See query. I consultedthe Master, but he had other fish to fry.
Apparently there are strategies to play, based on things like gender of opponent, and (if played) the last throw, and experience level. Knowing these can greatly increase your chances, making it not a true game of chance. Here’s one such cite: How to Win at Rock, Paper, Scissors: 10 Best Strategies
Assuming however totally random throws I would say it’s even money, for each throw there seems to be a equal win,lose or tie.
What do you mean, exactly?
With random picks, it’s obviously 1/3 win, 1/3 lose, 1/3 draw.
Furthermore, it’s equal chances if at least one player is playing randomly, no matter what the other one does.
You’ve hit the nail on the head with “if at least one player is playing randomly” which rarely occurs. Someone who is skilled in tells and reads (e.g. a professional poker player), could probably anticipate the other player’s next play after a few rounds. Since we played it so often in grade school, some players (myself included) were banned because they would always play the same pattern.
Bit off topic, I once worked with a guy who was an expert at the shell game. One day we played “just for fun”. Instead of choosing the shell which I “knew” concealed the pea, I would choose one or the other two at random. I was winning 50/50 until he caught on. We had a good laugh about that.
Here’s a variant with five choices instead of three:
(1,2,3) Ordinary rock, scissors, paper play as usual among themselves; collectively they are called the ‘Paper collective.’
(4) Super-scissors beats everything in the paper collective but loses to Super-rock.
(5) Super-rock beats Super-scissors but loses to everything in the paper collective.
What strategy is guaranteed to break even on average, in this variant?
If N is the number of choices, I went from N=3 to N=5 skipping 4.
For the mathematicians:
Devise a plausible set of constraints for such games to be “interesting” and prove that the constraints cannot be fulfilled when N is even.
If all players have access to superweapons, then nobody ever has any reason to play a non-super version of those weapons, and so the game is exactly equivalent to normal RPS. If you wish an asymmetric game, then you can give one player access to a superweapon but not the other.
So you play Super-rock and Super-scissors 50% each? I alternate Rock and Super-rock, winning 50% and drawing 25%. Next contestant!
And, no: Asymmetric games forbidden! 
Is this saying what you intend to say? I’d love to play someone who always plays the same pettern because it is then predictable. Hmmm. Scissors then two rocks; he’s gonna throw paper before going back to another rock.
Yeah, that doesn’t make any sense. Playing patterns is what makes RPS “breakable.”
Here’s a fun computer simulation of RPS that has been going on for years. With almost a million and a half rounds played, the computer has a winning percentage of 61% (Ties are not counted for the win percentage. If counting ties, it breaks down as win 45%, tie 27%, lose 28% for the computer.)
In some sense you are right, but you need to remember to call the Paper collective Super-paper and not forget to pick it.
This game has a best-2-out-of-three-rounds game of RPS with “Super-scissors” allowed Rock, Paper, Scissors, Double Scissors! | FiveThirtyEight - but if Scissors (not Superscissors) is the winning throw in any round the person who played it wins the whole game, so there is still reason to use scissors.
I think this is just equivalent to ordinary R-P-S, but with a sub-game that can sometimes resolve ties if both players choose Paper (that is, ‘super rock’ beats ‘super-scissors’ beats ‘paper collective’ beats ‘super rock’, and if both players choose ‘paper collective’ then the tie is resolved by what ‘ordinary’ R/P/S they chose).
Unbeatable strategy is just randomly choosing: 1/3 Super-rock, 1/3 super-scissors and 1/9 each of the paper collective variants.
Which, against any strategy, should be something like win [(1/3) + (1/9) *(1/3)], lose [(1/3) + (1/9) (1/3)], tie [(2/9) + (1/9)(1/3)]
When the other person begins throwing his rock-paper-scissor I go with roshambo for the win.
What odds was he giving you? A shell game should pay 2-to-1 (but they probably don’t). If you were able to find the pea half the time you should have made a tidy profit.
Yes. This might be the simplest symmetric RPS-type game with no dominated strategies which does NOT have a simple [1/N, 1/N, … 1/N] optimal strategy.
Now, prove there is no satisfactory such game with even N.
Ah, I misunderstood the rules. I was thinking that you meant that super-scissors beat paper and scissors, but lost to all rocks, and super-rock beats normal rock and any scissors, but loses to paper.
More interesting challenge: Given that septimus considers plausible and interesting conditions to be unfulfilable for even N, determine what conditions he considers “plausible and interesting”. Because I can come up with games with even N which seem sufficiently interesting to me.
Example one: Arbitrarily designate one player as “even” and one player as “odd” (you’re probably objecting to a pre-game arbitrary designation, but I contend that it’s fine, so long as even and odd have equal chances of winning (which they do), and so there’s no reason to prefer one over the other). Each player chooses either 1 or 2. If the sum of both numbers is even, the even player wins, otherwise the odd player wins. This game has the added benefit of never tying.
Example 2: Each player chooses North, South, East, or West. If both players choose the same direction, or opposite directions, the game is a tie (you’re probably objecting to players tying with different throws, but I contend that this is no more problematic than tying with the same throw). Otherwise, the player who’s clockwise of the other is the winner (i.e., E beats N, S beats E, W beats S, and N beats W).
As an aside, there’s also a three-player game which picks one winner, with four options, full symmetry, and only a 1/16 chance of a tie.
No one mentioned rock paper scissors lizard spock?
I was going to, but you beat me to it.
Kingdom of Loathing has 5 “elements” of magic (Hot, Cold, Sleaze, Stench, and Spooky) that each have weaknesses and strengths against the others in the same way as RPSLS. The way it’s laid out often obscures this, but it’s obvious that the structure is the same because everything is good against two things and bad against two things, and I’m almost certain there isn’t more than one pattern to do that besides just switching labels since each vertex of the directed graph is identical.