GingerAle

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UsernameGingerAle
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 #4
avatar+1841 
+1

Happy New Year, JB. My biggest fansmiley

 

I originally wrote that post with three scenes. I deleted the funniest scene because the Shanghai’d Post Production Supervisor or Shanghai’d Technical Consultant would have factored it to invisibility. laugh 

I’ll send you a copy of the deleted scene.

 

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This post is one in a set that I’ve periodically monitored for view counts.    

https://web2.0calc.com/questions/express-big-integers-into-smallest-possible-values

Creation date and time: 2019-01-10 @10:14:51UTC

In its lifetime of 346 days, the post collected 388 views with 13 views in the 15 days prior to its hiding by the Administrator on 12/21/19 @ ~ 14:12:52 UTC.

Posted question is class category (1) –informal, simple question.  

 

Subjective Question classes:

*Category (0) –Mr. BB Interest Rate questions or arcane questions optimally answered with Monty Carlo style or tabulated, logical resolution computer code.  

*Category (1) – Informal, simple non-homework, or curiosity based math question.

*Category (2) – Formal, simple homework.

*Category (3) – Formal, intermediate homework.

*Category (4) – Formal, advanced homework.

*Category (5) – Formal, AoPS (or related) homework.

*Category (6) – Formal, basic and intermediate physics, engineering, or chemistry.    

*Category (7) – Formal, advanced physics, engineering, or chemistry.    

*Category (8) – Formal, advanced collegiate mathematics or specialized math. (These are very rare on this forum.)

*Category (9) – Social or non-math, Off Topic posts (includes posts with random characters).

 

View count statistics for all visible questions (posts) dated 1/10/2019:

Mean: 205.88

Median: 175

Mode(s): 317, 138, 141, 182, 191, 328, 152

Minimum: 105

Maximum: 461

Range: 356

Count: 50 (Number of visible posted threads)

At least (10) posts were hidden by one or more moderators (including the auto-moderator). Several of the hidden posts were by a (now banned) member named Synth.  

 

Comparative analysis:

The view counts for the post are 85.2% higher than average for all visible questions posted on 01/10/19. This post is about 366% higher than the sample mean for all (Cat 1) non-homework, informally asked questions.  This post has 3.8% of all views for all visible posts made on 1/10/2019.

 

Other interesting stats, contrast the reasons for this.

 

GA

Dec 31, 2019
 #2
avatar+1841 
+1

The solution(s) presented in the prior posts have one or more errors; most notably an incorrect value for (p) –the probability of a single (successful) event

 

To solve: Find (p), the probability of a single event.

 

Statement:  If I shoot 5 pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability \(\frac {211}{243}\)

Equivalent statistical statement:

\(\Pr(X\geq 1)= 1- \left(\sum_ \limits {i=0}^{1} \binom{5}{i} p^{i}(1-p)^{5-i}\right)=\frac {211}{243}\)

\(\text {Equivalent statement: $Pr(X\geq 1)= \Pr(X > 0) = (\frac {211}{243})$  |  Binomial events are discreet integers. }\)

\(\text {Complementary statement $Pr(X < 1) = \Pr(X = 0) =  \frac {32}{243}$ | Probability of zero (0) successes in five attempts. } \)

\(\Pr(X = 0) =  \binom{5}{0} p^{5}(1-p)^{0}=\frac {32}{243} \)

\(\rightarrow p^5=\frac {32}{243}  \text { |  Solve complementary statement for (p)} \)

\( p = \sqrt [\leftroot{-1}\uproot{1}5]{\frac {32}{243}}\rightarrow p = \frac {2}{3} \text { |  Probability of failure on a single throw}\\ \)

\(1-\frac {2}{3} =\frac {1}{3}  \text { |  Complement of failure = success }\\ \)

\(\text {Use probability of success on a single throw  $(\frac {1}{3})$  to solve:} \\ \) 

If I shoot 6 pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin

Equivalent statistical statement:

\(\Pr(X\geq 2)= 1- \left(\sum_ \limits {i=0}^{1} \binom{6}{i}(\frac {1}{3})^{i}(1-\frac {1}{3})^{6-i}\right)=64.88\% \)

 

\({}\)

 

GA

Dec 14, 2019
 #2
avatar+1841 
+1

Solution:

One method for visualizing a solution is to view the cards in a 4 X 4 matrix, where the columns represent the players and the rows represent the cards in the players' hands. Valid conditions for success occur when the players have single “A” cards located in any one of the four positions of their respective hands.

 

\(\left[ {\begin{array}{cccc} A & A & A & A \\ X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {First arrangement of “A} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{cccc} A & A & A & X \\ X & X & X & A \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{cccc} A & A & X & A \\ X & X & A & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Fifth arrangement} \\ \left[ {\begin{array}{cccc} X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ A & A & A & A \\ \end{array} } \right] \text {Last arrangement} \)

 

 

From this, it’s easy to see there are (4^4) = 256 arrangements of success.

 

Dividing the number of successes by the total number of combinations

(44) / (nCr(16,4)), gives (64/445) ≈ 0.1406593.

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Here are two other solution methods for this question:

 

https://web2.0calc.com/questions/probability-help_11#r5

Here, Melody presents the total arrangements without the target cards then adds the arrangements of the target cards.

 

https://web2.0calc.com/questions/probability-help_11#r13

Here, Rom presents the arrangements of each hand without the target cards then adds the arrangements of the target cards to each hand in sequence.

 

 

GA

Dec 7, 2019