Math-Prof HOME Probability Table of Contents Ask A Question PREV NEXT

Win the Lottery
Dependent Events

 

 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

Math-Prof HOME Probability Table of Contents Ask A Question PREV NEXT