My Math Forum Rock Paper Scissors Probability Question

August 12th, 2014, 05:02 PM   #11
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Quote:
 Originally Posted by CRGreathouse 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 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.

 August 12th, 2014, 05:32 PM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra 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.
August 12th, 2014, 05:34 PM   #13
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Quote:
 Originally Posted by v8archie 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.

 August 12th, 2014, 06:04 PM #14 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra 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 06:19 PM.
 August 13th, 2014, 03:00 PM #15 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 623 Thanks: 85 Well 1,718,259 is close to the expected value of 1,719,665.
 August 13th, 2014, 03:34 PM #16 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Yes, my original Monte Carlo of 10 million came to a similar number.

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