In a two-player game of rock paper scissors against a random opponent, you win 1/3 of the time, lose 1/3 and tie 1/3, so your expected score is exactly zero. The strategy that cannot be beaten in the long run is to throw rock, paper and scissors each with probability 1/3; that mix is the game's Nash equilibrium.
That is the whole answer in two sentences. The rest of this page shows the working, extends it to groups of three or more players, and ends with a classroom lesson plan, because this is one of the best games there is for teaching probability.
The payoff table
Score a win as +1, a loss as -1 and a tie as 0. From your point of view (rows are your throw, columns are theirs):
| You / Them | Rock | Paper | Scissors |
|---|---|---|---|
| Rock | 0 | -1 | +1 |
| Paper | +1 | 0 | -1 |
| Scissors | -1 | +1 | 0 |
Every row and every column contains one +1, one -1 and one 0. That symmetry is the source of everything below. It also makes the game zero-sum: whatever you gain, your opponent loses.
Probability of each outcome
There are 3 x 3 = 9 equally likely pairs of throws if both players choose at random. Three are ties (rock-rock, paper-paper, scissors-scissors), three are wins for you and three are losses.
- P(win) = 3/9 = 1/3
- P(tie) = 3/9 = 1/3
- P(loss) = 3/9 = 1/3
Ignoring ties, you win exactly half of decided rounds.
Expected value
The expected value of a throw is the sum of each payoff multiplied by its probability. Suppose your opponent throws rock with probability r, paper with p and scissors with s, where r + p + s = 1. Reading across the payoff table:
- EV(rock) = 0·r - 1·p + 1·s = s - p
- EV(paper) = 1·r + 0·p - 1·s = r - s
- EV(scissors) = -1·r + 1·p + 0·s = p - r
If your opponent plays r = p = s = 1/3, all three are zero. Nothing you do helps or hurts.
If they lean on rock, say r = 0.40, p = 0.35, s = 0.25, then EV(paper) = 0.40 - 0.25 = +0.15 per throw. Over 100 throws that is an expected 15 points. This is exactly why real-world habits matter; our own throw frequency data shows people do lean on rock.
Why one-third each is the Nash equilibrium
A Nash equilibrium is a pair of strategies where neither player can improve by changing only their own strategy.
Look at the EV formulas again. If your opponent throws any option more than 1/3 of the time, at least one of your throws has a positive expected value and you should shift toward it. The only opponent mix that leaves you with no profitable response is r = p = s = 1/3. By symmetry the same holds for them. So the unique equilibrium is both players randomizing evenly, and the value of the game is 0.
Two consequences surprise people:
- The equilibrium does not win. It guarantees you break even against anyone. It is a defence, not an attack.
- Beating a real person requires leaving equilibrium. You profit only by exploiting their bias, which opens you up to being exploited in turn. Our game theory page goes further into this trade-off.
How long until someone wins?
Each two-player round is decisive with probability 2/3. The number of rounds until a decision follows a geometric distribution, so the expected number of rounds is 1 / (2/3) = 1.5. The chance of three ties in a row is (1/3)^3, about 3.7%.
Ties with more than two players
With n players all throwing at random, a round is decisive only if exactly two of the three throws appear. If all three appear, or everyone throws the same thing, there is no winner. For any particular pair of throws, the number of outcomes that use both and nothing else is 2^n - 2. There are 3 ways to choose the pair and 3^n total outcomes. So:
P(decisive) = 3(2^n - 2) / 3^n, and P(tie) = 1 - 3(2^n - 2) / 3^n
| Players (n) | P(tie) |
|---|---|
| 2 | 3/9 ≈ 33.3% |
| 3 | 9/27 ≈ 33.3% |
| 4 | 39/81 ≈ 48.1% |
| 5 | 153/243 ≈ 63.0% |
| 6 | 543/729 ≈ 74.5% |
| 10 | about 94.8% |
This is why a big group trying to throw at once rarely gets a result. For groups larger than three, run a bracket of head-to-head games instead. Our free event hosting tool does exactly that.
Teaching probability with rock paper scissors
I like this lesson because students already know the game and have opinions about it, which gives you a prediction to test.
- Predict. Ask the class which throw wins most. Record the vote.
- Enumerate. In pairs, list all 9 outcomes and mark each as win, loss or tie. Derive the 1/3 figures.
- Collect data. Each pair plays 30 rounds and tallies throws and results. Pool the class data on the board.
- Compare. Is the observed win rate close to 1/3? Is rock over-thrown? Discuss sampling variation versus real bias.
- Extend. Older students can compute expected value against the class's actual throw frequencies and find the best counter-throw.
- Groups. Have groups of 3, 4 and 5 throw simultaneously and count how often a round is decisive, then check against the table above.
The activity covers sample spaces, equally likely outcomes, relative frequency, expected value and, for advanced classes, the idea of an equilibrium. The rules page is a handy handout if anyone needs a refresher.
FAQ
What are the odds of winning rock paper scissors?
Against a random opponent: 1/3 win, 1/3 tie, 1/3 loss. Excluding ties, 50%.
Is there a mathematically best move?
No single move. The best defensive strategy is a random one-third mix, which cannot lose in expectation.
What are the odds of a tie with three players?
Exactly 1/3, the same as with two. From four players up, ties quickly become more likely than not.
Is rock paper scissors pure luck?
Between two perfect randomizers, yes. Between humans, no, because people have predictable habits that the math lets you exploit.