算法找到好的，可靠的球员 [英] Algorithm for Finding Good, Reliable Players

问题描述

我已经以下球员，每个值对应于一个结果在正确答案的在给定游戏百分比

I've the following players, each value corresponds to a result in percentage of right answers in a given game.

$players = array
(
    'A' => array(0, 0, 0, 0),
    'B' => array(50, 50, 0, 0),
    'C' => array(50, 50, 50, 50),
    'D' => array(75, 90, 100, 25),
    'E' => array(50, 50, 50, 50),
    'F' => array(100, 100, 0, 0),
    'G' => array(100, 100, 100, 100),
);

我希望能够拿起最好的球员，但我也想考虑到球员如何可靠的是（少熵=更可靠），到目前为止，我想出了下面的公式：

I want to be able to pick up the best players but I also want to take into account how reliable a player is (less entropy = more reliable), so far I've come up with the following formula:

average - standard_deviation / 2

不过我不知道这是否是一个最优的方案，我想听听你的想法在此。 我一直在思考一些关于这个问题，我想出了一个略微不同的公式，这是修改后的版本：

average - standard_deviation / # of bets

这个结果将被加权为下一个即将到来的投票，所以例如从播放器C上新建一个赌注只能算半个赌注。

我不能进入细节这里但是这是一个项目用的的群众理论智慧和德尔菲法 而我的目标是predict尽可能的下一个结果从几个球员加权过去的赌注。

I can't go into specifics here but this is a project related with the Wisdom of Crowds theory and the Delphi method and my goal is to predict as best as possible the next results weighting past bets from several players.

我AP preciate所有输入，谢谢。

I appreciate all input, thanks.

推荐答案

首先，我不会用标准差，如果你的数据阵列只有几个条目。使用诸如中位数绝对偏差（MAD）更可靠的统计方法，同样，你可能想测试使用中位数而不是平均。

First off, I would not use Standard Deviation if your data arrays have only a few entries. Use more robust statistical measures like Median Absolute Deviation (MAD), likewise you might want to test using the Median instead of the Average.

这是由于这样的事实，如果玩家的投注你的知识仅限于几样，你的数据将通过异常值为主，即玩家是幸运/不幸的。统计方法可能是在这些情况下完全不合适的，你可能想使用某种形式的启发式方法。

This is due to the fact that, if your "knowledge" of players' bets is limited to only a few samples, your data is going to be dominated by outliers, i.e. the player being lucky/unlucky. Statistical means may be entirely inappropriate under those circumstances and you may want to use some form of heuristic approach.

我也是从你的链接假设，你其实不打算挑选最好的球员，而是根据玩家下一组回答A要predict一组正确答案C，由称重根据玩家的previous记录A。

I also assume from your links, that you do not in fact intend to pick the best player but rather based on the players next set of answers "A" want to predict the correct set of answers "C" by weighing "A" based on the players' previous track record.

当然，如果有一个很好的解决这个问题，你可以大赚一笔的股市;-)上（即没有人做的其实应该是一个迹象，以这种解决方案的存在）。

Of course if there were a good solution to this problem, you could make a killing on the stock market ;-) (The fact that no-one does, should be an indication as to the existence of such a solution).

不过又回到了排名的球员。你的主要问题是，你（有什么打算？）采取正确答案为均匀分布，从0--100％的比例。如果测试包含多个问题，这肯定不是这样的。我想看看什么是完全随机的球员R分数测试，并建立了一个相对信心数量的基础上如何更好地/不如R给定的真正的球员。

But getting back to ranking the players. Your main problem is that you (have to?) take the percentage of right answers as evenly distributed from 0--100%. If the test contains multiple questions this is certainly not the case. I would look at what a completely random player "R" scores on the test and build up a relative confidence number based on how much better/worse than "R" a given real player is.

说，每轮游戏的产生一百万随机播放器，并期待在分数的分布。使用分配的权重为玩家真实分数。然后，使用MAD结合加权得分，并计算中位数 - MAD /一些数字，比如你已经建议

Say, for each round of the game generate a million random players and look at the distribution of scores. Use the distribution as a weight for the players' real scores. Then combine the weighted scores using MAD and calculate the Median - MAD / some number, like you already suggested.

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