Summary of Results

Mean Number of Fixed Points per Game = 1

Mean Number of Cycles of Length k = 1/k

Expected Value of Number of Cycles =  

Expected Value of Length of a cycle =  

Probability that there are k fixed points approx = 1/(k!e)

Probability that there are no cycles of length k approx e^-1/k

Probability that a particular student is in a cycle of length k = 1/n

Recursions for derangements:  and  

Recursions for Stirling numbers:  

Recursions for permutations with no 1 or 2-cycles:   

Number of different games when cycle notations are not equivalent =