Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Lottery Numbers
Name: ken
Status: student	
Age:  N/A
Location: N/A
Country: N/A
Date: N/A 


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?



Replies:
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



Click here to return to the Mathematics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory