## 2009年9月22日 星期二

### 洗幾次牌先「夠」？

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Below is for more advanced math students. Only people who have some background on Markov Chain theory should continue reading:

For those who know about Markov Chain theory, you may easily see that it is a Markov process. As every element in the symmetric group has an inverse, it is easy to observe that the eigenvector with eigenvalue 1 is indeed the "uniform distribution vector".

The period of the corresponding stochastic matrix must be 1, since there is non-zero probability that the inverse GSR shuffle remains the deck unchanged. Hence by Perron-Frobenius theorem, all other eigenvectors must be with eigenvalue of modulus strictly less than 1. So after shuffling sufficiently many times, no matter how the intial distribution is, it must converge to the unifrom distribution.

In other words, you may say converging to uniform distribution is an immediate result of Perron-Frobenius theorem. Bayer and Diaconis were analyzing how fast the convergence is.

#### 1 則留言:

Kahoo 提到...

http://www.mathdb.org/module/probability/7/7.htm