# What are some strategies for this number game?

Asked by

PhiNotPi (

12647)
February 4th, 2012

I just recently read about this number game, and I am interested in what sort of strategies might exist. I have simplified the game slightly.

There are five players. Each round, all of the players choose a number 1–10 without knowing the choices of the other players. Then, each player’s guess is revealed, and the winner is the person with the lowest number that *no other player chose*. The winner of the round gets one point. The winner of the game is the person with the most points after any given number of rounds.

What are some strategies that could be used to increase the odds of winning this game?

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## 13 Answers

Choose 1.1. You never said it had to be an integer.

By 1–10, I assume you mean it has to be an integer. I would always choose 1.

I would pick 1 or 2 or 3 at random.

This would be really interesting to model with a program: simulate players with different choosing strategies.

Obvious: never pick a number more than the number of players

I think I would choose 2 or 3 for the first round (because most will choose 1 until they figure out how the game works), and definitely 1 for the next couple of rounds (because everyone will be afraid to choose 1), At some point, people will start choosing 1 just to keep me from winning, so I’d have to go higher, probably depending on what numbers others are choosing.

I am only vaguely familiar with game theory. I wonder if it could be applied in this case. Without further analysis, I would randomly choose a number from 1 to 5.

It is possible to cheat at this game. Two people could use an algorithm for choosing a random number. One would use the random number and the other would use the number plus one mod 5. This would give them a slight edge, because their numbers would never match one another.

The game seems to invite collusion among players. Suppose someone adopted the simple minded strategy of choosing 1 each time. There is nothing that a single player can do to take advantage of this. However, the other 4 players could arrange things so that they took turns choosing 1, with the others choosing numbers from 2 to 5.

Three players could conspire so that one of them would have to win, by adopting a strategy by which they always chose the numbers 1, 2 and 3. There is nothing the other two players could do against this.

I’ve read about game theory but have never seen this problem. The nub of it seems to be that a rational strategy is normally adopted uniformly by all players, making the payoff matrix symmetrical about the diagonal. But in this game there’s an advantage for uniqueness (or a loss for ties) that maybe makes the whole thing unstable?

I am assuming this is limited to integers 1–10.

If you use @HungryGuy‘s strategy (*always choose 1*) then everybody would choose 1 and lose. So then everybody should avoid 1. But then an independent rogue will choose 1 anyway and win. And so on, endlessly reversing & ambiguous. By the same logic the strategy for choosing the same number every time is similarly ambiguous or unstable. Something like @flutherother‘s randomizing among 1, 2, 3 makes sense.

Or perhaps some other range of random values? With a uniform probability distribution? This bears further analysis…

If you reduce this to 2 players, the best strategy is to always pick 1, because the player who doesn’t pick 1 will always lose.

If there are 3 players, then the game becomes complicated. If two players adopt the same strategy, then the third player will always win. Perhaps the best strategy for this is to always chose 1 if there is no colluding between players allowed. This is because no other player will want to choose 1 because they would tie and the third player would get the point. If there is colluding, then I don’t know what might happen.

Randomizing among 1, 2, 3 does make sense. In any round, either there will be a hole in this range where no one guessed and that you could have won by guessing there, or the winner would have his guess inside this range. I am sort of answering my own question as I figure stuff out.

5 players choose from 10 numbers, so there are 100,000 different ways the game may be played. For each of these a point is awarded to the winner, if any. This is begging for a computer program that would play all possible games in a few milliseconds to determine long-run payoffs for each choice 1–10, from which a weighted set of probabilities would be assigned to each number, scaled to add up to 100%

This still assumes that players make choices without knowing what the others choose, which rules out collusion and other social complications. It might be possible to devise an adaptive strategy based on other players’ performance. In other words, if somebody plays 1 more often than expected, you might reduce the probability of choosing a 1 yourself, etc. If the other players’ choices are truly random, however, then history should be irrelevant.

A two person game reminds me of the Prisoner’s Dilemma – http://en.wikipedia.org/wiki/Prisoner's_dilemma In such a game, I would start out by alternating 10 and 1 and hope that the other player catches on and plays 1 to my 10 and 10 to my 1. If, after a few rounds, the other player does not cooperate, I would resort to all 1’s.

There is a problem if two of us use my strategy. We may find ourselves matching with 10 and 1. We need a way of getting in sync. In that case, I would alternate my strategy to randomly alternate between 1 and 10 until such time as we mismatch, at which point I return to the original strategy.

Using this strategy, I might not win a single game, but if we had repeated rounds against other players using either the alternating strategy or the all 1’s strategy, those of use using the alternating strategy would accumulate the most total points.

If one player chooses 1 a lot, then not choosing 1 will allow that player to win. The only way to score a point against player that always chooses 1 is to get another player to guess 1 also (at least for that particular round). I suppose that given a large enough number of players, then there will usually be at least two people that pick 1.

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