[Marxism] Fwd: Re: Calculating Octopus (Was: A question to mathematicians)

johnaimani johnaimani at earthlink.net
Mon Jul 12 16:25:01 MDT 2010


<<On Jul 12, 2010, at 12:01 AM, Néstor Gorojovsky wrote:
>
> Can anyone calculate the probabilities that Paul the octopus has
> reached his results by chance?

1/28  =  . 0039

Statistically significant at the 99.6 % level.

Sheldrake strikes again!>>


Shane Mage >>



Since there is a disagreement between the number above and my number 
below, I will demonstrate my reasoning.

Say the winner is always heads (H):

     With 1 throw the possibilities are HT and TH or 1/2 probability = 
                                         2 to the first power            
             2

     With 2 throws the possibilities are HH HT TH TT or 1/4 
probability=                                    2 to the second power 
                 4

     With 3 throws the possibilities are HHH      HTH        HHT     HTT
                                                                         
                                            probablity 1/8 =             
2 to the third power                     8
                                                                TTT    
     THT        TTH      THH

     With 4 throws the possibilities are HHHH     HHHT    HHTT    HTTT
                                                                
HTHH      HTHT    HTTH    HHTH
                                                                         
                                                    Probability 1/16=  2 
to the 4th power                         16
                                                                 TTTT   
     TTTH    TTHH    THHH
                                                                 THTT    
    THTH    THHT    TTHT

Etc...Leading to the conclusion that the possibility of 9 straight 
passes = 2 to 9th power = 512 with the chance of this happening 1/512
or 511 to 1.


 From the internet note this on 7 throws:

<<With an "unbiased" coin your odds of a single toss are 1/2.

With two tosses, it becomes 1/2*2.

With 7 tosses, it is 1/2^7 (that 2 to the 7th power) = 1/128.

Not impossible, but it's a very low percentage (less than 1% of the tries).

Best regards,

Omnivorous-GA>>  http://answers.google.com/answers/threadview/id/526284.html

7 throws ends with 128 different combinations and 128 is 2 to the 7th power.



Then there is this with 10 throws

     <<0 heads:   1/1024 = 0.0009765625
       1 head:   10/1024 = 0.009765625
       2 heads:  45/1024 = 0.0439453125
       3 heads: 120/1024 = 0.1171875
       4 heads: 210/1024 = 0.205078125
       5 heads: 252/1024 = 0.24609375
       6 heads: 210/1024 = 0.205078125
       7 heads: 120/1024 = 0.1171875
       8 heads:  45/1024 = 0.0439453125
       9 heads:  10/1024 = 0.009765625
      10 heads:   1/1024 = 0.0009765625>>   http://mathforum.org/library/drmath/view/56660.html

The importance of this lies in the number "1024" which is 2 to the 10th power.

And 1/512 (which is 2 to the 9th power) = .002% even more impressive than the comrade's assertion above


On 11:59 AM, johnaimani wrote:
>  2 to the 9th power or 511-1.
>
> Now 2 to the 9th is 512 but there is one chance that the octopus 
> selects correctly.
>
>



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