Mixing
Calculation of probability with arithmetic mean of random variables for a total of 3 times
Thanks to @Ben W for his answer to my previous question, now I can calculate and have 3 equal probabilities:
$P(X_1) = P(X_2) = P(X_3) approx 0.000148646896$
with $X_1 = X_2 = X_3 = 1620$
Following up, once that the 4 people got 405 as the arithmetic mean of the number on their cards, then they repeat the drawing 2 more time (3 in total). So, I would like to calculate the probability that the arithmetic mean of the number on their cards is 405 for a total of 3 times.
How to make that?
Some explanation is welcome.
probability statistics
asked Nov 20 '18 at 2:19
Backo
1184
|
show 10 more comments
Thanks to @Ben W for his answer to my previous question, now I can calculate and have 3 equal probabilities:
$P(X_1) = P(X_2) = P(X_3) approx 0.000148646896$
with $X_1 = X_2 = X_3 = 1620$
Following up, once that the 4 people got 405 as the arithmetic mean of the number on their cards, then they repeat the drawing 2 more time (3 in total). So, I would like to calculate the probability that the arithmetic mean of the number on their cards is 405 for a total of 3 times.
How to make that?
Some explanation is welcome.
probability statistics
asked Nov 20 '18 at 2:19
Backo
1184
1
"I would like to calculate the probability of the event that 'the above 3 probabilities are the same'" Uh.... I don't understand your question. If you give as a hypothesis that $P(X_1)=P(X_2)=P(X_3)$... then the probability that they are equal is $1$... since you already told us that they are equal. If things are equal then they are equal. A tautology is a tautology...
– JMoravitz
Nov 20 '18 at 2:22
1
I think you misunderstood me. Let me see if I can explain better. Can you describe the physical situation? Are there now twelve players, in three groups of four? What event are you asking to compute the probability of? Etc.
– Ben W
Nov 20 '18 at 2:32
1
You can answer your own question.
– Ben W
Nov 20 '18 at 2:48
1
There is a big difference between asking for the probability that $Pr(X_1=405)=Pr(X_2=405)=Pr(X_3=405)$ and asking for the probability $Pr(X_1=X_2=X_3=405)$.
– JMoravitz
Nov 20 '18 at 2:55
2
Thank you but I don't really keep track nor care about my rep points. I'm in it for the love of the game ; )
– Ben W
Nov 20 '18 at 2:58
|
show 10 more comments
Thanks to @Ben W for his answer to my previous question, now I can calculate and have 3 equal probabilities:
$P(X_1) = P(X_2) = P(X_3) approx 0.000148646896$
with $X_1 = X_2 = X_3 = 1620$
Following up, once that the 4 people got 405 as the arithmetic mean of the number on their cards, then they repeat the drawing 2 more time (3 in total). So, I would like to calculate the probability that the arithmetic mean of the number on their cards is 405 for a total of 3 times.
How to make that?
Some explanation is welcome.
probability statistics
asked Nov 20 '18 at 2:19
Backo
1184
Thanks to @Ben W for his answer to my previous question, now I can calculate and have 3 equal probabilities:
$P(X_1) = P(X_2) = P(X_3) approx 0.000148646896$
with $X_1 = X_2 = X_3 = 1620$
Following up, once that the 4 people got 405 as the arithmetic mean of the number on their cards, then they repeat the drawing 2 more time (3 in total). So, I would like to calculate the probability that the arithmetic mean of the number on their cards is 405 for a total of 3 times.
How to make that?
Some explanation is welcome.
probability statistics
probability statistics
asked Nov 20 '18 at 2:19
Backo
1184
asked Nov 20 '18 at 2:19
Backo
1184
edited Nov 20 '18 at 3:25
asked Nov 20 '18 at 2:19
Backo
1184
asked Nov 20 '18 at 2:19
Backo
1184
asked Nov 20 '18 at 2:19
Backo
1184
1184
1
"I would like to calculate the probability of the event that 'the above 3 probabilities are the same'" Uh.... I don't understand your question. If you give as a hypothesis that $P(X_1)=P(X_2)=P(X_3)$... then the probability that they are equal is $1$... since you already told us that they are equal. If things are equal then they are equal. A tautology is a tautology...
– JMoravitz
Nov 20 '18 at 2:22
1
I think you misunderstood me. Let me see if I can explain better. Can you describe the physical situation? Are there now twelve players, in three groups of four? What event are you asking to compute the probability of? Etc.
– Ben W
Nov 20 '18 at 2:32
1
You can answer your own question.
– Ben W
Nov 20 '18 at 2:48
1
There is a big difference between asking for the probability that $Pr(X_1=405)=Pr(X_2=405)=Pr(X_3=405)$ and asking for the probability $Pr(X_1=X_2=X_3=405)$.
– JMoravitz
Nov 20 '18 at 2:55
2
Thank you but I don't really keep track nor care about my rep points. I'm in it for the love of the game ; )
– Ben W
Nov 20 '18 at 2:58
|
show 10 more comments
1
"I would like to calculate the probability of the event that 'the above 3 probabilities are the same'" Uh.... I don't understand your question. If you give as a hypothesis that $P(X_1)=P(X_2)=P(X_3)$... then the probability that they are equal is $1$... since you already told us that they are equal. If things are equal then they are equal. A tautology is a tautology...
– JMoravitz
Nov 20 '18 at 2:22
1
I think you misunderstood me. Let me see if I can explain better. Can you describe the physical situation? Are there now twelve players, in three groups of four? What event are you asking to compute the probability of? Etc.
– Ben W
Nov 20 '18 at 2:32
1
You can answer your own question.
– Ben W
Nov 20 '18 at 2:48
1
There is a big difference between asking for the probability that $Pr(X_1=405)=Pr(X_2=405)=Pr(X_3=405)$ and asking for the probability $Pr(X_1=X_2=X_3=405)$.
– JMoravitz
Nov 20 '18 at 2:55
2
Thank you but I don't really keep track nor care about my rep points. I'm in it for the love of the game ; )
– Ben W
Nov 20 '18 at 2:58
1
1
"I would like to calculate the probability of the event that 'the above 3 probabilities are the same'" Uh.... I don't understand your question. If you give as a hypothesis that $P(X_1)=P(X_2)=P(X_3)$... then the probability that they are equal is $1$... since you already told us that they are equal. If things are equal then they are equal. A tautology is a tautology...
– JMoravitz
Nov 20 '18 at 2:22
"I would like to calculate the probability of the event that 'the above 3 probabilities are the same'" Uh.... I don't understand your question. If you give as a hypothesis that $P(X_1)=P(X_2)=P(X_3)$... then the probability that they are equal is $1$... since you already told us that they are equal. If things are equal then they are equal. A tautology is a tautology...
– JMoravitz
Nov 20 '18 at 2:22
1
1
I think you misunderstood me. Let me see if I can explain better. Can you describe the physical situation? Are there now twelve players, in three groups of four? What event are you asking to compute the probability of? Etc.
– Ben W
Nov 20 '18 at 2:32
I think you misunderstood me. Let me see if I can explain better. Can you describe the physical situation? Are there now twelve players, in three groups of four? What event are you asking to compute the probability of? Etc.
– Ben W
Nov 20 '18 at 2:32
1
1
You can answer your own question.
– Ben W
Nov 20 '18 at 2:48
You can answer your own question.
– Ben W
Nov 20 '18 at 2:48
1
1
There is a big difference between asking for the probability that $Pr(X_1=405)=Pr(X_2=405)=Pr(X_3=405)$ and asking for the probability $Pr(X_1=X_2=X_3=405)$.
– JMoravitz
Nov 20 '18 at 2:55
There is a big difference between asking for the probability that $Pr(X_1=405)=Pr(X_2=405)=Pr(X_3=405)$ and asking for the probability $Pr(X_1=X_2=X_3=405)$.
– JMoravitz
Nov 20 '18 at 2:55
2
2
Thank you but I don't really keep track nor care about my rep points. I'm in it for the love of the game ; )
– Ben W
Nov 20 '18 at 2:58
Thank you but I don't really keep track nor care about my rep points. I'm in it for the love of the game ; )
– Ben W
Nov 20 '18 at 2:58
|
show 10 more comments
1 Answer
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Just multiply the probabilities together.
You have $0.000148646896^3 ≈ 3.2844869e-12$
Credit @Ben W
answered Nov 20 '18 at 2:54
Backo
1184
add a comment |
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Just multiply the probabilities together.
You have $0.000148646896^3 ≈ 3.2844869e-12$
Credit @Ben W
answered Nov 20 '18 at 2:54
Backo
1184
add a comment |
Just multiply the probabilities together.
You have $0.000148646896^3 ≈ 3.2844869e-12$
Credit @Ben W
answered Nov 20 '18 at 2:54
Backo
1184
add a comment |
Just multiply the probabilities together.
You have $0.000148646896^3 ≈ 3.2844869e-12$
Credit @Ben W
answered Nov 20 '18 at 2:54
Backo
1184
Just multiply the probabilities together.
You have $0.000148646896^3 ≈ 3.2844869e-12$
Credit @Ben W
answered Nov 20 '18 at 2:54
Backo
1184
answered Nov 20 '18 at 2:54
Backo
1184
answered Nov 20 '18 at 2:54
Backo
1184
answered Nov 20 '18 at 2:54
Backo
1184
1184
add a comment |
add a comment |
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1
"I would like to calculate the probability of the event that 'the above 3 probabilities are the same'" Uh.... I don't understand your question. If you give as a hypothesis that $P(X_1)=P(X_2)=P(X_3)$... then the probability that they are equal is $1$... since you already told us that they are equal. If things are equal then they are equal. A tautology is a tautology...
– JMoravitz
Nov 20 '18 at 2:22
1
I think you misunderstood me. Let me see if I can explain better. Can you describe the physical situation? Are there now twelve players, in three groups of four? What event are you asking to compute the probability of? Etc.
– Ben W
Nov 20 '18 at 2:32
1
You can answer your own question.
– Ben W
Nov 20 '18 at 2:48
1
There is a big difference between asking for the probability that $Pr(X_1=405)=Pr(X_2=405)=Pr(X_3=405)$ and asking for the probability $Pr(X_1=X_2=X_3=405)$.
– JMoravitz
Nov 20 '18 at 2:55
2
Thank you but I don't really keep track nor care about my rep points. I'm in it for the love of the game ; )
– Ben W
Nov 20 '18 at 2:58