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I sometimes play a guessing game with a friend. So fx we might try to guess how many people live in Russia. So person A says 100 milion and then person B says either 'more' or 'less'. If B says 'more' and in fact more than 100 milion live in Russia, then B wins and conversely. Is this game fair? I have an intuition that B probably has an advantage, but I'm not sure how to formally analyse the situation.

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    $\begingroup$ What do you mean by fair? $\endgroup$
    – Herr K.
    Mar 18 '20 at 23:26
  • $\begingroup$ @Herr K . I'm pretty sure that the OP means: "Does each person have an equal chance of winning ? I'm not sure how to formalize but I don't think so because, for it to be fair, each person would have to know the expectations of other person. The chances of that being true are slim. $\endgroup$
    – mark leeds
    Mar 18 '20 at 23:34
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    $\begingroup$ Well I'm not really sure what I mean. Specifying what is fair probably would do a lot of the work of formalizing the situation. I guess what I am going for is something like this: The person who is epistemically superior should always win the game. If the correct answer is 101 milion and B thinks that the correct answer is 200 milion, then A is in some sense more right but B will win. But I would like to see what analytical tools exists to analyse situations like these. $\endgroup$
    – Caspar
    Mar 18 '20 at 23:50
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This is an interesting question!

It would seem that the game is not fair: Player A is at a pretty serious disadvantage.

Let's imagine each player has their own probability distribution of beliefs that is a sampling distribution of some population distribution centered at the true value. Both players are aware of this, and know their own distribution but have no further information on the other's belief distribution.

The optimal strategies are for player A to guess at the center of their own distribution and for B to guess "More" if the center of B's distribution is above A's guess and "Less" if it's below.

Player A will win if their guess is between the true value and Player B's guess.

-Since there is a 50/50 shot for either guess to be above or below the true value, 50% of the time the guesses will lie on either side of the true value and player B will win, regardless of whose guess is closer.

Via symmetry:

-25% of the time their guesses will be on the same side of the true value and player B's will be closer. Player B will also win here (deservedly).

-25% of the time their guesses will be on the same side of the true value and player A's will be closer. Here player A will win (deservedly).

Hence, player A will win only 1/4 of the time.

However the game also becomes increasingly complicated once you allow each player to have first order beliefs about the other player's beliefs (ie. specific information about the shape and center of the other's belief distribution), because then there's the opportunity for say, A to exploit their belief that B's guess is below the true value by guessing slightly lower than they would otherwise. And with second-order beliefs things get even more complex, and so on and so on. I'm not sure whether player A's disadvantage is robust to these nuances, but it's still quite striking to me that a game that at first glance seems fair is actually under normal conditions pretty lopsided!

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  • $\begingroup$ "each player has a guess that they are 100% confident in" What exactly does this mean? Are they irrational? If not, how are they 100% confident in a guess? $\endgroup$
    – Giskard
    Mar 19 '20 at 9:02
  • $\begingroup$ "We also assume each player has no information about the other player's beliefs. The optimal strategies are..." How do players know what is optimal? This seems to depend on the other's decision, which depends on their beliefs. Otherwise interesting answer. $\endgroup$
    – Giskard
    Mar 19 '20 at 9:02
  • $\begingroup$ Fair. I guess the only model that formally makes sense is the one where they each have their own probability distribution and are guessing ar its center, and they know the other player is in the same situation but have no idea what the others distribution looks like $\endgroup$
    – H Rogers
    Mar 19 '20 at 10:52
  • $\begingroup$ This will yield the same result. I'll edit my answer in a bit. $\endgroup$
    – H Rogers
    Mar 19 '20 at 10:53
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    $\begingroup$ I have edited my original answer to account for your feedback. Thanks! $\endgroup$
    – H Rogers
    Mar 19 '20 at 12:07
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A great question! While I won't attempt to provide a complete analysis, here are a few observations:

Observation 1: In this game, Player 1 lacks a dominant strategy.

To see this, suppose that player 2 chooses the (rather unwise) strategy of saying 'more' regardless of what number player 1 proposes. In that case, player 1 will obviously want to declare a very high number (to make sure that player 2 is wrong). Conversely, if player 2 says 'less' pretty much all the time, then player 1 will want to say a very low number. So what player 1 wants to do depends on player 2 strategy, which makes the situation game theoretic.

To make more progress, let us suppose that each player's beliefs about the unknown quantity can be represented using a smooth probability density function over some subset of the real line (i.e. we will view the quantity to be estimated as a continuous variable). To make things simple, let us further suppose that both probability distributions are symmetric and centred around the true value. We then have:

Observation 2: If players have the same beliefs, then there is an equilibrium in which each player's chance of winning is 50%.

More concretely, suppose that the players use the following strategies:

  • Player 1 guesses the median of the distribution (we can speak of 'the' distribution since the beliefs of the players, which is what the distribution represents, are the same.)
  • Player 2 says 'higher' if Player 1 guesses weakly less than the median of the distribution. Player 2 says 'lower' if Player 1 guesses strictly more than the median.

In the equilibrium, player 1 guesses the median and player 2 says 'higher' (since the guess is weakly lower than the median). This is true 50% of the time, so each player has a 50% chance of winning. It is clear, moreover, that this is indeed an equilibrium:

  • If Player 1 were to increase her guess, then Player 2 would say 'lower'. But Player 2 would be right more than 50% of the time, so Player 1 would have reduced her chance of winning. Similarly, if Player 1 were to decrease her guess, Player would (still) say 'higher', but Player 1 would now be correct more than 50% of the time.
  • Given Player 1's strategy, the true number is equally likely to be higher or lower than the guess. So Player 2 can do no better than say that it is 'higher', thereby winning 50% of the time. (Saying 'lower' would be equally good, but not a strict improvement.)

Finally, let us turn to the more realistic case in which player beliefs can differ. In general, we can view this as a Bayesian game in which player 1 chooses a rule specifying a guess for every possible belief (i.e. distribution) that she could have; and player 2 specifies a rule specifying whether she says 'higher' or 'lower' depending on (i) player 1's guess (ii) player 2's prior belief (distribution). Unfortunately, finding the equilibria of this game seems a little involved. We can, however, observe the following:

Observation 3: In any equilibrium, player 2 must win with probability of at least 50%.

To see this, suppose that player 2 follows the same strategy as above: say 'higher' if the guess is lower than the median of her (player 2's) distribution, and say 'lower' otherwise. In general, this strategy is not optimal since it disregards any information that might be conveyed by player 1's guess. However, even using this simple strategy, player 2 will win at least half of the time (since we have assumed her distribution to be centred around the true value).

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  • $\begingroup$ Observation 3 can also be proved by player 2 flipping a coin :) $\endgroup$
    – Giskard
    Mar 22 '20 at 17:31
  • $\begingroup$ @Giskard Re dominant strategies: my argument shows that player 1 doesn't have a dominant strategy. (I have revised the 'observation' to this weaker claim!) For player 1, the concept of strategies and actions coincide $\endgroup$
    – user17900
    Mar 23 '20 at 12:53
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    $\begingroup$ As it happens, I also don't think player 2 has a dominant strategy, but haven't shown this in the text -- hence, the revision $\endgroup$
    – user17900
    Mar 23 '20 at 12:54
  • $\begingroup$ @Giskard can you explain your point about coin flipping? Apologies if I am being a bit slow here! $\endgroup$
    – user17900
    Mar 23 '20 at 12:55
  • $\begingroup$ Player 2 can follow the strategy of disregarding all available information and choosing "less" or "more" depending on the flip of a coin. As this is a feasible strategy, her equilibrium strategy must win with at least as big a probability. $\endgroup$
    – Giskard
    Mar 23 '20 at 13:06
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The game is not fair. B has an advantage.

The definition of the word "information" in the below analysis means knowledge of the subject area of the game.

Consider the following scenarios.

Scenario one: The players have equal information.

If the game were fair in this scenario, both players should have an equal likelihood of winning. We can see this is the case. Assume both players have zero information. Then A has a 50/50 chance of being too high or too low. And B has a 50/50 chance of correctly calling the result. Both players have even odds of winning.

Scenario two: A has more information than B; but A's information is not perfect.

In this case, "fair" should mean that A has a better chance of winning than B. But we can see this is not true. Assume B has zero information and their answer comes by flipping a coin. B's odds of winning would still be 50/50. B could essentially flip a random coin for their answer and expect to win the game 50% of the time.

Scenario three: B has more information than A; but B's information is not perfect.

In this case, we would expect fair to mean that B's odds of winning would be greater than A's odds. We can see this is the case. Assume, for example, A has zero information and, therefore, their guess is made by choosing a random number. If B has any information at all, their odds of winning are significantly greater than 50%.

Since the game yields a fair outcome in scenarios one and three but the outcome is weighted in favor of B in scenario two, we must conclude the game is not fair and B has an advantage.

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  • $\begingroup$ If you see my answer, even in the case where the players have the same level of information, B will still have an advantage. $\endgroup$
    – H Rogers
    Mar 19 '20 at 10:51
  • $\begingroup$ Actually sorry, I guess this depends on how you choose to formally model "information" $\endgroup$
    – H Rogers
    Mar 19 '20 at 11:08
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The previous answers suggesting B has an advantage are correct under several tacit assumptions. However, in the general case it is impossible to say who has an advantage. This depends on the question and on the precision of the information the players possess.

Example: Consider the question: How many legs does a beetle have? Obviuosly, player A has an advantage here.

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  • $\begingroup$ Nice point, though of course A wouldn't have an advantage (in your example) if B were also allowed to say 'exactly that number' (in addition to 'more' and 'less'). $\endgroup$
    – user17900
    Mar 20 '20 at 16:21
  • $\begingroup$ I think in most cases it is fair to assume that the probability of A getting the number exactly right is 0. $\endgroup$
    – Caspar
    Mar 20 '20 at 18:47
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I think this holds up in the case where we assume each player draws from a normal distribution where the actual answer is the mean (on average they guess correctly). If Player A makes some guess G1 then the probability that the guess of the second person is on the same side as the mean (and therefore encompasses the mean) is greater the the probability it is not.

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