent of $N$ so you would expect it to apply in the limit $N to infty$
answered Jan 28 at 3:51
WW1WW1
7,3501712
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I see your dilemma. On the one hand we have:
Argument 1
It does not matter whether you pick all three numbers at once or one at a time. For example, suppose we pick three numbers A,B,D all at once, but reveal only two of A and B. And, also suppose that after we have revealed A and B, we pick a new number C.
Now, it makes little sense to think that the probability of D being between A and B would be any different than the probability of C being between A and B: the only difference is that one number was picked before the reveal, and the other after the reveal, and of course the picking of a number is not going to be affected by the reveal.
Also, as the first question makes clear, the probability that D is between A and B is $frac{1}{3}$ .. so therefore the probability that C is between A and B is also $frac{1}{3}$
But on the other hand we have:
Argument 2
Whatever A and B are, their difference is finite, and hence any third random number has a $0$ probability of being between them.
... so ... maybe this is a reductio ad absurdum against the very assumption that we can randomly pick numbers from all real numbers with equal likelihood?
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
$endgroup$
$begingroup$
Yes, randomly choosing a real number is not well defined. A uniform distribution requires a finite width. There do exist probability functions on the reals, so if you specify one of those then your question will be answerable.
$endgroup$
– dbx
Jan 28 at 2:40
$begingroup$
There are infinitely many pairs A and B to be integrated over.
$endgroup$
– WW1
Jan 28 at 3:27
add a comment |
$begingroup$
I see your dilemma. On the one hand we have:
Argument 1
It does not matter whether you pick all three numbers at once or one at a time. For example, suppose we pick three numbers A,B,D all at once, but reveal only two of A and B. And, also suppose that after we have revealed A and B, we pick a new number C.
Now, it makes little sense to think that the probability of D being between A and B would be any different than the probability of C being between A and B: the only difference is that one number was picked before the reveal, and the other after the reveal, and of course the picking of a number is not going to be affected by the reveal.
Also, as the first question makes clear, the probability that D is between A and B is $frac{1}{3}$ .. so therefore the probability that C is between A and B is also $frac{1}{3}$
But on the other hand we have:
Argument 2
Whatever A and B are, their difference is finite, and hence any third random number has a $0$ probability of being between them.
... so ... maybe this is a reductio ad absurdum against the very assumption that we can randomly pick numbers from all real numbers with equal likelihood?
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
$endgroup$
$begingroup$
Yes, randomly choosing a real number is not well defined. A uniform distribution requires a finite width. There do exist probability functions on the reals, so if you specify one of those then your question will be answerable.
$endgroup$
– dbx
Jan 28 at 2:40
$begingroup$
There are infinitely many pairs A and B to be integrated over.
$endgroup$
– WW1
Jan 28 at 3:27
add a comment |
$begingroup$
I see your dilemma. On the one hand we have:
Argument 1
It does not matter whether you pick all three numbers at once or one at a time. For example, suppose we pick three numbers A,B,D all at once, but reveal only two of A and B. And, also suppose that after we have revealed A and B, we pick a new number C.
Now, it makes little sense to think that the probability of D being between A and B would be any different than the probability of C being between A and B: the only difference is that one number was picked before the reveal, and the other after the reveal, and of course the picking of a number is not going to be affected by the reveal.
Also, as the first question makes clear, the probability that D is between A and B is $frac{1}{3}$ .. so therefore the probability that C is between A and B is also $frac{1}{3}$
But on the other hand we have:
Argument 2
Whatever A and B are, their difference is finite, and hence any third random number has a $0$ probability of being between them.
... so ... maybe this is a reductio ad absurdum against the very assumption that we can randomly pick numbers from all real numbers with equal likelihood?
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
$endgroup$
I see your dilemma. On the one hand we have:
Argument 1
It does not matter whether you pick all three numbers at once or one at a time. For example, suppose we pick three numbers A,B,D all at once, but reveal only two of A and B. And, also suppose that after we have revealed A and B, we pick a new number C.
Now, it makes little sense to think that the probability of D being between A and B would be any different than the probability of C being between A and B: the only difference is that one number was picked before the reveal, and the other after the reveal, and of course the picking of a number is not going to be affected by the reveal.
Also, as the first question makes clear, the probability that D is between A and B is $frac{1}{3}$ .. so therefore the probability that C is between A and B is also $frac{1}{3}$
But on the other hand we have:
Argument 2
Whatever A and B are, their difference is finite, and hence any third random number has a $0$ probability of being between them.
... so ... maybe this is a reductio ad absurdum against the very assumption that we can randomly pick numbers from all real numbers with equal likelihood?
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
edited Jan 28 at 2:00
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
answered Jan 28 at 1:44
Bram28Bram28
63.9k44793
63.9k44793
$begingroup$
Yes, randomly choosing a real number is not well defined. A uniform distribution requires a finite width. There do exist probability functions on the reals, so if you specify one of those then your question will be answerable.
$endgroup$
– dbx
Jan 28 at 2:40
$begingroup$
There are infinitely many pairs A and B to be integrated over.
$endgroup$
– WW1
Jan 28 at 3:27
add a comment |
$begingroup$
Yes, randomly choosing a real number is not well defined. A uniform distribution requires a finite width. There do exist probability functions on the reals, so if you specify one of those then your question will be answerable.
$endgroup$
– dbx
Jan 28 at 2:40
$begingroup$
There are infinitely many pairs A and B to be integrated over.
$endgroup$
– WW1
Jan 28 at 3:27
$begingroup$
Yes, randomly choosing a real number is not well defined. A uniform distribution requires a finite width. There do exist probability functions on the reals, so if you specify one of those then your question will be answerable.
$endgroup$
– dbx
Jan 28 at 2:40
$begingroup$
Yes, randomly choosing a real number is not well defined. A uniform distribution requires a finite width. There do exist probability functions on the reals, so if you specify one of those then your question will be answerable.
$endgroup$
– dbx
Jan 28 at 2:40
$begingroup$
There are infinitely many pairs A and B to be integrated over.
$endgroup$
– WW1
Jan 28 at 3:27
$begingroup$
There are infinitely many pairs A and B to be integrated over.
$endgroup$
– WW1
Jan 28 at 3:27
add a comment |
$begingroup$
You can not choose a real number at random with uniform probabilities. However, there are various probability distributions you can choose from. Assuming you choose one of these PDFs, then for question one, all the possible orderings are equally probable because you are choosing them independently or all at the same time. For the second question the probability of A < C < B completely depends on the PDF.
answered Jan 28 at 2:53
SomosSomos
14.7k11337
$endgroup$
add a comment |
$begingroup$
You can not choose a real number at random with uniform probabilities. However, there are various probability distributions you can choose from. Assuming you choose one of these PDFs, then for question one, all the possible orderings are equally probable because you are choosing them independently or all at the same time. For the second question the probability of A < C < B completely depends on the PDF.
answered Jan 28 at 2:53
SomosSomos
14.7k11337
$endgroup$
add a comment |
$begingroup$
You can not choose a real number at random with uniform probabilities. However, there are various probability distributions you can choose from. Assuming you choose one of these PDFs, then for question one, all the possible orderings are equally probable because you are choosing them independently or all at the same time. For the second question the probability of A < C < B completely depends on the PDF.
answered Jan 28 at 2:53
SomosSomos
14.7k11337
$endgroup$
You can not choose a real number at random with uniform probabilities. However, there are various probability distributions you can choose from. Assuming you choose one of these PDFs, then for question one, all the possible orderings are equally probable because you are choosing them independently or all at the same time. For the second question the probability of A < C < B completely depends on the PDF.
answered Jan 28 at 2:53
SomosSomos
14.7k11337
answered Jan 28 at 2:53
SomosSomos
14.7k11337
answered Jan 28 at 2:53
SomosSomos
14.7k11337
answered Jan 28 at 2:53
SomosSomos
14.7k11337
14.7k11337
add a comment |
add a comment |
$begingroup$
Let A, B and C be uniformly distributed on $(-N,N)$
Then each has pdf = $frac 1{2N}$
The probability $A<C<B$ is given by
$$ begin{eqnarray} P = & &frac 1 {(2N)^3}int_{-N}^Nint_A^Nint_A^B dC ; dB ; dA
\=&&frac 1 {(2N)^3}int_{-N}^N int_A^N (B-A); dB ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N ( frac 12 N^2-AN)-(frac 12 A^2-A^2 ) ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N (frac 12 A^2-AN+ frac 12 N^2 ) ; dA
\= & &frac 1 {(2N)^3} (frac 13 N^3 +N^3) \ = & &frac 16 end{eqnarray} $$
This result is independent of $N$ so you would expect it to apply in the limit $N to infty$
answered Jan 28 at 3:51
WW1WW1
7,3501712
$endgroup$
add a comment |
$begingroup$
Let A, B and C be uniformly distributed on $(-N,N)$
Then each has pdf = $frac 1{2N}$
The probability $A<C<B$ is given by
$$ begin{eqnarray} P = & &frac 1 {(2N)^3}int_{-N}^Nint_A^Nint_A^B dC ; dB ; dA
\=&&frac 1 {(2N)^3}int_{-N}^N int_A^N (B-A); dB ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N ( frac 12 N^2-AN)-(frac 12 A^2-A^2 ) ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N (frac 12 A^2-AN+ frac 12 N^2 ) ; dA
\= & &frac 1 {(2N)^3} (frac 13 N^3 +N^3) \ = & &frac 16 end{eqnarray} $$
This result is independent of $N$ so you would expect it to apply in the limit $N to infty$
answered Jan 28 at 3:51
WW1WW1
7,3501712
$endgroup$
add a comment |
$begingroup$
Let A, B and C be uniformly distributed on $(-N,N)$
Then each has pdf = $frac 1{2N}$
The probability $A<C<B$ is given by
$$ begin{eqnarray} P = & &frac 1 {(2N)^3}int_{-N}^Nint_A^Nint_A^B dC ; dB ; dA
\=&&frac 1 {(2N)^3}int_{-N}^N int_A^N (B-A); dB ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N ( frac 12 N^2-AN)-(frac 12 A^2-A^2 ) ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N (frac 12 A^2-AN+ frac 12 N^2 ) ; dA
\= & &frac 1 {(2N)^3} (frac 13 N^3 +N^3) \ = & &frac 16 end{eqnarray} $$
This result is independent of $N$ so you would expect it to apply in the limit $N to infty$
answered Jan 28 at 3:51
WW1WW1
7,3501712
$endgroup$
Let A, B and C be uniformly distributed on $(-N,N)$
Then each has pdf = $frac 1{2N}$
The probability $A<C<B$ is given by
$$ begin{eqnarray} P = & &frac 1 {(2N)^3}int_{-N}^Nint_A^Nint_A^B dC ; dB ; dA
\=&&frac 1 {(2N)^3}int_{-N}^N int_A^N (B-A); dB ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N ( frac 12 N^2-AN)-(frac 12 A^2-A^2 ) ; dA
\=& & frac 1 {(2N)^3}int_{-N}^N (frac 12 A^2-AN+ frac 12 N^2 ) ; dA
\= & &frac 1 {(2N)^3} (frac 13 N^3 +N^3) \ = & &frac 16 end{eqnarray} $$
This result is independent of $N$ so you would expect it to apply in the limit $N to infty$
answered Jan 28 at 3:51
WW1WW1
7,3501712
edited Jan 28 at 4:04
answered Jan 28 at 3:51
WW1WW1
7,3501712
answered Jan 28 at 3:51
WW1WW1
7,3501712
answered Jan 28 at 3:51
WW1WW1
7,3501712
7,3501712
add a comment |
add a comment |
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$begingroup$
What are your thoughts on this?
$endgroup$
– Bram28
Jan 28 at 1:23
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It seems the answer to the first question is yes and the answer to the second question is 0 since there is only a finite distance between A and B.
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– user637421
Jan 28 at 1:24
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@michael35 Do you think it matters for the order the 3 numbers will be in whether we pick the 3 numbers at the same time or one after the other? Or whether we pick all 3 at once but reveal them one at a time?
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– Bram28
Jan 28 at 1:27
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That's basically my question.
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– user637421
Jan 28 at 1:28
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This question doesn't make sense. You can't choose real numbers uniformly at random.
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– Matt Samuel
Jan 28 at 1:40