Chance Variability
I. Box Models
We can analyze the chance variability involved in a probability experiment by
constructing an appropriate box model and using it to simulate the experiment. We
will represent each of the possible outcomes of the experiment by a number. We will
write the number on a ticket and put the ticket into a box; the probability of a
particular outcome will determine the number of tickets of that kind in the box. We
will begin the experiment by randomly selecting a ticket from the box, recording the
number on this ticket, and then replacing the ticket. We will then randomly draw
another ticket from the box, add the number on this ticket to the number from the first
draw, and again replace the ticket. We will continue this procedure for a specified
number of draws. When we have finished, then we will have the sum of all the draws.
If someone else were to perform the same experiment – same box model and number
of draws – then he would probably get a different sum for the draws. By analyzing the
probability of getting different sums of the draws, we are analyzing the chance
variability for that probability experiment.
Example 1: Construct a box model for playing RED in American roulette 100 times.
The probability of winning with RED is
The probability of losing is
18
; if you bet $1, then you would win $1.
38
20
; if you bet $1, then you would lose $1. One
38
possible box model is:
18
1
20
1
100 Draws
Note : In simulating games of chance, we use negative numbers to represent a
loss.
Example 2: Construct a box model for playing a single number (1 – 36, 0, 00) in
American roulette 25 times.
The probability of winning with a single number is
would win $35. The probability of losing is
lose $1. One possible box model is:
1
1
; if you bet $1, then you
38
37
; if you bet $1, then you would
38
1
35
1
37
25 Draws
Example 3: Construct a box model for making a PASS bet in craps 400 times.
244
(this would be
495
given to you); if you bet $1, then you would win $1. The probability of
251
losing is
; if you bet $1, then you would lose $1. One possible box model is:
495
The probability of winning in craps with a PASS bet is
244
1
251
1
400 Draws
Example 4: Construct a box model for flipping a fair coin 100 times.
1
;
2
we will put “1” on the tickets that represent getting HEAD since we are counting
1
HEAD’s. The probability of getting TAIL is ; we will put “0” on the tickets
2
that represent getting TAIL since we do not want to count TAIL’s. One possible
box model is:
Suppose that we wish to count HEAD’s. The probability of getting HEAD is
1
1
1
0
100 Draws
Note: We will use a “0 – 1” box for counting.
Example 5: Construct a box model for rolling a fair die 60 times and counting the
number of

‘s.
1
, and we will use “1” on the tickets that
6
5
represent this outcome. The probability of not getting  is , and we will
6
use “0” on the tickets that represent this outcome. One possible box model is:
The probability of getting
1
1

is
5
0
2
60 Draws
II. Expected Value and Standard Error for the Sum of Draws
Consider the following box model:
1
2
3
2 Draw
There are 9 possible sums for 2 draws:
S1  1  1  2
S2  1  2  3
S3  1  3  4
S4  2  1  3
S5  2  2  4
S6  2  3  5
S7  3  1  4
S8  3  2  5
S9  3  3  6
The average of these 9 sums is S  4 and the standard deviation is  S  4 . The
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average of all the possible sums of the draws from a box model with replacement is
called the expected value for the sums of the draws, and the standard deviation for
these possible sums of draws is called the standard error for the sums of the draws.
Instead of having to find all the possible sums of draws, we can calculate the expected
value (EV) and the standard error (SE) with the following formulas:
EV = (AVE OF BOX)(NUMBER OF DRAWS)
SE = (SD OF BOX)( NUMBER OF DRAWS )
For the box model in the example above, AVE OF BOX = 2 and SD OF BOX =
 
2 .
3
Thus, EV = (2)(2) = 4 and SE =  2  2  4 and this checks with the previous
3
 3
results.
Example 1: Find the EV and SE for playing RED in American roulette 100 times
betting $1 each time.
18
1
AV OF BOX =
SD OF BOX =
20
1
100 Draws
18(1)  20(1)
 .053
38
18(1) 2  20(1) 2
 (.053) 2  .999
38
3
EV  (.053)(100)  5.30
SE  (.999)( 100 )  .999(10)  9.99
Note: For box models which simulate games of chance, the number of draws is the
the number of plays, the average of the box is the expected amount won (usually
lost) on any one play, the standard deviation of the box is the variability for one
play, the expected value for the sum of the draws is the expected amount won
(usually lost) for all the plays, and the standard error is the variability for all the
plays. In the previous example, on any play of RED in American roulette, betting
$1 on each play, you would expect to lose about $.053 give or take about $1.00;
for the 100 plays, you would expect to lose about $5.30 give or take about $10.
Note: When the tickets in the box show only two different numbers, the SD of the
box equals
 BIGGER SMALLER  FRACTION  OF  TICKETS FRACTION  OF  TICKETS




 NUMBER NUMBER  WITH  BIGGER  NUMBER WITH  SMALLER  NUMBER
For the box model in the previous example,
18 20

 .999
38 38
SD  [1  (1)]
Example 2: Find the EV and SE for playing a single number in American roulette 25
times, betting $1 each time.
1
AV 
35
37
1
1(35)  37(1)
2

 .053
38
38
SD  [35  (1)]
1 37

 5.76
38 38
EV  (.053)( 25)  1.33
SE  (5.76) 25  28.80
4
25 Draws
Example 3: Find the EV and SE for making a PASS bet in craps 400 times, betting $1
each time.
244
AV 
1
251
1
400 Draws
244(1)  251(1)
 .014
495
EV  (.014)( 400)  5.60
244 251

 1.00
495 495
SD  [1  (1)]
SE  (1.00) 400  20.00
Example 4: Find the EV and SE for flipping a fair coin 100 times and counting
the numbers of HEAD’s.
1
AV 
1
1
0
1(1)  1(0)
 .5
2
SD  (1  0)
100 Draws
EV  (.5)(100)  50
1 1
  .5
2 2
SE  (.5) 100  5
Note: In the previous example, if a fair coin is tossed 100 times, you would expect to
get about 50 HEAD’s, give or take about 5 HEAD’s.
Example 5: Find the EV and SE for rolling a fair die 60 times and counting the
number of  ’s.
1
AV 
1
5
1(1)  5(0) 1

6
6
SD  (1  0)
1 5
  .373
6 6
0
60 Draws
EV 
1
(60)  10
6
SE  (.373) 60  2.89
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III. The Normal Approximation for Probability Histograms
Example 1: If you play RED in American roulette 100 times, betting $1 each time,
what is the probability of your winning between $10 and $20?
(1) Construct the appropriate box model for the probability experiment and find
the EV and SE for the sum of the draws.
18
AV 
1
1
20
18(1)  20(1)
 .053
38
SD  [1  (1)]
100 Draws
EV  (.053)(100)  5.30
18 20

 .999
38 38
SE  (.999) 100  9.99
(2) Determine the z-score for the significant numbers in the question using the
following facts:
(i) The probability histogram for the sum of the draws from a box model with
replacement will follow the normal curve provided a reasonably large
number of draws is taken. This is the Central Limit Theorem.
(ii) The EV locates the center of this probability histogram, and the SE
measures the spread.
99%
87%
6%
-2.53
z10 
-1.53
-5.30
0
10  (5.30)
 1.53
9.99
10
1.53
z 20 
20
2.53
20  (5.30)
 2.53
9.99
(3) Using the areas from  z to z for these z-scores, determine the area of the
shaded region. This area is the probability.
99%  87% 12%

 6%
2
2
Thus, the probability of winning between $10 and $20 is 6%.
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Example 2: If you play a single number in American roulette 25 times, betting $4
each time, what is the probability of your losing more than $10?
1
AV 
140
4
37
1(140)  37(4)
 .21
38
SD  [140  (4)]
EV  (.21)( 25)  5.25
1 37

 23.05
38 38
48.5%
25 Draws
SE  (23.05)(5)  115.25
3%
-10
-.04
48.5%
-5.25
0
z ( 10 ) 
 10  (5.25)
 .04
115.25
The probability of your losing more than $10 is about 48.5%.
Example 3: A fair coin is flipped 100 times. What is the probability of getting
exactly 53 HEAD’s?
1
AV 
1
1
0
1(1)  1(0)
 .5
2
SD  [1  0]
100 Draws
EV  (.5)(100)  50
1 1
  .5
2 2
SE  (.5)(10)  5
51.61%
38.29%
6.6%
-.7
-.5
50
0
52.5
.5
7
53
53.5
.7
z 52 .5 
52.5  50
 .5
5
z 53 .5 
53.5  50
 .7
5
Thus, the probability of getting exactly 53 HEAD’s is about 6.6%.
Note: P(53H ,47T ) 
100!
(.5) 53 (.5) 47  6.66% !!!
(53!)( 47!)
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Practice Sheet – Chance Variability
1. A fair coin is flipped 100 times. What is the probability of getting more than 58
heads?
2. A loaded coin, P(HEAD) = 0.6, is tossed 400 times. What is the probability that
the number of tails is between 150 and 165?
3. A fair die is rolled 64 times. What is the probability that the sum of the rolls is less
than 215?
4. A fair die is rolled 60 times. What is the probability of getting exactly 12  ’s?
Check your answer using the binomial formula.
5. Play RED in roulette 100 times, betting $1 each time. What is the probability of
winning more than $10? What is the probability of losing more than $10?
6. Play 4 numbers (payoff is 8 to 1) in roulette 100 times, betting $1 each time. What
is the probability of winning more than $10?
7. Play the “HOUSE SPECIAL” (5 numbers, payoff is 6 to 1) in roulette 100 times,
betting $1 each time. What is the probability of winning more than $10?
8. Play a single number (payoff is 35 to 1) in roulette 100 times, betting $1 each time.
What is the probability of winning more than $10?
9. Make a PASS bet (probability of winning is 0.493, payoff is 1 to 1) in craps 100
times, betting $1 each time. What is the probability of winning more than $10?
10. In CHUCK-A-LUCK, a dollar is bet that one of the six numbers {1, 2, 3, 4, 5, 6},
say 5, is a winner. Three dice in a cage are tossed and the number of 5’s is counted.
Your dollar is returned, plus $1 for each 5 that shows on the dice; if there are no 5’s,
you lose your dollar. Play CHUCK-A-LUCK 100 times. What is the probability of
winning more than $10?
11. Guess the answers on a 100-question, multiple choice test in which each question has
4 choices. You get 1 point for a correct answer and you lose 1 point for a wrong
3
answer. What is the probability of making at least 10 on the test?
12. Play RED in roulette 25 times, betting $4 each time. What is the probability of
winning more than $10?
13. Play RED in European roulette (same as American except there is no 00) 100 times,
betting $1 each time. What is the probability of winning more than $10?
Solution Key for Chance Variability
1. 5.5%
2. 54%
3. 25%
4. 11%
5. 6.5%, 32%
6. 29%
7. 22.5%
8. 40%
9. 13%
10. 5%
11. 4.25%
12. 22.5%
13. 10.5%
9
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