Lotto numbers generated with Bingo balls

[Stig Holmquist]

A toy Bing ball set is well suited for unbiased drawings of simulated
lotto games including Powerball and Mega Millions.

Drawings can be performed in two different ways, either with or
without replacements between each game.Let's consider
drawing a 5/55 type game without replacements and thus not
a simulation of Powerball.

Find a Bingo ball set and select balls numbered 1-55 and mix them
till they are fully randomnized as far as possible.

Now draw five, record their numbers and calculate their sum.
Do not replace these five but repeat the process ten times
till all 55 bals have been drawn and thus 11 sets of fibe numbers
have been generated with eleven sums. Now calculate the
variance for these sums.

This process of generating eleven sets of five numbers can be
repeated endlessly, but since there are about 3.48 million
different combinations of five numbers, there would be only
316251 unique sets of eleven.

My problem now is: suppose one were able to generate
all 316251 sets of mean sums of 140 and their variances from 140,
what kind of distribution would one get for the variances?

I can visualize the unique situation of each set of five numbers
being consecutive numbers, viz. 1-2-3-4-5......51-52-53-54-55,
which would have the sums 15-40-65-90----190-215-240-265,
with a variance =6875 but what would be the lowest value?
Might it come from sets of nearly equal sums such as
1-2-28-54-55(140), 3-4-27-52-53 (139),5-6-29-50-51(141)....
all sums equal to 140+1or-1? I generated one set of eleven
and got a variance of 80.

I'm particularly interested in the formula if any for the
distribution and its mean. Could it be a Chi squared?

Stig Holmquist

[Stig Holmquist]

Correction:

The set I generated had a variance of 1995 not 90.
The most uniform set of eleven 5-number sets might be
1-2-34-35-33 (105)
3-4-37-38-32 (112)
5-6-38-39-31 (119)
-
-
-
-
17-18-50-51-25 (161)
19-20-52-53-24 (168)
21-22-54-55-23 (175)

These yield a variance of 539.

The mean of the max and min =3707.

Stig Holmquist

[Old Earl]

Post by Stig Holmquist
The set I generated had a variance of 1995 not 90.
The most uniform set of eleven 5-number sets might be
1-2-34-35-33 (105)
3-4-37-38-32 (112)
5-6-38-39-31 (119)
-
-
-
-
17-18-50-51-25 (161)
19-20-52-53-24 (168)
21-22-54-55-23 (175)
These yield a variance of 539.
The mean of the max and min =3707.
Stig Holmquist

Hi!
Now the holidays are passed, I have the time to respond. You posted a
problem that is not really
easy to answer with a simple formula analytically defined. So I
resorted to a simulation to
explore the data from sample runs. First, however, I ran a simple
counting porgram to determine
the expected variance estimate. The variance of the sum of five of
fifty five numbers turned out
to be 1166.6666...

I set up my simulation, and made several runs with the following
results:
No. Trials Var.Est. Std.Dev. of Est.
100 1216.256 481.18
1K 1168.615 483.415
10K 1169.877 468.119
20K 1170.204 467.584
100K 1169.597 469.11

The next question concerned the probability distribution of the
estimate of the variance. This
is a bit messier. The distribution for the 10K run was nowhere near
either a Chi-Square
or a normal distribution. Just to be sure, I also tested the square
root of the variance estimate, and found a very nearly perfect fit to a
normal distribution. To explain a bit better, when I ran the
simulation, I also kept a histogram of the variance estimates for each
run (acutallly the sum before dividing by eleven to maintain integers).
The 10K run found 6140 occupied bins, for an
average of about 1.6hits per bin. I wrote the data to a file, and then
used Excel to look at the
data. I plotted the data against a normal probability scale, and found
the square root of the
variance estimate was linear, with a mean of 33.515, a standard
deviation of 6.8658, and a
regression coefficient of .9993

Thus, with a fair degree of confidence, the probability that a given
estimate of the variance, x, is
not exceeded can be said to be P(sqrt(x),m,sig), where P(v,m,s) is the
normal probability
function.

Very neat, but very hard to explain. It gets even worse, I fear. Note
that 6.8658^2/33.515 equals
the square root of 2 to within .27%! If that is not enough, note that
1169.877/468.119 equals
the square root of 2*pi to within .3%

I really love studies that result in more questions than answers.

Earl