



A
lottery works something like this.



There are
36 ping pong balls in a box. 


They are
numbered thru 36. 


The
lottery people choose 5 of he ping pong balls 


The
numbers on those ping pong balls are the winning numbers. 





Before
the winning numbers are chosen, 


people
pay a dollar or two to buy a ticket with 5 numbers on it. 


You can
chose your own numbers, 


or
have the machine chose them for you. 


If your
ticket has the 5 numbers chosen on it, 


you win
big bucks. 


If your
ticket has most of the 5 numbers chosen on it, 


you win a
small prize. 





The
first, and most important, question is 


What is
the chance that a ticket has the winning numbers on it 





Here's
how it works. 


A ticket
has 5 of 36 numbers on it. 











Five
numbered ping pong balls are drawn out of the container. 


Once
drawn, a ping pong ball is not replaced in the container. 


That
means the number of ping pong balls in the box 


keeps
getting smaller. 


The
chance of a certain yet to be drawn number being drawn 


depends
on how many numbers are left in the box. 


Because
of that, the chance of choosing any particular number 


DEPENDS
on how many numbers are left in the box. 


That
makes lottery number choice different from dice rolls. 


Lottery
number choices are DEPENDENT EVENTS. 








The
chance of the first winning number drawn, 


being on
of the 5 numbers on the ticket, is ... 











So say
the first number IS on the ticket. 


Great,
now what. 


There are
35 ping pong balls left to chose from. 


(They
don't put back the ones they choose.)



There are
4 more numbers on the ticket 


that have
not been chosen yet. 


That
means there are 4 of 35 numbers 


that
will keep the ticket in the game. 





The
chance of the second number chosen



also
being on the ticket are ... 











Now,
let's start putting things together. 


What is
the chance that the first 


AND the
second number chosen are on the ticket?



Since
we used AND, we need to multiply these two together ... 











This
fraction can be simplified to ... 











One thing
to be sure you see here. 


The
chance of having both of the first two winning numbers, 


is much
less than the chance of having either 


of
the two first winning numbers individually. 





Now say
the ticket does happen to have the first two winning numbers,



what is
the chance that it ALSO has the third winning number? 





There
are three unchosen numbers left on the ticket. 


There are
34 numbers still left to chose from. 


The
chance that one of the three unchosen numbers on the ticket 


will be
chosen next is ... 











OK, stop
and look at something. 


The
chance of the ticket having the first winning number 


was about
14%. 


The
chance of the ticket having the second winning number 


if it had
the first, was about 11%. 


The
chance of the ticket having the third winning number 


if it had
the first two, was about 9% 


The
chance of the ticket having each additional winning number gets
smaller. 


The
chance of the ticket having all the first three 


winning
numbers selected. 


That
is, the first AND the second AND the third numbers, is ... 











This
simplifies to ... 











Say
the first three numbers were on the ticket. 


What is
the chance that the fourth number is also on the ticket? 


There are
two numbers left on the ticket. 


There are
33 numbered ping pong balls left in the lottery box. 


The
chance that the next number chosen 


is
on of those on the ticket is ... 











The
chance of the ticket having all 


of the
first 4 numbers drawn is ... 











120
out of 1,545,720 simplifies to ... 











OK,
last time through. 


The
ticket has all of the first 4 winning numbers. 


What
is the chance of it having the last one too? 





There is
1 number left on the ticket. 


There are
32 numbers left in the box. 


The
chance of the number on the ticket 


being
chosen from the box is ... 











So
the chance of all 5 winning numbers 


being
on the ticket is ... 











If
you divide out that big nasty fraction, you get ... 











So
that's it. 


There
is less than a 1 in 400,000 chance 


of
picking the right 5 numbers out of 36. 





If
you had to pick more than 5, or if there were more than 36 numbers, 


the
chance of having all the right numbers would be even less. 





It
would be nice to have a formula that worked on problems like this. 


That
way you wouldn't have to go through this long process 


if
we changed the problem just a little bit. 





The good
news is, there is one. 


Take a
look at the multiplication we ended up with ... 











The
term on top is a factorial. 


The
term on the bottom is the start of a factorial. 





We
can turn the term on the bottom 


into
something made of factorials with a little trickiness. 


We
get ... 











36! will
have all the numbers from 1 to 36 multiplied together. 


31! will
have all the numbers from 1 to 31 multiplied together. 


Since
everything is multiplied, you can cancel. 


That
leaves you with 36 × 35 × 34 × 33 × 32 in the denominator. 





We could
stop here, but that denominator just looks too nasty. 


We can
clean up this fraction a bit, by multiplying it by another name for
1 ... 











Look
at what we have left, 


if
we say the total numbers selected is "r" 


and all
of the numbers there are is "n" 


we can
write this as ... 











Example: 





What
are the odds of a lottery ticket having all of the winning numbers, 


if
there are 6 numbers selected as winning numbers from the numbers 1 
45? 

















Factorials
as big as 39! or 45! will be very big numbers when multiplied out. 


If
you do this one on your calculator as it sits, 


you
will see these numbers shown in scientific notation. 


Since
everything is multiplied together, 


the
factors of 45! from 39 down to 1 in the denominator 


can be
cancelled with the 39! in the numerator. 


That will
leave ... 











Multiplying
this out we get ... 











This
is about 1 in eight million. 


Notice
that we just raised the number of ping pong balls in the box by a
few, 


and
raised the numbers selected from 5 to 6, 


and
it is now 20 TIMES less likely that a ticket has all the winning
numbers. 





Logic
Riddle (in
two parts): 





Part
1: In the above example, there was about a 1 in eight
million chance 


that all
6 winning numbers would be on any ticket. 


How many
different unique sets of 6 numbers are possible? 


(Number
order does not matter.) 





Part
2: What is a formula
for calculating 


the
number of possible ticket number combinations? 


(Number
order does not matter.) 








Logic
Riddle Answer ... 





Part
1: Selecting the 6 winning numbers is really no different 


than
selecting 6 numbers on any ticket. 


If the
chance of having the winning numbers on a ticket 


are about 1 in
eight million, that is
because there are 


about 8 million different number
combinations. 


Notice
that this is the inverse of the odds that all the winning numbers 


are
on any one ticket. 





Part
2: If the number of
possible number combinations is the inverse 


of
the odds that any one ticket has all 6 winning numbers, 


then
it should not be a surprise that the formula 


for
the number of possible number combinations 


is the
inverse of the formula for all the winning numbers 


being on
any particular ticket. 


That
is ... 











Where
n is the number of ping pong balls in the box 


and
r is the number selected. 





Bulletin: 


This
formula, in both variations, is used A LOT in probability 





copyright 2005 Bruce Kirkpatrick 
