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Toni Marquard
2007 All rights reserved
Will I play the lottery? Probably.
Will I win the lottery? Probably not.
What do we really mean when we say “probably,” “most likely,” or “chances
are?” What exactly is probability?
The purpose of this lesson is to help you understand probability. To do that, we
are going to study games of chance and compare our chances of winning among
the various games.
Ready to roll? You bet!
We’ll begin by studying the simplest games of chance first.
One of the simplest games of chance is tossing a coin.
As long as the coin is “fair,” and the toss is “fair,”
there is an equal chance of the coin landing
with heads facing up as with tails facing up.
Those are the only two possibilities that can happen.
Let’s suppose that you and I play a simple game. You flip a coin and I will call
“heads” or “tails.” If the coin comes up as I call it, I win. If not, you win.
What is the probability that you will win?
Ready? I call “heads.” Now you toss the coin. . .
Heads! I win!
But the question was, “What is the
probability that you will win?
Well, now that the coin has been
tossed, there is 0 probability that you
will win. But before the coin was
tossed, tails could have come up with
the same probability as heads.
There were two possible outcomes for the toss, and after I called “heads,” you
had one chance out of the two possible outcomes of winning the toss.
So the probability that you would win was ½. That means that for every two
times the coin is tossed, you would probably win one of those two times.
Now let’s toss two coins. . .
In this game, let’s suppose you “win” whenever BOTH coins come up “heads.”
Now what is the probability that you will win?
Careful! You must think about this before you answer. There are 4 possible
outcomes for the two coins – both could come up heads, the first could come up
heads and the second tails, the first could come up tails and the second heads, or
they could both come up tails.
H
H
H
T
T
H
T
T
Note that two heads is one possibility out of four.
So the probability that you win is ½ on the first coin and ½ on the second coin.
Thus, the total probability that you would win is ¼. In other words, if the coins
were tossed 4 times, you would probably win one of those 4 times.
Coin tossing is an independent event; that is, the tossing of one coin does not
depend on the outcome of the toss of the other coin.
Sometimes the outcome of one event depends on the other. For example,
suppose we have eight disks in the four colors shown below. And suppose
that we place the 8 disks in a bag and have you draw 2 disks from the bag
without being able to see the disks as you draw them. What is the
probability that both disks that you draw will be yellow?
Now you might think that the probability that both disks are yellow is simple –
There are 2 yellow disks out of 8 disks, so isn’t the probability 2/8 or ¼?
No! To see why it’s not, it helps to think about drawing the disks one at a time.
On the first draw, there are indeed 2 yellow disks out of 8 in the bag and so the
probability of drawing a yellow disk on the first draw is ¼. However, on the
second draw, assuming you drew a yellow disk on the first draw, there is now
only one yellow disk in the bag out of 7 remaining disks. So this time the
probability of drawing a yellow disk is 1 out of 7 or 1/7.
However, if the fist disk you draw is NOT yellow, then there are still 2 yellow
disks in the bag and the probability of drawing a yellow disk on the second
draw is now 2/7.
Now let’s think about it
this way How many different pairs
could you draw?
If the fist disk
you draw is
yellow, these
are the
possible
outcomes.
If the fist
disk you
draw is pink,
these are the
possible
outcomes.
If the fist disk
you draw is
green, these
are the
possible
outcomes.
If the fist
disk you
draw is blue,
these are the
possible
outcomes.
There’s a 1 in 4 chance of getting yellow on the first draw, but it looks as
though there is a 1 in 16 chance of drawing two yellow disks.
This DEPENDENT stuff gets confusing, so for now, let’s just stay with
INDEPENDENT events like tossing a coin or rolling a die.
We’ll come back to dependent events later.
What is the probability of rolling a one on a
singe die?
Well, there are 6 faces on a die, so the probability of
rolling a one would be 1 in 6.
What if you had 2 dice? What is the
probability of rolling a one on both? This
is called “snake eyes” in the gambling
world.
Obviously, the 2 dice roll independently, so the probability of each one coming
up with a one is 1/6 for each die. But now there are many more possible
outcomes than 6. Each die has 6 possible outcomes. So this time there are 36
possible outcomes all together. Therefore, there is a 1/36 chance of getting
“snake eyes.”
Now let’s play the Lottery! We’ll start with the simplest lottery game first –
the 3-digit number. Let’s say that you have a favorite 3-digit number – 123.
What is the probability that your number will “hit”? In case you don’t know
how the 3-digit lottery works, there are three machines each holding 10 balls
numbered 0 through 9. The balls are constantly jumbled so that when a
trigger is released, one ball comes out of each machine at random. Three balls
are independently chosen, so each number has a 1 in 10 chance of popping up.
So the probability of your number coming up is as easy as 1-2-3!
A 1/10 chance that the first ball will be a 1,
A 1/10 chance that the second ball will be a 2,
A 1/10 chance that the third ball will be a 3.
But how many possible outcomes are there altogether? LOTS! An easy
way to compute the answer when events are independent is to multiply the
separate probabilities together. So we have 1/10 times 1/10 times 1/10, or
1/1,000. Your chances of winning are one in a thousand. That’s if you play
it straight. . .
The 3-digit lottery allows you to play a number straight or boxed. If you play
it straight, it must come out in the order you have selected; in this case, 123.
If you play it boxed, you win no matter what order the numbers come out,
although you do not win as much as when you play it straight. If you play 123
boxed, you win if 123 comes up, or if 231 comes up, or if 321 comes up, or if
213 comes up, or if 132 comes up, or if 312 comes up. So your chances of
winning are now increased to 6 in 1,000. That’s 6 times better than playing it
straight.
Straight
Boxed
Now what if you played a triple number such as 444?
Would there be any advantage to boxing it this time? Obviously not. So if
you’re going to play a triple number, play it straight. But what is the
probability of winning with a triple?
There’s still a 1/10 probability of each ball coming up a 4, but this case is
similar to playing it boxed since the order obviously doesn’t matter. The only
difference is that there are only 3 ways it could come up boxed instead of 6
since there is only one digit in the number instead of 3 different digits.
4
4
4
So every triple has the same probability of coming up – 3/1000. In other
words, if you played the number 444 a thousand times, you would probably
win 3 of those times.
So now what’s the probability of winning
on the 4-digit number?
Applying the same principles, there’s a 1/10 chance on each individual number
coming up, so the probability of the number 1234 coming up is 1/10 x 1/10 x
1/10 x 1/10, or this time, 1/10,000.
That’s 10 times less likely than winning the 3-digit number. Playing any 3digit number straight, you would probably win once if you played that number
a thousand times. Or, looking at it another way, if you played a thousand
different numbers straight, 123 would probably come up just once. Using a 4digit number, you would have to play that number 10,000 times before you’d
be likely to win. Or again, if you played 10,000 different 4-digit numbers
straight, 1234 would probably come up once.
What about a quadruple coming up such as 4444?
If a triple has the probability of 3/1000, a quadruple should have the
probability of 4/10,000.
Next, what about the 6-digit number?
Ah, here’s where the game takes a nasty twist because the rules change.
Instead of using 10 balls numbered 0 through 9, the 6-digit lottery uses 30
or more balls. Also, all 6 balls
are chosen from the same machine,
not 6 different machines.
Let’s start this game
with 30 balls numbered 1 – 30.
Let’s choose the following
6-digit number as “our” number: 5, 10, 15, 20, 25, 30
What is the probability that we will win?
OK, this time we’re choosing 6 balls out of 30. Is this an independent
event or a dependent event?
Because the balls are chosen from the same machine, this is now a dependent
event. Looks like we just can’t avoid those tricky dependent events when
dealing with probability!
The first ball is easy to deal with. Order is not important, so on the first ball
we have 6 chances in 30 of one of our six balls coming up. So far so good.
Now the next ball depends on what happened with the first ball. There are two
different scenarios:
First Scenario – one of our six
balls comes up:
Second Scenario – the first ball is not
one of our six choices:
Now on the next ball we have 5
chances out of 29 balls
remaining.
So now we still have 6 chances out of
29 remaining balls; however, keep in
mind that there are now only 5 more
balls to be drawn.
Let’s carry this another step further:
Each of the two preceding scenarios creates two more scenarios:
First ball was one of our
numbers.
First ball was not one of our
numbers.
Second ball
is also one of
our numbers.
Second ball is
not one of our
numbers.
Second ball is
one of our
numbers.
Second ball is
not one of our
numbers.
Now we
have 4
chances out
of 28
remaining
balls.
Now we have
5 chances out
of 28
remaining balls
but there are
only 4 more
balls to be
drawn.
Now we have
5 chances out
of 28
remaining balls
but there are
only 4 more
balls to be
drawn.
Now we still
have 6
chances but
only 4 more
balls will be
drawn.
Next, each of the 4 scenarios above will produce two more scenarios in a
like manner. This pattern will continue until 6 balls are drawn. So what is
the probability that we will win?
Well, since all 6 balls must come up in order to win, we actually have to
follow only that first case scenario strand at each step of the process.
Recall that the probability of one of our six numbers coming up on the first
ball was 6/30. Then the probability of one of our six numbers also coming
up on the second ball was 5/29. Then the probability of one of our six
numbers also coming up on the third ball was 4/28. We said this pattern
would continue, so the probability of one of our six numbers also coming
up on the fourth ball would be 3/27, and the fifth ball would be 2/26, and
finally the sixth ball would therefore be 1/25.
Now also recall that the simplest way to combine these probabilities is to
multiply them all together. We saw that that is how it worked in the
simplest cases. So then the probability of ALL SIX of our chosen numbers
coming up is: 6
5
4
3
2
1
720
x
x
x
x
x

30 29 28 27 26 25 427,518,000
This is equal to 0.000001684. So our chances of winning such a six-digit
lottery are slightly better than 1 in a million!
Now consider the fact that most State Lottery games actually use more than 30
balls, and you begin to understand just how slim your chances are of winning
the really big money.
OK, now let’s play cards!
Before we get started, there are a few facts you need to know about a standard
deck of cards. First, there are 52 cards in a deck. There are 4 different suits –
hearts, spades, diamonds, and clubs. Each suit consists of an ace, a jack, a
queen, a king, and cards numbered 2 through 10. That’s 13 cards per suit.
The first game we’re going to play is actually not a game at all, but just an
exercise to make you think about probability.
Suppose I deal you 5 cards at random from a shuffled deck. What is the
probability that one of the 5 cards you are dealt is the Ace of Spades?
This is a very easy question as there is only one Ace of
Spades in a deck, and a deck has 52 cards. Now you were
dealt 5 cards, so you have 5 chances in 52 of getting the
Ace of Spades. Simple.
Now what is the probability that you would have been dealt five cards of
the same suit, say spades?
Let’s think – there are 13 spades in the deck, and you are being dealt 5 cards.
The chance of getting a spade would be 13 out of 52, and being dealt 5 cards
would increase your chances of getting a spade to 5 times that. But the
question asked for the probability that ALL 5 cards are spades.
This sounds kind of like the lottery number game, so we must ask whether
the spade question represents a dependent or an independent event. Since
we are drawing from the same deck and not replacing the cards as we draw
them, you guessed it – this is a dependent event.
So let’s approach it the same way as the number game. The probability of
getting a spade on the first card dealt would be 13/52 since there are 13 spades in
the deck. The probability of the next card also being a spade is now 12/51. The
next card then has a probability of 11/50, then the next card 10/49, and finally
the last card, 9/48.
13 12 11 10 9
154,440
x x
x
x

52 51 50 49 48 311,875,200
This is equal to 0.000495 rounded to
millionths place. So the chances are
495 in a million of being dealt 5 cards
of the same suit. (Note: in Poker this is
called a “flush.”)
Here’s an interesting observation – that’s hundreds of times better than winning
the six-number lottery!
In poker, a royal flush is a ten, jack, queen, king, and ace of the same suit.
What are the chances of getting a royal flush? Try answering this question
yourself. If you need to see the solution with an explanation, click on these
words:
ROYAL FLUSH
Before proceeding any further, you may have noticed a pattern when dealing
with dependent events. Recall that the calculation for figuring out the
probability of all six of our chosen numbers coming up in a 30-ball lottery
game was: 6
5
4
3
2
1
720
x
x
x
x
x

30 29 28 27 26 25 427,518,000
Then the calculation for figuring out the probability of being dealt a 5-card
hand of all spades was: 13 12 11 10 9
154,440
x x
x
x

52 51 50 49 48 311,875,200
Do you see a pattern? The pattern is rather obvious, but describing it may not
be. Let’s start describing the pattern this way:
Looking at both calculations, we can say that the series of denominators
begins with the total number of elements in the game; 30 balls in the case of
the lottery, and 52 cards in the card game.
6
30
x
5
29
x
4
28
x
3
27
x
2
26
x
1
25

720
13
427,518,000
52
x
12
x
11
51 50
x
10
49
x
9
48

154,440
311,875,200
Then we can say that the number of fractions to be multiplied seems to be
determined by the number of balls selected or cards drawn. Six balls are
selected in the lottery game and there are six fractions in that calculation; five
cards are dealt in the card game and there are five fractions in that calculation.
Notice that the playing card calculation starts with 13 which is also the number
of spades in the deck. There is no such similar comparison for the lottery balls
as they are numbered successively from 1 to 30 so each ball is different in that
sense.
Keep these patterns in mind as we proceed.
Perhaps one more example would help to clarify this pattern we are
looking for. In a five-card hand, what is the probability of being dealt
four of a kind in any suit? That means you could have 4 twos or 4
threes or 4 fours or 4 anything as long as they are in the same suit.
The fifth card could be anything in any suit.
The first card could, of course, be anything. Probability enters once the first
card is dealt. Whatever it is, three of the four remaining cards to be dealt must
match that first card. Now since the second card dealt must match the first,
there are exactly three cards left in the remaining deck of 51 that qualify. So the
probability that the second card will match the first is 3/51. For the third card,
there are now only two cards that will match the first out of the remaining deck
of 50. So the probability that the third card will match is 2/50. Next, there is
only one card remaining out of the 49 cards left that could match, so the
probability of this card matching is 1/49. What about the fifth card? Since the
fifth card could be anything, the probability of it “matching” the first card can
be thought of as 1/1; that is, no matter what comes up, it wins. All in all, we
need only 4 cards to match. So let’s put it all together:
3 2
1
6
x
x

51 50 49 124,950
This represents the probability of being dealt
four-of-a-kind.
How does this last example help clarify the pattern we are looking for? Well,
remember that we said the series in the denominator begins with the total
number of elements in the game. We start with 51 this time instead of 52
because that first card could have been anything. So there actually are only 51
cards to start with when we begin considering probability.
Next, recall that we said the number of fractions in the calculation seems to be
determined by the number of cards drawn. Although we are dealing 5 cards,
this time we had a card to “play with” in that only 4 of the 5 cards dealt had to
match. Also, the first card could have been anything at all. So we really are
only concerned about 3 of the 5 cards dealt when we consider probability.
Finally, the series of numerators is the trickiest part. Our last two examples
indicated that it starts with the number of things in the subgroup. Thirteen
spades in the flush of spades, one ball of each kind in the lottery game, and this
time, four-of-a-kind. So why start with 3 and not 4?
You probably guessed it – the first card could be anything, so there were
actually only 3 possible cards in the deck that could make us “win.”
Is this beginning to make sense? See if you can try this next example on your
own to test whether or not you really understand the pattern:
What is the probability of being dealt a straight flush? A straight flush consists
of 5 cards all of the same suit in chronological order.
Here is an example:
Click on the following words to see the solution:
STRAIGHT FLUSH
So far we’ve flipped coins, tossed dice, played the lottery, and played some
poker. What’s next?
Let’s discuss one last “game” before we wind up. Let’s suppose that you are at
a party and there are 23 people including yourself at this party. What is the
probability that at least two of you will have the same birthday?
This is actually a very famous problem in probability. It
turns out that 23 is a magic number. If there are fewer than
23 people at the party, the probability of two people having
the same birthday is unlikely. With 23 or more people,
however, it is likely that two people will have the same
birthday. Let’s see why.
Let’s think this through by starting with you and figuring out the probability
that no one else at the party will have the same birthday as you or anyone else.
You can have any of 365 possible birthdays. The second person at the party
can then have any of 364 possible birthdays. The third person, any of 363
possible birthdays, and so on.
So the probability that no two people at the party will have the same birthday
can be calculated as follows:
365 364 363
343
x
x
x x
 0.4927
365 365 365
365
Do you see why? Starting with only yourself, the probability is guaranteed,
that is, 1/1 or 365/365, that no two people have the same birthday. As the
second person enters, that person has a 364/365 chance of not having the same
birthday as you. The next person then has a 363/365 chance of not matching
birthdays with you or the second person. This pattern continues down to the
23rd person who would have a 343/365 chance of not matching any other
birthday in the room. Thus, there is a slightly less than 50-50 chance that no
two birthdays would match. Therefore, there is a slightly greater than 50-50
chance that they would match. This means that when there are 23 people in
the room, it is slightly more likely that two of them will have the same
birthday than not. Do you see that if there were fewer than 23 people in the
room, the probability would tend toward none of them having the same
birthday? So the next time you are in a group, count the people, and if there
are at least 23, bet someone that two of you have the same birthday. Chances
are you’ll win the bet!
Are you wondering why we computed the probability that no two birthdays
would match instead of computing the probability that they would? If so, think
about this – any two people could match birthdays. It would not be too difficult
to calculate the probability that someone else in the room has the same birthday
as any one person, say yourself, but then you would have to account for every
other person in the room. There’s a difference between the probability that
someone else in the room has the same birthday as you and the probability that
two people have the same birthday.
Probability definitely makes you think, but it can be
lots of fun. Here’s a thought-provoking question for
you: Is probability an exact science? In other
words, if the probability of getting “heads” when
tossing a coin is ½, if you toss 10 coins, will exactly
half of them always come up “heads”? Try it and
see. Most likely, any time you toss 10 coins, they
will probably NOT come up with exactly 5 heads
and 5 tails. So what are we saying?
Are we saying that probability is not an exact science and therefore it is useless
in predicting outcomes of events such as games of chance? Obviously not.
Even though we may not get exactly 5 heads and 5 tails each and every time
we toss 10 coins, it is still true that upon tossing 10 coins, half of them will
probably come up heads and half tails. Here’s the point – the more coins we
toss, the more likely it will be that half are heads and half tails. Prove it by
tossing more and more coins. The more coins you toss, the closer you will
come to exactly half of them being heads and half tails.
This is something called “theoretical probability.”
Here’s a good definition of theoretical probability:
Theoretical probability is the ratio of the number of
ways the event can occur to the total number of
outcomes.
When tossing a coin, what are the total number of outcomes? Two, of course –
heads or tails. So the probability of getting the event called “heads” is ½.
When rolling a die, what are the total number of outcomes? There are 6
possible outcomes – you could roll a one, two, three, four, five, or six.
What if you wanted to know the probability of rolling an odd number – then
what are the total number of outcomes? Is it three because there are only three
odd numbers – 1, 3, or 5? Or is it still six?
To be accurate with our definition of theoretical probability, we need to
distinguish “outcomes” from the set of all possible outcomes. Mathematicians
call the set of all possible outcomes the “sample space.” Notice that our
definition of theoretical probability said it is the ratio between the number of
ways the event can occur to the total number of outcomes, which is the sample
space. So the probability of rolling an odd number on a single die would be
3/6, or ½. There are 3 ways the event called “odd” can happen, and the
sample space for a die consists of six possible outcomes.
Let’s take one more look at playing cards. What is the probability of drawing a
black card? According to the definition of theoretical probability, there are 26
ways the event called “black card” can occur since half the cards in a deck of
cards are black. The sample space is 52 since there are 52 cards in a deck. So
the probability of drawing a black card is 26/52 or ½.
Theoretical probability is not such a hard concept to understand. What other
kind of probability is there? Good question! The other kind of probability is
called “empirical probability.” Here is a good definition of empirical
probability:
Empirical probability is an estimate that an event will happen based on how
often the event occurs after collecting data or running an experiment in a large
number of trials.
So how is this different from theoretical probability? Another good question!
After all, didn’t we say that when tossing 10 coins, even though exactly half
may not actually come up “heads” as theoretically predicted, the probability is
still ½ that 5 of them will come up “heads”?
To better understand, let’s conduct an experiment. Obtain 100 coins of the same
denomination. Create a chart like the one below:
Number of Coins
Number of Heads
Number of Tails
Ratio of
heads/tails
5
10
15
20
25
.
.
.
100
Continue your chart to 100. If you prefer to count by tens instead of fives to
save time, you may. The goal is to notice that the more coins you toss, the
closer the number of heads to tails should be. The ratio of heads to tails
should get closer and closer to 1. Therefore, we can say that with the
collection of more data, the empirical probability of a particular outcome
approaches the theoretical probability.
If you couldn’t take the time to conduct the last experiment or if you just don’t have
100 coins, you could do one of two things now:
You could either toss the same coin over and over again, or you could click on the
following link which will take you to a simulated penny toss. There you can collect
the data you need without having to actually toss a coin in order to fill in the chart on
the previous page.
http://academic.pgcc.edu/~ssinex/excelets/flipping_pennies.xls
You may have to minimize this presentation in order to see the simulation application.
If it does not open automatically, copy and paste the link into your web browser.
It is highly recommended that you take the time to conduct this experiment, either in
actuality or simulation, in order to learn an important concept regarding theoretical
versus empirical probability. The point is that the greater the number of trials you
conduct, the closer the experimental or empirical probability will come to the
theoretical probability.
Now that you’ve learned something about probability, it may not improve
your chances of winning when you play games of chance, but you should
certainly understand more clearly why you lost. And if you do win, you
should now understand just how lucky you are!
May you have plenty of luck!
For extra practice with some interesting probability
problems, click HERE.
ROYAL FLUSH
A royal flush is a ten, jack, queen, king, and ace of the same suit.
To calculate the probability of being dealt a royal flush, let’s consider these
facts:
There are 4 different suits.
There are 13 cards in a suit.
Each suit contains one ten, one jack, one queen, one king, and one ace.
The first card you are dealt determines the suit for the royal flush.
First, what is the probability of getting the ten, jack, queen, king, or ace in
any suit with the first card?
Since we can have any one of these 5 cards come up in any suit on the first
card, there are 20 “good” cards out of 52, giving us a 20/52 probability on
the first card.
Continue to next slide. . .
After the first card, we now have a problem like the 30-ball lottery game in
that the probability of these 5 particular cards coming up is similar to the 6
particular numbers coming up. Since our first card had to be a ten, jack,
queen, king, or ace, we now have only 4 “good” cards out of 51. So the
second card has a 4/51 chance of giving us a royal flush. The next card
would then have a 3/50 chance. The fourth card would be 2/49, and the
last card 1/48. Putting it together, the probability of a royal flush would
be:
20 4 3
2
1
480
x x x
x

 0.000001539
52 51 50 49 48 311,875,200
Click here to return to the lesson.
Straight Flush
As in the previous games, the first card can be anything. Subsequent cards
must then be of the same suit and all 5 cards must be sequential. It helps
if we use a particular card to start with. Let’s suppose the first card dealt
is the 5 of diamonds. What are the possible cards that would have to
come up next?
Any one of these 4 cards,
or any one of these 4,
or any one of these 4,
or
or any one of these 4,
or any one of these 4.
So there is a total of 20 cards which could come up on the second card dealt
that would enable us to “win.” What about the third card dealt? Well if
there was a 20/51 chance of a “win” on the second card, the third card must
also be one of these 20, but since we already were dealt one of the 20 on the
second card, the third draw must have a 19/50 chance of a “win.”
Continuing the pattern, we have: 20 19 18 17
116,280
x
x
x

 0.01939
51 50 49 48 5,997,600
Check to see that this also follows the “pattern” of the other hands we’ve
studied.
Click here to return to the lesson.