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I am reading a book about statistics and probability. In one chapter, the author use roulette as an example to explain the use of central limit theorem. He says for ordinary roulette, there are 38 numbers (including one 0 and 00). The chance of occurrance of 17 is 1/38. But for a wrapped wheel (assume extreme warp in section 17), the change of getting 17 will be increased. So statistically, how to tell if a wheel is wrapped? The following is excerpted from the book

In order to such a decision you must account for chance variation. One way to do this is by computing the standard deviation and using the central limit theorem.It turns out that even for this extremely warped wheel you need to observe about 2000 spins to get enough separation to effectively rule out change. In 2000 spins, the three-SD (three times of standard deviation) range around the average number of 17's for the warped wheel (average=105.26; 3SD=29.96) does not overlap the three-SD range for an ordinary wheel (average=52.63; 3SD=21.48). In other words, the number of occurrences of 17 with this extremely warped wheel will most likely be outside the three SD range of an ordinary wheel, so you could confidently rule out chance variance.

Sorry for the long information. Here are what I confuse

1) Why we have to rule out the "chance variation"? My opinion is to make sure the biased outcome of roulette is because of the wrap not because the chance, is that right?

2) I don't understand why the author say because of no overlap of two normal distribution, we could confidently rule out chance variance? How to understand that? What happen if two curves do overlap?

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# About probability of playing roulette

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