August 9th, 2014, 07:40 PM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 630 Thanks: 85  Rock Paper Scissors Probability Question
Two people play rock paper scissors until there are 5 trials where the players picked different items (which has probability 2/3 on each trial). What is the probability that at some point there will be at least 3 consecutive times where the players pick the same item (which has probability 1/3 * 1/3 * 1/3 = 1/27 on any 3 consecutive trials)?

August 10th, 2014, 11:26 AM  #2 
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 
Let's simplify the game to choosing "S" with probability 1/3 or "D" with probability 2/3. There's just one way to finish in 5 rounds: DDDDD. There are 5 ways to finish in 6: SDDDDD, DSDDDD, DDSDDD, DDDSDD, and DDDDSD. In general there are $$ {n1\choose4} $$ ways for the game to take $n$ rounds, since the last round needs to end with D. Each of the ways has probability $2^d/3^n$ where $d$ is the number of Ds. It's clear, I think, that if n > 15 then there must be some three consecutive S. So it suffices to check up to 15 and count. Not very elegant but doable on a computer. Does anyone have a nicer solution? 
August 10th, 2014, 02:01 PM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 630 Thanks: 85 
Having 3 consecutive S requires at least 8 rounds, so I need to calculate how many it could happen for the eight amounts of rounds, 8, 9, 10, 11, 12, 13, 14, and 15. Is the following correct for 8 trials with 5 D and 3 S: 7 nCr 4 = 35 * 2^5 / 3^8 (the 2^d you gave me is always 2^5 = 32 if d always = 5) with a probability of about 0.1707 Using my calculator for the eight probabilities, I got an answer of 0.4275481052 (of course rounded). I would have guessed a lower probability. 
August 10th, 2014, 08:23 PM  #4 
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 
That's the total number of ways to finish in 8 through 15 rounds. But not every finish of those lengths includes a 3inarow of S, and not every 3inarow of S takes 8 through 15 rounds  for one thing, every finish of 16+ rounds does as well.

August 10th, 2014, 09:23 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,557 Thanks: 2558 Math Focus: Mainly analysis and algebra  You need to work out $1P(\text{game has no SSS})$ (an SSSS includes SSS  twice in fact). I should think. P(game has no S) is simple: DDDDD = $\Pr(D)^5 = (\tfrac{1}{3})^5$ P(game has no SS) is: a pick of five events A or B where A=D and B = SD. $\Pr(B) = \tfrac{2}{9}$ and $\Pr(A) = \tfrac{1}{3}$. The probability that we get a sequence of 5 As and/or Bs is $\Pr(A \text{ or } B)^5 = (\Pr(A) + \Pr(B))^5 = (\tfrac{1}{3} + \tfrac{2}{9})^5 = (\tfrac{5}{9})^5$. We can repeat this analysis for P(game has no SSS). In this case we have events A, B and C where C=SSD and A and B are as before. $\Pr(C) = \tfrac{4}{27}$. In this case we need a run of five events A, B and C with probabiltity $\Pr(A \text{ or } B \text{ or } C)^5 = (\Pr(A) + \Pr(B) + \Pr(C))^5 = (\tfrac{1}{3} + \tfrac{2}{9} + \tfrac{4}{27})^5 = (\tfrac{19}{27})^5 = 0.17256$ to 5d.p. So $\Pr(\text{game has SSS  or more}) = 1  0.17256 = 0.827$ to 3d.p. The beauty of this solution is that it extends easily and it gets the binomial theorem to do the combinatorics for you. Last edited by v8archie; August 10th, 2014 at 09:33 PM. 
August 11th, 2014, 01:23 PM  #6  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 630 Thanks: 85  Quote:
Thank you v8archie. I'm surprised the probability is that large. I can do trials myself with dice where 1 and 2 count as the same number, 3 and 4 count as the same number, and 5 and 6 count as the same number. Here are my results of ten trials: Trial 1: 6 pairs of rolls, Most consecutive S was 1 5 and 6: S 1 and 3: First D 4 and 6: Second D 4 and 5: Third D 2 and 5: Fourth D 1 and 4: Fifth D Trial 2: 5 pairs or rolls, No S 1 and 4: First D 3 and 5: Second D 3 and 5: Third D 1 and 4: Fourth D 4 and 6: Fifth D Trial 3: 5 pairs or rolls, No S 1 and 4: First D 2 and 5: Second D 2 and 4: Third D 4 and 6: Fourth D 2 and 6: Fifth D Trial 4: 10 pairs of rolls, Most consecutive S was 2 4 and 6: First D 1 and 1: S 1 and 5: Second D 1 and 2: S 3 and 4: Second consecutive S 2 and 3: Third D 4 and 5: Fourth D 3 and 3: S 6 and 6: Second consecutive S 1 and 4: Fifth D Trial 5: 6 pairs of rolls, Most consecutive S was 1 2 and 3: First D 2 and 6: Second D 3 and 4: S 2 and 4: Third D 2 and 4: Fourth D 3 and 5: Fifth D Trial 6: 8 pairs of rolls, Most consecutive S was 2 1 and 6: First D 3 and 5: Second D 2 and 2: S 3 and 6: Third D 3 and 3: S 1 and 2: Second consecutive S 2 and 3: Fourth D 1 and 4: Fifth D Trial 7: 7 pairs of rolls, Most consecutive S was 1 1 and 1: S 1 and 6: First D 1 and 3: Second D 1 and 1: S 2 and 3: Third D 2 and 6: Fourth D 2 and 5: Fifth D Trial 8: 6 pairs of rolls, Most consecutive S was 1 3 and 6: First D 4 and 5: Second D 3 and 4: S 4 and 6: Third D 1 and 4: Fourth D 1 and 3: Fifth D Trial 9: 9 pairs of rolls, Most consecutive S was 2 2 and 2: S 2 and 4: First D 1 and 6: Second D 5 and 6: S 1 and 2: Second consecutive S 2 and 6: Third D 3 and 3: S 3 and 5: Fourth D 2 and 3: Fifth D Trial 10: 8 pairs of rolls, Most consecutive S was 4 1 and 6: First D 1 and 4: Second D 1 and 4: Third D 2 and 2: S 3 and 4: Second consecutive S 4 and 4: Third consecutive S 4 and 4: Fourth consecutive S 1 and 6: Fourth D 3 and 6: Fifth D Therefore I got at least 3 consecutive S in only 1 out of 10 trials. 50 pairs of rolls were D and 20 pairs of rolls were S. Using a probability of S of 0.333 because I found website that can't handle fractions and 1/3 cannot be written as a terminating decimal, the probability of exactly 20 S out of 70 trials is 7.32%, the probability of more than 20 S is 75.97%, and the probability of fewer than 20 S is 16.72% (it adds up to 100.01% due to rounding).  
August 11th, 2014, 01:31 PM  #7 
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  
August 11th, 2014, 04:05 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,557 Thanks: 2558 Math Focus: Mainly analysis and algebra 
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. 
August 11th, 2014, 04:35 PM  #9 
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 
That probability seems right  or rather, it matches my Monte Carlo simulation to within expected deviation.

August 11th, 2014, 07:11 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,557 Thanks: 2558 Math Focus: Mainly analysis and algebra 
Yeah, I rather foolishly transposed the probabilities of S and D, so it's not surprising I got an odd answer. Of course, the A, B, C, ... have probabilities determined by a geometric distribution with parameter $\tfrac{2}{3}$. 

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