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Probability of rock-paper-scissors

MrDudeMan

Lifer
Jan 15, 2001
15,063
90
91
Background:

I am driving to a college campus today with a friend from work to stand in as an industry expert to talk to students about their career paths. We were talking about who would drive and naturally used rock-paper-scissors to decide. The first game was one of the best I've ever played. It went like this:

  1. SS
  2. SS
  3. SS
  4. SS
  5. SS
  6. RR
  7. SS
  8. SS
  9. PR (I threw paper)

It got me thinking about probability and I realized I'm not sure how to think about this problem. Granted, I never did well in statistics so I'm not surprised.

Would these be considered independent or dependent events if you want to know the probability of getting 8 ties in a row followed by a win? It seems obvious, but I started to confuse myself because, while another round is only played if the previous was a tie, the outcome of the previous doesn't change the probabilities of the same outcomes on the next throw. I may have already said the magic words to know the answer, but I'm simply not able to convince myself either way.

If they are independent, then is the probability of this happening (1/3)^8*(1/2)? If it is dependent, I have no clue how to solve it I think.

Thanks for any insight you can offer. I am going to post this on buzz as soon as I am able to come up with a solution to brag about winning this round (I won the next round as well but it was more typical. 3-4 fights before I claimed victory).
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,148
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1/3^9

Your chance was 1/3 for each specific occurance, win lose or tie.

To have that specific sequence happen (ignoring any influence on the seeding in the first place) you need to hit 1 out of 3 nine times.
 

Biftheunderstudy

Senior member
Aug 15, 2006
375
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I think the real kicker here is that you can only do that sort of probability analysis on truly random events. Rock-paper-scissors is something you can get good at, which right away means your chances of win/lose/tie are not 1/3.

If instead you each rolled a die and the larger number won, then you could do that sort of analysis. While playing the game, each game round would be independant of the last.

If you look at the final result and ask what are the chances of tieing 6 games in a row then its a different question...
 

MrDudeMan

Lifer
Jan 15, 2001
15,063
90
91
Thanks for your replies. Unfortunately I remember very little from statistics.

It makes sense now that the answer is (1/3)^9. I said 1/2 as the last term as I had confused myself with the train of thought that the actual probability to win a round is 1/2 since a tie doesn't actually change the win or loss (ignoring the human element that skews the probabilities).
 

iCyborg

Golden Member
Aug 8, 2008
1,237
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If they are independent, the probability for the exact above sequence is (1/3)^18, there are 18 symbols. They aren't really idependent though, but modeling human mind is beyond our capabilities at the moment. The best you could do is a statistical study with a lot of people...

Anyway, where I grew up, we didn't have rock-paper-scissor, I only learned about them from US movies. We had something we called even-odd: one person chooses odd, the other even, then they simultaneously in the same R-P-S way show a certain number of their fingers (generally 1-5 using one hand). If the sum is odd, the person choosing odd wins, if it's even, the one choosing even wins. Very simple, and no ties ever.
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,148
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iC, the RPS game is a 1 of 3. You can only put out 3 symbols, regardless of what the other person puts out. You have a 1 in 3 chance, each round, of getting a particular result (win, lose or tie).
 

an_89

Junior Member
Oct 21, 2010
8
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Probability of each event is independent, if we see the event as purely random, that i.e. is debatable in a rock-scissor-paper game played between two humans.
It is just unusual to get a long streak of same result, but that does not matter on the next event, unless the humans gets influenced and plays following the trend they seem recognizing in previous events, it is quite common both in human and animal behavior.
 

Pulsar

Diamond Member
Mar 3, 2003
5,225
306
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Probability of each event is independent, if we see the event as purely random, that i.e. is debatable in a rock-scissor-paper game played between two humans.
It is just unusual to get a long streak of same result, but that does not matter on the next event, unless the humans gets influenced and plays following the trend they seem recognizing in previous events, it is quite common both in human and animal behavior.
First, IF The probability of each event is random, then the calculation has already been shown above (1/3)^18.

However, it is not random. Humans have memory, and as such they remember the last game and infer things (incorrectly, but that's human nature). That is why there is a rock/paper/scissors world championship each year and certain players do, in fact, do better than others. There IS a real strategy to it, but it involves understanding who you're playing against and what patterns they might have. (For instance, do they turn their hand on the way down on the third stroke differently depending on what they're throwing?)

Therefore, solving this particular problem quantitatively is pretty much impossible. It would be more constructive to say it's incredibly unlikely to ever see that again, then move on.
 

DrPizza

Administrator Elite Member Goat Whisperer
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Don't get too carried away with the (1/3)&18 calculation though. EVERY 9 rounds has that same probability of occurring. (assuming the letter to the left is the person to the left and the letter to the right is the person to the right.)
 

iCyborg

Golden Member
Aug 8, 2008
1,237
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Actually I just noticed he was asking for "probability of getting 8 ties in a row followed by a win" and not this particular instance. That's (1/3)^9 all right...
 

uclabachelor

Senior member
Nov 9, 2009
448
0
71
Thanks for your replies. Unfortunately I remember very little from statistics.

It makes sense now that the answer is (1/3)^9. I said 1/2 as the last term as I had confused myself with the train of thought that the actual probability to win a round is 1/2 since a tie doesn't actually change the win or loss (ignoring the human element that skews the probabilities).
1/3^9 IF each round are independent, meaning that you did not take into consideration the previous rounds to make the decision for the current round.

If you did take into consideration previous rounds then the answer is much, much less than 1/3^9.
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,148
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No, it isn't.

Guys, it is (1/3)^9.

That is is, stop thinking so much. If the other guy's choices are not dependant on anything, truly random, than you have a 1 in 3 chance of getting a particular result.

If you are asking how likely it is to, say, tie 9 times it works like this:

Your opponent has chosen 1 of 3 things. You choose yours. You have a 1 in 3 chance of getting the same. He will have R P or S, you will chose one of three. being independent each round will have a 1 in 3 chance.

The more you stack, the more it multiplies. We are not talking about the chances of getting a specific combination, just a specific "win, lose or tie" sequence.

If you wanted to do the other (specific sequence) your possible combinations are:

RR
RP
RS
PR
PP
PS
SR
SP
SS

So, if you do not count who throws what, you have this probability of getting:

RR, PP, SS - 1 in 9 (each)
RP/PR - 2 in 9
RS/SR - 2 in 9
PS/SP - 2 in 9

You then multiply those probabilities for your total is you are not concerned with your order.

If you are concerned about who throws what and you declare a specific (unknown to the throwers) set of combos, you have a 1 in 9 of each happening. So the chances of having that SPECIFIC SEQUENCE THROWN IN A SPECIFIC ORDER BY SPECIFIC PEOPLE, it would be (1/9)^9

Hope this clears it all up.
 

MrDudeMan

Lifer
Jan 15, 2001
15,063
90
91
Thanks everyone. So is it correct to say the permutation is (1/9)^9, and the combination is (1/3)^9? That was actually the motivation for asking this question. Right after I graduated from college I was on jury duty and the judge asked us, in the middle of the trial, if anyone knew how many different ways the jury of 13 could sit in 14 seats. I knew the answer right off the top of my head (I had statistics in my last semester) but that was a while ago and I couldn't remember how to apply those terms.
 

an_89

Junior Member
Oct 21, 2010
8
0
0
That is why there is a rock/paper/scissors world championship each year and certain players do, in fact, do better than others. There IS a real strategy to it, but it involves understanding who you're playing against and what patterns they might have.
Agree.
What I meant is that if it was a math homework, I would have responded with the math calculations shown in the thread.
Otherwise I would had responded "does not compute!" because the only possible math solution has the logic fallacy of setting by hypothesis the events as true random, which they are not.
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,148
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Thanks everyone. So is it correct to say the permutation is (1/9)^9, and the combination is (1/3)^9? That was actually the motivation for asking this question. Right after I graduated from college I was on jury duty and the judge asked us, in the middle of the trial, if anyone knew how many different ways the jury of 13 could sit in 14 seats. I knew the answer right off the top of my head (I had statistics in my last semester) but that was a while ago and I couldn't remember how to apply those terms.
Isn't it "13!" (13 x 12 x 11 x 10......)

As for the somantics, I am not sure, athough "permutations" sounds right for the specific order (each possible combo being "different"). The combination? I am not sure. The 1/3 ^9 would probably just be called a probility of occurance. It is not a special generic situation. It is a specific set of conditions that need to be satisfied.....
 

Evadman

Administrator Emeritus<br>Elite Member
Feb 18, 2001
30,990
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Rock-Paper-Scissors is not random since there is a thought process involved. I have proven that I can destroy people that I play more often than probability would state. (I would estimate I have roughly a 75&#37; win rate)

Players tend to pick Scissors first, then they will play whatever would beat what the opponent threw the last time. For example, if I throw a rock and beat your scissors, you will have a very high chance of throwing paper next time. Even when I tell people that I know what they will pick with this knowledge, they change their process in a predictable way so I can still beat them.

Rock-Paper-Scissors is a mental game, not a probability game.
 
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TecHNooB

Diamond Member
Sep 10, 2005
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Rock-Paper-Scissors is not random since there is a thought process involved. I have proven that I can destroy people that I play more often than probability would state. (I would estimate I have roughly a 75% win rate)

Players tend to pick Scissors first, then they will play whatever would beat what the opponent threw the last time. For example, if I throw a rock and beat your scissors, you will have a very high chance of throwing paper next time. Even when I tell people that I know what they will pick with this knowledge, they change their process in a predictable way so I can still[b/] beat them.

Rock-Paper-Scissors is a mental game, not a probability game.


probability encapsulates all of that though. it's called conditional probability.
 

Evadman

Administrator Emeritus<br>Elite Member
Feb 18, 2001
30,990
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Probability theory is based on an initial random event tho. If a human can think and change the initial probability, then probability theory goes out the window. I understand what you are saying on event 1 influencing event 2, but probability theory is based upon event 1 being random. Maybe I am just being a semantic nazi or something, but I am sure that RPS can not be solved by probability theory.

Let's use an example; cards & poker. Probability theory is used to determine which cards have what probability of being dealt. There is exactly a 1/52 chance that I will get a certain card if it is the first out of the deck. Every player has exactly the same probability of receiving a given card. This can be expounded to all the cards dealt in a hand. So the cards you get are determined by probability theory. However, probability theory does not model who will win the hand. There is a large human part to the game, such as reading opponents and bluffing, raise amounts, price to buy the pot, etc. If who would win the hand was based only on probability theory, it would be almost impossible for the same person to win 2 tournaments given a large number of entries (1 divided by total entries). However, what is seen in the real world is that there are a consistent few winners (or at least the same few at the final table).

Probability theory just doesn't allow for the human factor of skill in any game, so probability theory can not accurately predict the outcome of the game. If it was a computer randomly picking rock, paper or scissors, then it would be a different story. In that case, probability theory would be the correct mathematics branch to predict the outcomes.
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,148
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Eva, knowing what the other person is more likely to pull is not eliminating probability, it just adds additional rules and constraints.

Life is nothing but a complex equasion!

Hari Seldon said so!
 

Evadman

Administrator Emeritus<br>Elite Member
Feb 18, 2001
30,990
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Eva, knowing what the other person is more likely to pull is not eliminating probability, it just adds additional rules and constraints.
Encyclopedia Britannica said:
http://www.britannica.com/EBchecked/topic/477530/probability-theory

Branch of mathematics that deals with analysis of random events.
Wikipedia said:
http://en.wikipedia.org/wiki/Statistical_randomness

A numeric sequence is said to be statistically random when it contains no recognizable patterns or regularities; sequences such as the results of an ideal die roll, or the digits of pi exhibit statistical randomness

  • The frequency test, was very basic: checking to make sure that there were roughly the same number of 0s, 1s, 2s, 3s, etc.

  • The serial test, did the same thing but for sequences of two digits at a time (00, 01, 02, etc.), comparing their observed frequencies with their hypothetical predictions were they equally distributed.
  • The poker test, tested for certain sequences of five numbers at a time (aaaaa, aaaab, aaabb, etc.) based on hands in the game poker.
  • The gap test, looked at the distances between zeroes (00 would be a distance of 0, 030 would be a distance of 1, 02250 would be a distance of 3, etc.).
If a given sequence was able to pass all of these tests within a given degree of significance (generally 5%), then it was judged to be, in their words "locally random". Kendall and Smith differentiated "local randomness" from "true randomness" in that many sequences generated with truly random methods might not display "local randomness" to a given degree — very large sequences might contain many rows of a single digit. This might be "random" on the scale of the entire sequence, but in a smaller block it would not be "random" (it would not pass their tests), and would be useless for a number of statistical applications.
If someone can win rock paper scissors consistently over 50% of the time, then by definition, the sequence is not random. We can't use a process that is used to define random events to define a process that is not random.

Extreeme example: If you and I play rock, paper scissors 100 times, and you try to win, I can all but guarantee that I can get you to win at least 95% of the matches. I will just always pick paper. If it takes you more than 5 times to figure out that I am lazy and always picking paper, then you may have a different issue :p. If you can't beat me by more than whatever the expected statistical variation beyond 50% is in that instance, then I would be worried.
 

DominionSeraph

Diamond Member
Jul 22, 2009
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If someone can win rock paper scissors consistently over 50&#37; of the time, then by definition, the sequence is not random.
It's pseudorandom, which falls under probability theory when you don't know the underlying structure because it's not practically deterministic.

Three deterministic subsystems -- one in which rock is certain, one in which paper is certain, and one in which scissors is certain -- in deterministic system, does not negate that we may have to use probability.
 
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C1

Platinum Member
Feb 21, 2008
2,260
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They way everyone talks is as if they never heard of Bayes Theorem or conditional probability. Com'mon, get real.
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,148
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No, he is seeing how many times he can split the same hair.







All the bunnies are still hiding.
 

mathsman

Junior Member
Jan 6, 2011
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Interestingly, even if it is possible to improve your chances of winnng beyond 50&#37; by using a strategy, this is completely nullified if the person you are playing against is in fact making their choices completely randomly. It is only possible to have a meaningful strategy if the other person is also using some non-random choice strategy.

There is no strategy to beat random...!
 

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