也許是自打咀巴

星期日晚看見選舉結果，望見兩大選區（新界東／新界西）各泛民名單的得票如此相近，心想「噢，law of large number…」，然後跟據幾個模糊的概念，草草地寫了那篇
Fooled by randomness。

事實上該文很不準確。先是兆安的幾個留言提醒我對量子力學的理解有錯，然後是朋友S私底下指出這不可能是純綷隨機結果。我自己計了一次，對，如果純粹的隨機投票，五張名單的得票應該是極之接近（七成機會正負一百）。多謝賜正。關於量子力學，我的理解只是大學通識程度，不多提了。數學還有點點能力，下面是我的算法。對數學沒有興趣的，可跳至尾二段。

Given n balls and k urns. All balls are uniformly distributed among all k urns. We calculate the probability of the first urn getting not more than t balls.

The random variable is the number of balls in the first urn, it can be approximated by a Poisson Distribution.

where is the expected rate.

We can get the probability of the first urn getting not more than t balls by summing the probability of all events, since they are mutually excluded.

Begin lazy, we ask matlab for results with the following command,

poisscdf(t, u)

Suppose we have 204030 balls and 5 urns, the expected number of each urn is 40806. The probability of the first urn getting less than 99% of the expected number (i.e. 40398) is

On the other hand we can get the probability of getting more than 1% of the expected number (i.e. 41214) similarly,

In fact a more precise approximation scheme can be achieved by using Chernoff Bound, however it depends on a moment generating function that i am not familiar with. This article gives a loose approximation of how unlikely an urn to get 1% fewer(more) balls than the expected value.

Thank you S for the useful discussion in Probability.

把問題簡化，只考慮任何一張名單。上面的結果得出，假設有204030張選票和五張名單，若所以選民皆是隨機投票的話，平均每張名單應得40806張選票。然而，一張名單超過／低過平均數的百份之一的機會為百份之五；超過／低過平均數的百份之二更不足百份之零點零零零一。然而沒有一張名單的實際得票是平均正負百份之二內，意即不能以純粹隨機過程去解悉投票行為。

縱然該文錯漏百出，我仍然堅持該文的基本概念，每個人的喜惡不一，無可能準確配票，反而，隨機（於兩大選區）能容易地得出較平均的分配。