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August 12th, 2014, 04:02 PM   #11
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Quote:
Originally Posted by CRGreathouse View Post
It's easy to calculate. Either tackle the infinite sum directly (use the geometric series formula)
I found the geometric series formula on Wikipedia, but I don't know what numbers to put in where.

Quote:
Originally Posted by v8archie View Post
I have some individual probabilities wrong.

The overall probability of no SSS is $(\tfrac{2}{3}+\tfrac{2}{9} + \tfrac{2}{27})^5= (\tfrac{26}{27})^5$. That'll give answer much closer to 0.1.

So, right method, wrong answer.
Thank you. (26/27)^5 = about 0.828, so the probability of getting at least one SSS is about 0.172, which is much closer to 0.1 but is not 0.1 to the nearest tenth.

Using a random number generator for numbers from 1 to 3 with 3 being an S, I got 37 S out of 100 numbers. Therefore there were 63 D, which is enough for 12 trials, with the last 3 D and any S with them ignored. Counting four consecutive S as two sets of three and five consecutive S as three sets of three, I got 6 sets of at least three consecutive S from 12 trials, which is 0.5 per trial, which is much higher than the probability.
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August 12th, 2014, 04:32 PM   #12
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You shouldn't count sets of 3 consecutive S, but rather trials containing a set of 3. My point earlier about fours was to highlight that a set of four contains a set of three.
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August 12th, 2014, 04:34 PM   #13
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Quote:
Originally Posted by v8archie View Post
You shouldn't count sets of 3 consecutive S, but rather trials containing a set of 3. My point earlier about fours was to highlight that a set of four contains a set of three.
Well then 3 out of 12 trials got at least 3 consecutive S.
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August 12th, 2014, 05:04 PM   #14
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I've just run 10 million trials, of which 1,718,259 contained three consecutive draws.
Code:
use constant S => 1/3;
use constant D => 1 - S;
use constant T => 100000000;
use constant G => 5;
use constant C => 3;

my $t = 0;
while ( $t < T ) {
  my $g = 0;
  my $c = 0;
  while ( $g < G ) {
    if ( rand() < S ) {
      $c++;
    } else {
      $c = 0 if $c < C;
      $g++;
    }
  }
  $t++;
  $s++ if $c >= C;
}

printf STDOUT "Trials: %10d   Games per Trial: %2d   P(Draw): %1.3f  Consequetive Draws Sought: %2d\n", T, G, S, C;
printf STDOUT "Trials Completed: %10d   Trials Counted: %10d\n", $t, $s;
Output:
Code:
Trials:   10000000   Games per Trial:  5   P(Draw): 0.333  Consequetive Draws Sought:  3
Trials Completed:   10000000   Trials Counted:    1718259
Output for 100,000,000 (which took 5 minutes):
Code:
Trials:  100000000   Games per Trial:  5   P(Draw): 0.333  Consequetive Draws Sought:  3
Trials Completed:  100000000   Trials Counted:   17195001
Thanks from EvanJ

Last edited by v8archie; August 12th, 2014 at 05:19 PM.
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August 13th, 2014, 02:00 PM   #15
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Well 1,718,259 is close to the expected value of 1,719,665.
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August 13th, 2014, 02:34 PM   #16
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Yes, my original Monte Carlo of 10 million came to a similar number.
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