My Math Forum Rock Paper Scissors Probability Question

 August 9th, 2014, 06:40 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 619 Thanks: 83 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, 10:26 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 $${n-1\choose4}$$ 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, 01:01 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 619 Thanks: 83 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, 07:23 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 3-in-a-row of S, and not every 3-in-a-row of S takes 8 through 15 rounds -- for one thing, every finish of 16+ rounds does as well.
August 10th, 2014, 08:23 PM   #5
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Quote:
 Originally Posted by CRGreathouse Does anyone have a nicer solution?
You need to work out $1-P(\text{game has no SSS})$ (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 08:33 PM.

August 11th, 2014, 12:23 PM   #6
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Quote:
 Originally Posted by CRGreathouse That's the total number of ways to finish in 8 through 15 rounds. But not every finish of those lengths includes a 3-in-a-row of S, and not every 3-in-a-row of S takes 8 through 15 rounds -- for one thing, every finish of 16+ rounds does as well.
There's no way of calculating how many ways take 16+ rounds because hypothetically there's no limit.

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, 12:31 PM   #7
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Quote:
 Originally Posted by EvanJ There's no way of calculating how many ways take 16+ rounds because hypothetically there's no limit.
It's easy to calculate. Either tackle the infinite sum directly (use the geometric series formula) or compute the complement and subtract from 1.

 August 11th, 2014, 03:05 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 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, 03:35 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. Thanks from v8archie
 August 11th, 2014, 06:11 PM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 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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