# MMX Technology

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Expected Value
CSCE 115
Revised Nov. 29, 2004, May 2, 2005
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Probability
 Probability is determination of the chances
of picking a particular sample from a
known sample.

 Notation:  if A is some event, the
probability of A is
P(A)
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Probability
 Probability of  success (when the events are
equally likely) is
Number of successful outcomes
Number of possible outcomes
 Example: If 1 student is picked at random from
a class of 7 woman and 13 men, what is the
probability that the student is a woman?
           P(woman) = 7/(13+7) = 7/20
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Probability
 Non-example:      If you roll two dice and add
the spots the possible outcomes are 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12. What is the
probability of rolling a 2?
 Is the following correct?
There is 1 success (getting a two) out of 11
possibilities (2, 3, …, 12) so the probability
is 1/11
 Why?
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Independent events
 Two   events are independent if the way the
first event happens does not affect the way
the second event happens.
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Example: Independent events
 Put 3 red balls and 2 green balls in a bag. Event
1: select a ball at random from the bag and
determine its color. Put the ball back.
Event 2: Select a second ball at random.
Events 1 and 2 are ____________
 Event 3: select a ball at random and set it aside.
Event 4: select a second ball at random. Events
3 and 4 are ____ _____________
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First fundamental rule:
 The probability that something does not
happens is 1 - the probability it happens
            P(not A) = 1 - P(A)
 Example: The probability of picking a man
from the class of 7 women and 13 men is
     1 - (7/20) = 20/20 - (7/20) = 13/20
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Second fundamental rule
 Iftwo events are independent, the
probability that both A and B happen is
P(A and B) = P(A) * P(B)
 Example: We randomly select a ball from a
bag with 3 red and 2 green balls. We put it
back and draw again. The probability that
both balls are red is
 P(red, red) = (3/5) * (3/5) = 9/25
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Example: Role 2 dice
 Suppose   that we have a red die and a blue die. We
roll and sum. What are the possible outcomes?
    RB RB RB RB                 RB RB
1,1 1,2 1,3        1,4     1,5 1,6
2,1 2,2 2,3        2,4     2,5 2,6
3,1 3,2 3,3        3,4     3,5 3,6
4,1 4,2 4.3        4,4     4,5 4,6
5,1 5,2     5,3    5,4     5,5   5,6
6,1 6,2     6,3    6,4     6,5   6,6
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Example: Role 2 dice
 P(2) = 1/36               P(8) = 5/36
 P(3) = 2/36 = 1/18        P(9) = 4/36 = 1/9
 P(4) = 3/36 = 1/12        P(10) = 3/36 = 1/12
 P(5) = 4/36 = 1/9         P(11) = 2/36 = 1/18
 P(6) = 5/36               P(12) = 1/36
                P(7) = 6/36 = 1/6
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Probability of rolling 2 dice

0.2
Probability.

0.15
0.1
0.05
0
2   3   4   5   6   7   8   9 10 11 12
Sum of two dice
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A children's game with spinner
The spinner is used to
3    3
determine how far you      7            5
move. What is the
probability of each       3             2
move?
P(2) = 1/10                8            6
P(3) = 4/10
3    6
P(5) = 1/10
How far, on the average,
P(6) = 2/10
P(7) = 1/10        do you expect to move
P(8) = 1/10        each time you spin?
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A children's game with spinner
 Just averaging  the possible the possible outcomes
(2 + 3 + 5 + 6 + 7 + 8) = 31 = 5.1667
6               6
is _____ correct because the various values are
not equally likely.
 A correct way is
(2 + 3 + 3 + 3 + 3 + 5 + 6 + 6 + 7 + 8) = 46
10                     10
= 4.6
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A children's game with spinner
 (2 +3 + 3 + 3 + 3 + 5 + 6 + 6 + 7 + 8)
10
 = (2 +3*4 + 5 + 6*2 + 7 + 8)
10
= 2 * 1 + 3 * 4 + 5 * 1 + 6 * 2 + 7 * 1 + 8 * 1
10      10       10       10     10    10
 = 2*P(2) + 3*P(3) + 5*P(5) + 6*P(6) + 7*P(7)
+ 8 * P(8)
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Expected value
 Suppose that a certain experiment X could result
in the values of {a, b, c, …, k} and the
probabilities of these outcomes are P(a), P(b),
P(c), …, P(k). The expected value is
E(X) = a * P(a) + b * P(b) + c * P(c)
+ … + k * P(k)
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Example: Expected value
 Recall   the kid's spinner game
P(2)  = 1/10         P(3) = 4/10
P(5) = 1/10         P(6) = 2/10
P(7) = 1/10         P(8)= 1/10
E(spin) = 2 * .1 + 3 * .4 + 5 * .1
+ 6 * .2 + 7 * .1 + 8 * .1
          = 4.6
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Example: Expected value
 Experiment:    You flip a coin and get 1 point
of a head and 0 points for tail
 P(head) = .5,    P(tail) = .5
 E(flip) = 1 * .5 + 0 * .5 = .5
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Example: Expected value
 Experiment:   You roll one die and count
dots.
 P*(1) = 1/6, P(2) = 1/6, … P(6) = 1/6
 E(roll) = 1*(1/6) + 2*(1/6) + 3*(1/6) +
4*(1/6) + 5*(1/6) +6 *(1/6)
=      1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6
= 21/6 = 3.5
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Example: Expected value
 Experiment:  You roll two dice and count dots
 P(2) = 1/36      P(6) = 5/36       P(9) = 4/36
P(3) = 2/36      P(7) = 6/36       P(10) = 3/36
P(4) = 3/36      P(8) = 5/36       P(11) = 2/36
P(5) = 4/36      P(12) = 1/36
 E(two dice) =
2*(1/36) + 3*(2/36) + 4*(3/36) + 5*(4/36)
6*(5/36) + 7*(6/36) + 8*(5/36) + 9*(4/36)
10*(3/36) + 11*(2/36) + 12*(1/36)
 = 252/36 = 7
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Example: Expected value
 John proposes    that the charity bazaar sell tickets
for \$2. The player rolls 2 dice. The play wins \$12
if the roll is a 2 or a 8. On the average, how much
will the charity win each time a player rolls the
dice?
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Example: Charity bazaar
 Solution  1:
 Outcomes are P(2) = 1/36,      Prize(2) = \$12
P(8) = 5/36, Prize(8) = \$12
P(other) = 30/36, Prize(other) = \$0
E(Prize) = \$12*(1/36) + \$12*(5/36) +
\$0*(30/36) = \$72/36 = \$2
 Cost of ticket - Expected value of prize
= \$2 - \$2 = 0
 The charity does not expect to win any money with
this game.
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Example: Charity bazaar
 Solution 2:
 Outcomes   are charity wins \$2 or loses \$10
 P(\$2) = 30/36 = 5/6
P(-\$10) = 1/36 + 5/36 = 6/36 = 1/6
 E(winnings) = \$2 * (5/6) + (-\$10) * (1/6)
\$10/6 + (-\$10/6) = 0
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Fair Game
 A fair  game is a game where the expected value
of winning is 0
 Fair games are highly desirable when play with
 Fair games are not desirable for organizations
trying to earn money by offering games of chance.
 Casinos would go out of business if they had fair
games.
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Example: Charity bazaar
 John  proposes that the charity bazaar sell tickets for \$2.
The player rolls 2 dice. The player wins \$12 if the roll is
a 2 or a 11. They win nothing otherwise. On the
average, how much will the charity win each time a player
rolls the dice?
 P(-\$10) = 1/36 + 2/36 = 3/36 = 1/12
P(\$2) = 1- 1/12 = 11/12
 E(winnings) = (-\$10) * (1/12) + (\$2)*(11/12)
= (-\$10 + \$22)/12 = \$12/12 = \$1
 The game is ___ fair. This is desirable for _____.
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Odds
 Unfortunately   popular slang uses “odds” in at
least 3 different ways.
 It may indicate the payoff if you win a bet
 It may be a synonym for probability
– This is used by the Washington State Lottery
– It is used in many popular articles about odds
may indicate the ratio of the probability of
 It
winning to the probability of losing
– This is the definition we will use
– This the definition is the one normally sees in math and
statistics books
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Odds of Winning
 Suppose the  probability of winning is p and
probability of losing is q = 1 - p. Then the
odds of winning are p:q.
 We treat p:q as a fraction and normally
multiply and divide both parts to clear of
fractions and to remove common factors.
 The odds of losing are q:p.
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Example: Odds
 In therevised charity bazaar game, the
probability of a player winning is 1/12.
 The probability of losing is ____
 The odds of winning are 1/12:11/12 or ____
 The odds of losing are ______
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Example: Odds
 You rolltwo dice and win if you roll a 9.
 The probability of winning is 4/36
 The probability of losing is ____
 The odds of winning are 4/36:32/36
= 4:32 = ___ : ___
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Example: Odds
 The odds of winning a game are 5:31. What is the
probability of winning and losing?
 Suppose that you played 5+31 = 36 times. You
would expected to win 5 times and lose 31 times.
 The probability of winning is 5/36.
The probability of losing is 31/36.
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 Suppose   the odds of winning are given as a:b. We
want to the probability of winning.
 Algebraically, suppose p:q = p:(1-p) = a:b
Treat ":" like it was a "/“
 p/(1-p) = a/b
 bp = a(1-p)
 bp = a - ap
 ap + bp = a
 (a+b)p = a
 p = a/(a+b)
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Example: Odds
 The odds of winning first prize in a raffle are
1:1999. What is the probability of winning?
 Suppose that 1+1999 = 2000 tickets are sold. We
would expect to win 1 time out of 2000
 The probability of winning is 1/2000.
 Using the formula: a = ___, b = _____
 The probability is p = a/(a + b)
= 1/(1 + 1999)
= _________
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Example: Raffle
 The prize list for a raffle is
 Prize              Number       Value           Odds
New car (Kia)           1      \$10,000         1:9999
TV set                 10         \$300          1:999
Meal for two           20          \$50          1:499
 Determine:
– Expected value of a ticket
– Number of tickets sold
– If all of tickets costing \$2.50 each are sold and if there is an
addition cost of \$2000 for printing and advertising, write a
budget for the sponsors
– Solution: See