What are the odds on rock, paper, scissors?

Yep, that’s me!

I can’t find or recall the episode right now, but on a Korean variety show one of the hosts explained his strategy when playing rock, paper, scissors. It was along the lines of if they both showed the same thing, say rock, then his opponent would likely play scissors thinking he would play paper to mix it up. Therefore he would play rock again and win.

This challenge isn’t “more interesting” since it is an implicit part of the original challenge. But since the problem is quite non-trivial anyway, I’ll put you out of your misery. :slight_smile:

To be “RPS”-like the game must

  • be symmetric; i.e. payoff(A,B) + payoff(B,A) = 0;
  • have payoffs of only +1 and -1 (or zero when both players make the same move);
  • have no bad or dominated plays; i.e. the minimax mixed strategy selects each move with some positive probability.
    (Without the third constraint it’s trivial to turn N=3 RPS into a N=4 game. Simply introduce a fourth move, single finger, which loses to R, P and S.)

Many years ago, I had insomnia and turned on ESPN in the middle of the night. They were broadcasting the world championship of rock-paper-scissors, and Tom Arnold was doing the color commentary.

Many of the players wore elaborate, bizarre outfits, and had given themselves silly wrestler-type nicknames like The Minneapolis Mauler (I just made that up; I don’t recall any of the actual nicknames). They were engaging in lots of pre-game trash talk, trying to psych each other out. They’d taunt their opponents and act like they had already won. The actual games were played very fast and were over in the blink of an eye. It was clear from the player interviews that a lot of poker-like strategy was involved.

This was quite an eye-opener. RPS is a psychological game; there’s nothing random about it.

I’m not quite sure what the notation in this definition of ‘symmetric’ means exactly; is it enough to rule out Chronos’s odd-even game?

Yes. Calling the two players “George” and “Peter”, the game is symmetric if George and Peter can be interchanged with no effect. Chronos’ example is not symmetric: George wants “Even”, but if interchanged, he’d want “Odd.”

Simplest, I think, is to work with an explicit game payoff matrix M. The symmetry constraint for a zero-sum game is M[sub]ab[/sub] = - M[sub]ba[/sub].

Why do people play Tic Tac Toe (which I’m bad at too, too predictable) when “The only way to win, is not to play”

It’s because there’s the unpredictable (unless you’re me) human factor that no simulation can emulate. The best strategy is for the first play to always take the center square, but some players start elsewhere just to mix up the action. I’ve played games where the center or other square is off limits unless it’s only possible move.

Why is predictability a factor in Tic Tac Toe?

I have a game that meets all your constraints when N is not odd.[spoiler]Plays are rock, paper, super-rock, super-paper, super-super-rock, etc.

More-supers always beats paper with less supers, and always loses to rock with less supers. Same-supers paper beats rock. Same play ties.

Yes, yes, N is still not even. But it’s also not odd :).[/spoiler]

Like RPS, play enough times and you’ll notice the other player will repeat patterns or make certain faces or movements before making the next move. A sharp poker player will probably catch on to the tells within a few games.

I also stink at checkers and chess for this same reason. Out of xxxxxxxx+ moves, my repertoire is limited to xxx and I end up repeating those xxx number of moves if I play long enough.

:slight_smile: This is why I wrote “suitably constrained” rather than attempting to enumerate ALL the constraints. So I’ll add another constraint to the list:
The set of legal moves must be finite.
That N must be odd is “neat” and non-obvious. AFAIK, the simplest known proof isn’t real simple. If anyone wants to discuss further (or ask for hints) please start a Mundane thread to avoid cluttering GQ.

So what? Play alternates in tic tac toe. Your opponent makes his move, then it’s your turn. What difference does it make if you can predict what move your opponent will make ahead of time?

I’m going to predict that you already encoded a hint in there. Or am I being irrational?

https://www.youtube.com/watch?v=qg5g0bvxyes (warning, fast-forward a bit, past Itchy and Scratchy :slight_smile:

I am going to go out on a limb and claim that no mathematics beyond some basic linear algebra should be required. YMMV how not-simple that makes it.

This kind of mathematical reasoning about games goes back to von Neumann et al.

Player 1 in Tic-Tac-Toe can guarantee that he doesn’t lose by starting either in the center or in a corner, and cannot guarantee a win with either. No ability to read the opponent is required, only the ability to read the board. It’s so simple that you can literally build a computer capable of it out of Tinkertoys.

Yes. The proof, while non-trivial enough to earn a “Neato!”, is quite elementary and uses only very simple linear algebra.

It does seem that such a simple result — interesting RPS-like games cannot have an even number of options — might have an easy intuitive demonstration but AFAIK none is known.

OP here: dumbish question I’m not sure how to say in the correct words:

Is there a word for “the time [ie number of events] it takes for the outcome of this series of events to be statistically as close as damnit to the individual case.”

Another example of that way (the query) of thinking, which I had in mind for OP, also put in terms of betting strategy. In craps, eg, there are long-term, slow, crunching methods as well as martingale/double up ones, each of which demand a different number-of-events (as well as initial bet) to reveal the ultimate truth of the simple truths of the dice odds.

On a side-note (and because it recently came up with an American friend) over here the game is called Paper, Scissors, Stone.

Here being Northern Ireland.