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August 9th, 2014, 06:40 PM   #1
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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)?
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August 10th, 2014, 10:26 AM   #2
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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?
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August 10th, 2014, 01:01 PM   #3
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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.
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August 10th, 2014, 07:23 PM   #4
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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.
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August 10th, 2014, 08:23 PM   #5
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Quote:
Originally Posted by CRGreathouse View Post
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.
Thanks from EvanJ

Last edited by v8archie; August 10th, 2014 at 08:33 PM.
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August 11th, 2014, 12:23 PM   #6
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Quote:
Originally Posted by CRGreathouse View Post
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).
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August 11th, 2014, 12:31 PM   #7
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Quote:
Originally Posted by EvanJ View Post
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.
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August 11th, 2014, 03:05 PM   #8
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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.
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August 11th, 2014, 03:35 PM   #9
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That probability seems right -- or rather, it matches my Monte Carlo simulation to within expected deviation.
Thanks from v8archie
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August 11th, 2014, 06:11 PM   #10
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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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