[Ivan Pedroso]

atul said:

Two random numbers, x1 and x2, both with uniform distribution from zero to one. Each have an expected value of 0.5. But if you're asking what's the expected value of the _greater_ of two, that's 2/3!

You a right that if I roll a number of dies (in this case eight-sided) and then only write down the largest value every time, then the average of this "highest-value-thrown" will indeed be higher than the usual 4.5 that is the average value of a standard eight-sided die. But that is (if understand it correctly) not the situation at hand.

As I see it, we are dealing with:
Some dude rolls a bunch of eight-sided dies, and then write down how many ones he got, how many twos he got and so on. He then adds the numbers of the three most common results, and divides this number with the total number of dies rolled.

An example:
100 eight-sided dies are rolled, and the following is written down:
#1 : 15
#2 : 12
#3 : 19
#4 : 10
#5 : 9
#6 : 12
#7 : 10
#8 : 13
The three highest are added (i.e. #3,#1, and #8) and we get:
19+15+13 = 47
And get (the Duck_Number): 47/100 = 0.47

The observed frequencies of the different values of the above example are:
P(x=1) : 15/100 = 0.15
P(x=2) : 12/100 = 0.12
P(x=3) : 19/100 = 0.19
P(x=4) : 10/100 = 0.10
P(x=5) : 9/100 = 0.09
P(x=6) : 12/100 = 0.12
P(x=7) : 10/100 = 0.10
P(x=8) : 13/100 = 0.13
Which are not all that close to the 1/8 = 0.125 value that where used to generate this sample.

If you increase the number of rolled dies to a much larger number than 100, then these frequencies will be closer to 1/8. (Well the probability of getting a sample using say 1.000.000.000.000.000 dies that results in frequencies that deviate greatly from 1/8 will be extremely unlikely - that is why I say that they "will approach" 1/8)

In fact you could choose any two small positive numbers, epsilon >0 and delta >0, (could be 0.00000001 and 0.000000001) and it will then be possible to find a laaaarge number N that insures that:

If N dies are rolled then the probability of getting an observed frequency that deviates from 1/8 with more than the small number epsilon, is smaller than delta.

That is:
Probability( |"observed frequency" - 1/8| > epsilon ) < delta

And then adding up the three largest observed frequencies will then result in a value that is in the interval
[3/8 - 3*epsilon ; 3/8 + 3*epsilon]
with as close to one hundred percent certainty as you want (just choose epsilon and delta to be very small)

So yes - I do in fact claim that given N = some extremely large number, then the Duck_Number will (most most most likely) be (ever ever ever so close to) 3/8.

Sorry for all this dry boring stuff