Concrete mathematics : a foundation for computer science

8.2 MEAN AND VARIANCE 379
We estimate the average spot sum of these dice to be 7.4&2.1/m = 7.4~tO.7,
on the basis of these experiments.
Let’s work one more example of means and variances, in order to show
how they can be ca.lculated theoretically instead of empirically. One of the
questions we considered in Chapter 5 was the “football victory problem,’
where n hats are thrown into the air and the result is a random permutation
of hats. We showed fin equation (5.51) that there’s a probability of ni/n! z 1 /e
that nobody gets thle right hat back. We also derived the formula
P(n,k) = nl ’ ‘n (n-k)i = -!&$
. 0\
k
for the probability that exactly k people end up with their own hats.
Restating these results in the formalism just learned, we can consider the
probability space FF, of all n! permutations n of {1,2,. . . , n}, where Pr(n) =
1 /n! for all n E Fin. The random variable
Not to be confused F,(x) = number of “fixed points” of n , for 7[ E Fl,,
with a Fibonacci
number.
measures the number of correct hat-falls in the football victory problem.
Equation (8.22) gives Pr(F, = k), but let’s pretend that we don’t know any
such formula; we merely want to study the average value of F,, and its standard
deviation.
The average value is, in fact, extremely easy to calculate, avoiding all the
complexities of Cha.pter 5. We simply observe that
Hence
F,(n) = F,,I (7~) + F,,2(74 + + F,,,(d)
Fn,k(~) = [position k of rc is a fixed point] , for n E Fl,.
EF, = EF,,, i- EF,,z + . . . + EF,,,,
And the expected value of Fn,k is simply the probability that Fn,k = 1, which
is l/n because exactly (n - l)! of the n! permutations n = ~1~2 . . . n, E FF,
have nk = k. Therefore
EF, = n/n =: 1 , for n > 0. (8.23)
One the average. On the average, one hat will be in its correct place. “A random permutation
has one fixed point, on the average.”
Now what’s the standard deviation? This question is more difficult, because
the Fn,k ‘s are not independent of each other. But we can calculate the