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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.

Backo
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    "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 02:22
  • You haven't defined what $X_1, X_2, X_3$ are. – angryavian Nov 20 '18 at 02:24
  • $X_1 = X_2 = X_3 = 1620$ – Backo Nov 20 '18 at 02:25
  • I get the impression that you don't yet have a good handle on how to use the stats notation here. Maybe try explaining your question in plain English. – Ben W Nov 20 '18 at 02:25
  • @BenW, I know because I'm not an expert in the matter. Please, see the updated question and suggest improvements if it is not sufficiently descriptive. – Backo Nov 20 '18 at 02:32
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    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 02:32
  • @BenW, 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 know the probability that the arithmetic mean of the number on their cards is 405 for a total of 3 times. Is it more clear? BTW I'll also update the question. – Backo Nov 20 '18 at 02:45
  • Just multiply the probabilities together. You have $0.000148646896^3\approx3.2844869e-12$ – Ben W Nov 20 '18 at 02:47
  • @BenW, please post your answer so I can accept it. – Backo Nov 20 '18 at 02:48
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    You can answer your own question. – Ben W Nov 20 '18 at 02:48
  • I'd to reward merit. – Backo Nov 20 '18 at 02:50
  • "I would like to know the probability that the arithmetic mean of the number on their cards is $405$ for a total of $3$ times"... So... you have $X_1,X_2,X_3$ are random variables describing the average of the cards drawn in the first, second, third rounds respectively. You calculated in the previous question $Pr(X_1=405)$ to be the value mentioned above and the same logic is used to find $Pr(X_2=405)$ and $Pr(X_3=405)$ to be the same. You are not asking for the probability that the probabilities are the same... you are asking for the probability that the outcomes are all $405$. – JMoravitz Nov 20 '18 at 02:55
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    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 02:55
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    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 02:58
  • @JMoravitz, at the end everything should refer to number 405, in both the previous question where it is calculated the probability of the event that "the arithmetic mean of the number on the 4 cards is 405" and in this question where it is calculated the probability of the event that "the previous event repeated 2 more times (3 in total during the same observation) has again the arithmetic mean of the number on the 4 cards of 405". – Backo Nov 20 '18 at 03:15

1 Answers1

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Just multiply the probabilities together.

You have $0.000148646896^3 ≈ 3.2844869e-12$

Credit @Ben W

Backo
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