Ask A Scientist© Mathematics Archive

name         ken
status       other
age          50s

Question -   I am between jobs, while interviewing and waiting for the
"call", I have gotten hooked on what finding out what a formula would be
for the probability of choosing random sets of numbers, within some
degree of accuracy.  Example: What would be the next set of 3 numbers to
be chosen on the "Pick 3" lotto drawing if I would to buy 10 tickets.  I
have the historical database consisting of the 1634 games played.  As you
know, there three individual pools numbers are drawn from.  Each pool
consist of numbers ranging from 0 to 9.  I've seen definite patterns
emerge, and some numbers dropping more than others.  I know this would be
a "hit or miss" proposition, but there has to be a formula that I can use
to reduce the "odds".  Would I base the results from the historical
data?  Or use a formula?  How can I figure this out?
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It depends upon what the rules of the game are:
1. If you mean that you start out with numbers "in a hat" 0,1,2,...,9, the
probability of choosing one of the 10 digits is 1/10. If you do not put the
number back in the hat, the probability of picking the second digit is 1/9,
and the third 1/8... until finally there is only a single number in the has
so the probability of selecting it will be 1/1 since you already know the
value of the other 10 digits 0,1,2,...,9.

2.    If you replace the chosen digit to the hat and are asking what is the
probability that any three will be chosen, the problem is stated, " I have
10 digits 0,1,2,...,9. How many ways can I choose any particular three (in
no special order). That is given by: N!/p!(N-p)!, where 'N' is the number of
digits ( 10 in this case ) and 'n' is the number of digits chosen. In this
case 3 the symbol "!" means factorial, that is: 3! = 1*2*3, and 5!=
1*2*3*4*5. So the probability, P, is:    P!/n!*(P-n)! =
1*2*3*4*5*6*7*8*9*10* / (1*2*3) * (1*2*3*4*5*6*7) =
8*9*10/1*2*3 = 120. So the probability of choosing any three numbers (in any
order) is
1/120. Not very good odds.

3.    It is possible to see "patterns" in such an exercise, but in the long
run each digit has an equal opportunity. There are no short cuts.

Vince Calder
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