Two aspects of randomness
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0
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Consider a random sequence of integers
1, 4, 3, 8, 2, 5, 3, 8 ...
The only sufficient condition for the sequence to be random is its unpredictability ie. probability of any number coming next must be equal to $frac{1}{10}$.
Now consider that we are getting only numbers less than 5 in the sequence, it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more, this does not follow the randomness criteria as numbers are now in some form more predictable.
Do the two aspects of randomness contradict with each other?
Or am I wrong somewhere in this deductive thinking?
probability sequences-and-series statistical-inference random
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
asked Nov 16 at 8:07
mathaholic
492
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mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
Consider a random sequence of integers
1, 4, 3, 8, 2, 5, 3, 8 ...
The only sufficient condition for the sequence to be random is its unpredictability ie. probability of any number coming next must be equal to $frac{1}{10}$.
Now consider that we are getting only numbers less than 5 in the sequence, it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more, this does not follow the randomness criteria as numbers are now in some form more predictable.
Do the two aspects of randomness contradict with each other?
Or am I wrong somewhere in this deductive thinking?
probability sequences-and-series statistical-inference random
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
asked Nov 16 at 8:07
mathaholic
492
New contributor
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
So, when you write, "integer", what you actually mean is "digit"?
– Gerry Myerson
Nov 16 at 8:20
Your first sentence is contradictory. The second part refers to equiprobability, not randomness.
– Yves Daoust
Nov 18 at 16:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a random sequence of integers
1, 4, 3, 8, 2, 5, 3, 8 ...
The only sufficient condition for the sequence to be random is its unpredictability ie. probability of any number coming next must be equal to $frac{1}{10}$.
Now consider that we are getting only numbers less than 5 in the sequence, it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more, this does not follow the randomness criteria as numbers are now in some form more predictable.
Do the two aspects of randomness contradict with each other?
Or am I wrong somewhere in this deductive thinking?
probability sequences-and-series statistical-inference random
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
asked Nov 16 at 8:07
mathaholic
492
New contributor
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider a random sequence of integers
1, 4, 3, 8, 2, 5, 3, 8 ...
The only sufficient condition for the sequence to be random is its unpredictability ie. probability of any number coming next must be equal to $frac{1}{10}$.
Now consider that we are getting only numbers less than 5 in the sequence, it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more, this does not follow the randomness criteria as numbers are now in some form more predictable.
Do the two aspects of randomness contradict with each other?
Or am I wrong somewhere in this deductive thinking?
probability sequences-and-series statistical-inference random
probability sequences-and-series statistical-inference random
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
asked Nov 16 at 8:07
mathaholic
492
New contributor
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
asked Nov 16 at 8:07
mathaholic
492
New contributor
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
edited Nov 18 at 16:33
Mauro ALLEGRANZA
63.4k448110
63.4k448110
asked Nov 16 at 8:07
mathaholic
492
New contributor
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 8:07
mathaholic
492
asked Nov 16 at 8:07
mathaholic
492
492
New contributor
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mathaholic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
So, when you write, "integer", what you actually mean is "digit"?
– Gerry Myerson
Nov 16 at 8:20
Your first sentence is contradictory. The second part refers to equiprobability, not randomness.
– Yves Daoust
Nov 18 at 16:50
add a comment |
2
So, when you write, "integer", what you actually mean is "digit"?
– Gerry Myerson
Nov 16 at 8:20
Your first sentence is contradictory. The second part refers to equiprobability, not randomness.
– Yves Daoust
Nov 18 at 16:50
2
2
So, when you write, "integer", what you actually mean is "digit"?
– Gerry Myerson
Nov 16 at 8:20
So, when you write, "integer", what you actually mean is "digit"?
– Gerry Myerson
Nov 16 at 8:20
Your first sentence is contradictory. The second part refers to equiprobability, not randomness.
– Yves Daoust
Nov 18 at 16:50
Your first sentence is contradictory. The second part refers to equiprobability, not randomness.
– Yves Daoust
Nov 18 at 16:50
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
Welcome to MSE,
it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more.
This is not true. If every choice of digits is independent, there is no change in the probabilities for the next digit of the sequence.
You can take a look at this question, which is somehow close to yours.
Does the probability change if you know previous results?
If that's not what you are asking please provide us with more details.
answered Nov 16 at 8:27
Gâteau-Gallois
35819
isn't it expected from a series of coin tosses half of them must be head and half must be tails,ie. if i toss coin 1000 times i expect 500 heads and 500 tails, if after 800 tossed i have 700 heads doesn't it mean that to maintain 1:1 ratio tails are more expected, atleast mathematically.
– mathaholic
Nov 17 at 7:04
Precisely not. The tosses are independent. Consider the case with 3 tosses. You have $3^2 = 9$ possible outcomes (if you take care of the order of the tosses). Each outcome is obtained with probability $frac{1}{9}$. If you already had head twice, it means that you are either in the situation HHH or in the situation HHT. But both of those situations have the same probability to be obtained, hence you have the same probability to obtain H or T at the last toss.
– Gâteau-Gallois
Nov 17 at 17:41
then why do we get 50:50 heads and tails in a large sequence of tosses
– mathaholic
Nov 18 at 6:24
You're mixing two different things. If you have no previous informations about the sequence, and toss a coin $10^9$ times, you expect the number of heads to be close to the number of tails. This is basically the Law of Large Numbers. On the other hand, what the law of Large numbers say is just that it's very unlikely (but not impossible) to have for instance $10^9$ heads and 0 tails.
– Gâteau-Gallois
Nov 18 at 13:28
Now if you toss a coin 800 times and get 800 heads, you are just in one of those ``less likely'' scenarii. And you might end up in the scenario with 800-200 or with 1000-0. Both are actually equally likely if you already know that the first 800 tosses have given heads. And they are both part of those unlikely scenarii in the law of large numbers for which the ratio is not close to 1:1.
– Gâteau-Gallois
Nov 18 at 13:32
add a comment |
up vote
0
down vote
You confuse probability of the next drawing, which is always $dfrac1{10}$ (provided the distribution is uniform and the samples are independent), and the probability of certain sequences.
For instance, the probabilities of drawing $1111111111$ or $1234567890$ or $5369574581$ are all $10^{-10}$, i.e. they are extremely unlikely events.
Now you can consider other events, such as all digits appearing exactly once among $10$ drawings; this is $10!cdot 10^{-10}=0.00036288$ because you can permute $1234567890$ in $10!$ ways.
Or having as many digits below $6$ than above $5$ among $10$ drawings, which is $displaystylebinom{10}5cdot2^{-10}=0.24609375$.
And so on.
Randomness is indeed unpredictability. It does not require the outcomes to be equiprobable, nor independent of each other.
answered Nov 18 at 16:58
Yves Daoust
121k668216
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Welcome to MSE,
it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more.
This is not true. If every choice of digits is independent, there is no change in the probabilities for the next digit of the sequence.
You can take a look at this question, which is somehow close to yours.
Does the probability change if you know previous results?
If that's not what you are asking please provide us with more details.
answered Nov 16 at 8:27
Gâteau-Gallois
35819
isn't it expected from a series of coin tosses half of them must be head and half must be tails,ie. if i toss coin 1000 times i expect 500 heads and 500 tails, if after 800 tossed i have 700 heads doesn't it mean that to maintain 1:1 ratio tails are more expected, atleast mathematically.
– mathaholic
Nov 17 at 7:04
Precisely not. The tosses are independent. Consider the case with 3 tosses. You have $3^2 = 9$ possible outcomes (if you take care of the order of the tosses). Each outcome is obtained with probability $frac{1}{9}$. If you already had head twice, it means that you are either in the situation HHH or in the situation HHT. But both of those situations have the same probability to be obtained, hence you have the same probability to obtain H or T at the last toss.
– Gâteau-Gallois
Nov 17 at 17:41
then why do we get 50:50 heads and tails in a large sequence of tosses
– mathaholic
Nov 18 at 6:24
You're mixing two different things. If you have no previous informations about the sequence, and toss a coin $10^9$ times, you expect the number of heads to be close to the number of tails. This is basically the Law of Large Numbers. On the other hand, what the law of Large numbers say is just that it's very unlikely (but not impossible) to have for instance $10^9$ heads and 0 tails.
– Gâteau-Gallois
Nov 18 at 13:28
Now if you toss a coin 800 times and get 800 heads, you are just in one of those ``less likely'' scenarii. And you might end up in the scenario with 800-200 or with 1000-0. Both are actually equally likely if you already know that the first 800 tosses have given heads. And they are both part of those unlikely scenarii in the law of large numbers for which the ratio is not close to 1:1.
– Gâteau-Gallois
Nov 18 at 13:32
add a comment |
up vote
3
down vote
Welcome to MSE,
it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more.
This is not true. If every choice of digits is independent, there is no change in the probabilities for the next digit of the sequence.
You can take a look at this question, which is somehow close to yours.
Does the probability change if you know previous results?
If that's not what you are asking please provide us with more details.
answered Nov 16 at 8:27
Gâteau-Gallois
35819
isn't it expected from a series of coin tosses half of them must be head and half must be tails,ie. if i toss coin 1000 times i expect 500 heads and 500 tails, if after 800 tossed i have 700 heads doesn't it mean that to maintain 1:1 ratio tails are more expected, atleast mathematically.
– mathaholic
Nov 17 at 7:04
Precisely not. The tosses are independent. Consider the case with 3 tosses. You have $3^2 = 9$ possible outcomes (if you take care of the order of the tosses). Each outcome is obtained with probability $frac{1}{9}$. If you already had head twice, it means that you are either in the situation HHH or in the situation HHT. But both of those situations have the same probability to be obtained, hence you have the same probability to obtain H or T at the last toss.
– Gâteau-Gallois
Nov 17 at 17:41
then why do we get 50:50 heads and tails in a large sequence of tosses
– mathaholic
Nov 18 at 6:24
You're mixing two different things. If you have no previous informations about the sequence, and toss a coin $10^9$ times, you expect the number of heads to be close to the number of tails. This is basically the Law of Large Numbers. On the other hand, what the law of Large numbers say is just that it's very unlikely (but not impossible) to have for instance $10^9$ heads and 0 tails.
– Gâteau-Gallois
Nov 18 at 13:28
Now if you toss a coin 800 times and get 800 heads, you are just in one of those ``less likely'' scenarii. And you might end up in the scenario with 800-200 or with 1000-0. Both are actually equally likely if you already know that the first 800 tosses have given heads. And they are both part of those unlikely scenarii in the law of large numbers for which the ratio is not close to 1:1.
– Gâteau-Gallois
Nov 18 at 13:32
add a comment |
up vote
3
down vote
up vote
3
down vote
Welcome to MSE,
it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more.
This is not true. If every choice of digits is independent, there is no change in the probabilities for the next digit of the sequence.
You can take a look at this question, which is somehow close to yours.
Does the probability change if you know previous results?
If that's not what you are asking please provide us with more details.
answered Nov 16 at 8:27
Gâteau-Gallois
35819
Welcome to MSE,
it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more.
This is not true. If every choice of digits is independent, there is no change in the probabilities for the next digit of the sequence.
You can take a look at this question, which is somehow close to yours.
Does the probability change if you know previous results?
If that's not what you are asking please provide us with more details.
answered Nov 16 at 8:27
Gâteau-Gallois
35819
answered Nov 16 at 8:27
Gâteau-Gallois
35819
answered Nov 16 at 8:27
Gâteau-Gallois
35819
answered Nov 16 at 8:27
Gâteau-Gallois
35819
35819
isn't it expected from a series of coin tosses half of them must be head and half must be tails,ie. if i toss coin 1000 times i expect 500 heads and 500 tails, if after 800 tossed i have 700 heads doesn't it mean that to maintain 1:1 ratio tails are more expected, atleast mathematically.
– mathaholic
Nov 17 at 7:04
Precisely not. The tosses are independent. Consider the case with 3 tosses. You have $3^2 = 9$ possible outcomes (if you take care of the order of the tosses). Each outcome is obtained with probability $frac{1}{9}$. If you already had head twice, it means that you are either in the situation HHH or in the situation HHT. But both of those situations have the same probability to be obtained, hence you have the same probability to obtain H or T at the last toss.
– Gâteau-Gallois
Nov 17 at 17:41
then why do we get 50:50 heads and tails in a large sequence of tosses
– mathaholic
Nov 18 at 6:24
You're mixing two different things. If you have no previous informations about the sequence, and toss a coin $10^9$ times, you expect the number of heads to be close to the number of tails. This is basically the Law of Large Numbers. On the other hand, what the law of Large numbers say is just that it's very unlikely (but not impossible) to have for instance $10^9$ heads and 0 tails.
– Gâteau-Gallois
Nov 18 at 13:28
Now if you toss a coin 800 times and get 800 heads, you are just in one of those ``less likely'' scenarii. And you might end up in the scenario with 800-200 or with 1000-0. Both are actually equally likely if you already know that the first 800 tosses have given heads. And they are both part of those unlikely scenarii in the law of large numbers for which the ratio is not close to 1:1.
– Gâteau-Gallois
Nov 18 at 13:32
add a comment |
isn't it expected from a series of coin tosses half of them must be head and half must be tails,ie. if i toss coin 1000 times i expect 500 heads and 500 tails, if after 800 tossed i have 700 heads doesn't it mean that to maintain 1:1 ratio tails are more expected, atleast mathematically.
– mathaholic
Nov 17 at 7:04
Precisely not. The tosses are independent. Consider the case with 3 tosses. You have $3^2 = 9$ possible outcomes (if you take care of the order of the tosses). Each outcome is obtained with probability $frac{1}{9}$. If you already had head twice, it means that you are either in the situation HHH or in the situation HHT. But both of those situations have the same probability to be obtained, hence you have the same probability to obtain H or T at the last toss.
– Gâteau-Gallois
Nov 17 at 17:41
then why do we get 50:50 heads and tails in a large sequence of tosses
– mathaholic
Nov 18 at 6:24
You're mixing two different things. If you have no previous informations about the sequence, and toss a coin $10^9$ times, you expect the number of heads to be close to the number of tails. This is basically the Law of Large Numbers. On the other hand, what the law of Large numbers say is just that it's very unlikely (but not impossible) to have for instance $10^9$ heads and 0 tails.
– Gâteau-Gallois
Nov 18 at 13:28
Now if you toss a coin 800 times and get 800 heads, you are just in one of those ``less likely'' scenarii. And you might end up in the scenario with 800-200 or with 1000-0. Both are actually equally likely if you already know that the first 800 tosses have given heads. And they are both part of those unlikely scenarii in the law of large numbers for which the ratio is not close to 1:1.
– Gâteau-Gallois
Nov 18 at 13:32
isn't it expected from a series of coin tosses half of them must be head and half must be tails,ie. if i toss coin 1000 times i expect 500 heads and 500 tails, if after 800 tossed i have 700 heads doesn't it mean that to maintain 1:1 ratio tails are more expected, atleast mathematically.
– mathaholic
Nov 17 at 7:04
isn't it expected from a series of coin tosses half of them must be head and half must be tails,ie. if i toss coin 1000 times i expect 500 heads and 500 tails, if after 800 tossed i have 700 heads doesn't it mean that to maintain 1:1 ratio tails are more expected, atleast mathematically.
– mathaholic
Nov 17 at 7:04
Precisely not. The tosses are independent. Consider the case with 3 tosses. You have $3^2 = 9$ possible outcomes (if you take care of the order of the tosses). Each outcome is obtained with probability $frac{1}{9}$. If you already had head twice, it means that you are either in the situation HHH or in the situation HHT. But both of those situations have the same probability to be obtained, hence you have the same probability to obtain H or T at the last toss.
– Gâteau-Gallois
Nov 17 at 17:41
Precisely not. The tosses are independent. Consider the case with 3 tosses. You have $3^2 = 9$ possible outcomes (if you take care of the order of the tosses). Each outcome is obtained with probability $frac{1}{9}$. If you already had head twice, it means that you are either in the situation HHH or in the situation HHT. But both of those situations have the same probability to be obtained, hence you have the same probability to obtain H or T at the last toss.
– Gâteau-Gallois
Nov 17 at 17:41
then why do we get 50:50 heads and tails in a large sequence of tosses
– mathaholic
Nov 18 at 6:24
then why do we get 50:50 heads and tails in a large sequence of tosses
– mathaholic
Nov 18 at 6:24
You're mixing two different things. If you have no previous informations about the sequence, and toss a coin $10^9$ times, you expect the number of heads to be close to the number of tails. This is basically the Law of Large Numbers. On the other hand, what the law of Large numbers say is just that it's very unlikely (but not impossible) to have for instance $10^9$ heads and 0 tails.
– Gâteau-Gallois
Nov 18 at 13:28
You're mixing two different things. If you have no previous informations about the sequence, and toss a coin $10^9$ times, you expect the number of heads to be close to the number of tails. This is basically the Law of Large Numbers. On the other hand, what the law of Large numbers say is just that it's very unlikely (but not impossible) to have for instance $10^9$ heads and 0 tails.
– Gâteau-Gallois
Nov 18 at 13:28
Now if you toss a coin 800 times and get 800 heads, you are just in one of those ``less likely'' scenarii. And you might end up in the scenario with 800-200 or with 1000-0. Both are actually equally likely if you already know that the first 800 tosses have given heads. And they are both part of those unlikely scenarii in the law of large numbers for which the ratio is not close to 1:1.
– Gâteau-Gallois
Nov 18 at 13:32
Now if you toss a coin 800 times and get 800 heads, you are just in one of those ``less likely'' scenarii. And you might end up in the scenario with 800-200 or with 1000-0. Both are actually equally likely if you already know that the first 800 tosses have given heads. And they are both part of those unlikely scenarii in the law of large numbers for which the ratio is not close to 1:1.
– Gâteau-Gallois
Nov 18 at 13:32
add a comment |
up vote
0
down vote
You confuse probability of the next drawing, which is always $dfrac1{10}$ (provided the distribution is uniform and the samples are independent), and the probability of certain sequences.
For instance, the probabilities of drawing $1111111111$ or $1234567890$ or $5369574581$ are all $10^{-10}$, i.e. they are extremely unlikely events.
Now you can consider other events, such as all digits appearing exactly once among $10$ drawings; this is $10!cdot 10^{-10}=0.00036288$ because you can permute $1234567890$ in $10!$ ways.
Or having as many digits below $6$ than above $5$ among $10$ drawings, which is $displaystylebinom{10}5cdot2^{-10}=0.24609375$.
And so on.
Randomness is indeed unpredictability. It does not require the outcomes to be equiprobable, nor independent of each other.
answered Nov 18 at 16:58
Yves Daoust
121k668216
add a comment |
up vote
0
down vote
You confuse probability of the next drawing, which is always $dfrac1{10}$ (provided the distribution is uniform and the samples are independent), and the probability of certain sequences.
For instance, the probabilities of drawing $1111111111$ or $1234567890$ or $5369574581$ are all $10^{-10}$, i.e. they are extremely unlikely events.
Now you can consider other events, such as all digits appearing exactly once among $10$ drawings; this is $10!cdot 10^{-10}=0.00036288$ because you can permute $1234567890$ in $10!$ ways.
Or having as many digits below $6$ than above $5$ among $10$ drawings, which is $displaystylebinom{10}5cdot2^{-10}=0.24609375$.
And so on.
Randomness is indeed unpredictability. It does not require the outcomes to be equiprobable, nor independent of each other.
answered Nov 18 at 16:58
Yves Daoust
121k668216
add a comment |
up vote
0
down vote
up vote
0
down vote
You confuse probability of the next drawing, which is always $dfrac1{10}$ (provided the distribution is uniform and the samples are independent), and the probability of certain sequences.
For instance, the probabilities of drawing $1111111111$ or $1234567890$ or $5369574581$ are all $10^{-10}$, i.e. they are extremely unlikely events.
Now you can consider other events, such as all digits appearing exactly once among $10$ drawings; this is $10!cdot 10^{-10}=0.00036288$ because you can permute $1234567890$ in $10!$ ways.
Or having as many digits below $6$ than above $5$ among $10$ drawings, which is $displaystylebinom{10}5cdot2^{-10}=0.24609375$.
And so on.
Randomness is indeed unpredictability. It does not require the outcomes to be equiprobable, nor independent of each other.
answered Nov 18 at 16:58
Yves Daoust
121k668216
You confuse probability of the next drawing, which is always $dfrac1{10}$ (provided the distribution is uniform and the samples are independent), and the probability of certain sequences.
For instance, the probabilities of drawing $1111111111$ or $1234567890$ or $5369574581$ are all $10^{-10}$, i.e. they are extremely unlikely events.
Now you can consider other events, such as all digits appearing exactly once among $10$ drawings; this is $10!cdot 10^{-10}=0.00036288$ because you can permute $1234567890$ in $10!$ ways.
Or having as many digits below $6$ than above $5$ among $10$ drawings, which is $displaystylebinom{10}5cdot2^{-10}=0.24609375$.
And so on.
Randomness is indeed unpredictability. It does not require the outcomes to be equiprobable, nor independent of each other.
answered Nov 18 at 16:58
Yves Daoust
121k668216
edited Nov 18 at 17:16
answered Nov 18 at 16:58
Yves Daoust
121k668216
answered Nov 18 at 16:58
Yves Daoust
121k668216
answered Nov 18 at 16:58
Yves Daoust
121k668216
121k668216
add a comment |
add a comment |
mathaholic is a new contributor. Be nice, and check out our Code of Conduct.
mathaholic is a new contributor. Be nice, and check out our Code of Conduct.
mathaholic is a new contributor. Be nice, and check out our Code of Conduct.
mathaholic is a new contributor. Be nice, and check out our Code of Conduct.
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2
So, when you write, "integer", what you actually mean is "digit"?
– Gerry Myerson
Nov 16 at 8:20
Your first sentence is contradictory. The second part refers to equiprobability, not randomness.
– Yves Daoust
Nov 18 at 16:50