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Never-ending rock, paper, scissors?

Lego rock, paper, scissors cards

In the latest set of collectable Lego cards, each card includes a ‘rock’, ‘paper’ or ‘scissors’ icon in the corner, allowing you to play the classic game without having to think for yourself. Can such a game theoretically go on forever?

My son was desperate to play rock, paper, scissors with his Lego cards so, like a good mum, I dealt the cards out between us and started playing. We followed top-trumps rules:

  • the winner of a round keeps both cards from that round
  • if both cards are the same, they are put aside for the winner of the next round to keep
  • the last person to have any cards remaining is the overall winner.

We started with 20 cards each. After playing for 10 minutes or so our piles still looked about the same size, so we counted them up. He had 21, I had 19. This got me thinking: was the game destined to last forever? Could I prove that it would?

To simplify matters for this post, I assume that each player starts with three cards: one rock, one paper and one scissors. This means that there are six possible orders for each player’s hand and 36 possible outcomes for the first three rounds of play. These 36 outcomes fit into three categories.

Draw after three rounds

The simplest scenario is that both players have their cards in the same order. The game will then end in a draw after three rounds, when both players run out of cards. The probability of this is \(\frac{6}{36}=\frac{1}{6}\).

Win after three rounds

The next simplest scenario is that the hands are ordered such that one player will win all three rounds. For example, player one has rock, paper, scissors and player two has scissors, rock, paper. Rock blunts scissors, paper wraps rock and scissors cut paper, so player one wins after three rounds. The probability of this scenario is \(\frac{12}{36}=\frac{1}{3}\).

One draw, one win, one lose

The most interesting scenario for the first three rounds is that one round will be a draw and each player will win one round. This is the only other scenario; it’s not possible for a player to win two rounds and lose or draw one. Can you see why? The probability of this scenario is \(\frac{18}{36}=\frac{1}{2}\).

The thing that makes this scenario interesting is that it is the first time the players have some choice in how to play. The winner of each round might return the cards to their deck with the winning card on top or with the losing card on top. The cards from a draw might be placed above or below the cards from the winning round. Does this affect how long the game will take?

I wrote some code in Python to test some of the options and found the following results.

  • If cards are always replaced in the order winning card, losing card, draw cards, the game will always end by the fifth round.
  • If cards are always replaced in the order losing card, winning card, draw cards, the game will always end by the fifth round and the winner will be the same as in the previous case (i.e. consistently replacing the cards in a different order doesn’t change who wins).
  • If one player replaces cards in the order winning card, losing card, draw cards and the other player replaces them in the order losing card, winning card, draw cards then by the 12th round at the latest the players will have three cards each again. What’s more, the cards will once again be ordered so that one round will be a draw and each player will win one round. This means that the game will continue indefinitely with no overall winner.

Conclusion

There is clearly some further investigation to do here. The players needn’t replace the cards in the same order each time and there are orders of replacement other than the ones I’ve already described. However, one thing is clear: even if you start with only three cards each, it is possible to have a never-ending game. The moral of the story? Never promise ‘just one game’ before bedtime.

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About Sam Hartburn

Sam offers proofreading, copy-editing and answer checking for maths textbooks and digital resources at all levels and for popular and recreational maths books and content. Follow her on Twitter at @SamHartburn or find her on LinkedIn

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