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36 Odds, Expected Value, and Conditional
Probability
What’s the difference between probabilities and odds? To answer this ques- tion, let’s consider a game that involves rolling a die. If one gets the face 1 then he wins the game, otherwise he loses. The probability of winning is 16 whereas the probability of losing is 56. The odds of winning is 1:5(read 1 to 5). This expression means that the probability of losing is five times the probability of winning. Thus, probabilities describe the frequency of a favorable result in relation to all possible outcomes whereas the ”odds in favor” compare the favorable outcomes to the unfavorable outcomes. More formally,
odds in favor = (^) unfavorable outcomesfavorable outcomes
If E is the event of all favorable outcomes then its complementary, E, is the event of unfavorable outcomes. Hence,
odds in favor = n n((EE))
Also, we define the odds against an event as
odds against = unfavorable outcomesfavorable outcomes = n n((EE))
Any probability can be converted to odds, and any odds can be converted to a probability.
Converting Odds to Probability Suppose that the odds for an event E is a:b. Thus, n(E) = ak and n(E) = bk where k is a positive integer. Since E and E are complementary then
n(S) = n(E) + n(E). Therefore,
P (E) = n n((ES))
= (^) n(En)+(En)(E)
= (^) akak+bk = (^) aa+b
P (E) = n n((ES))
= (^) n(En)+(En)(E)
= (^) akbk+bk = (^) a+bb
Example 36. If the odds in favor of an event E is 5 to 4, compute P (E) and P (E).
Solution. We have
P (E) =
and
P (E) =
Converting Probability to Odds Given P (E), we want to find the odds in favor of E and the odds against E. The odds in favor of E are
n(E) n(E) =^
n(E) n(S) ·^
n(S) n(E)
= P P^ ((EE))
= 1 −P^ P(E (E))
and the odds against E are
n(E) n(E)
1 − P (E)
P (E)
Problem 36. On a tote board at a race track, the odds for Gameylegs are listed as 26:1. Tote boards list the odds that the horse will lose the race. If this is the case, what is the probability of Gameylegs’s winning the race?
Problem 36. If a die is tossed, what are the odds in favor of the following events? (a) Getting a 4 (b) Getting a prime (c) Getting a number greater than 0 (d) Getting a number greater than 6.
Problem 36. Find the odds against E if P (E) = 34.
Problem 36. Find P(E) in each case.
(a) The odds in favor of E are 3: (b) The odds against E are 7:
Expected Value A cube has three red faces, two green faces, and one blue face. A game consists of rolling the cube twice. You pay $ 2 to play. If both faces are the same color, you are paid $ 5(that is you win $3). If not, you lose the $2 it costs to play. Will you win money in the long run? Let W denote the event that you win. Then W = {RR, GG, BB} and
P (W ) = P (RR) + P (GG) + P (BB) =
Thus, P (L) = 1118 = 61%. Hence, if you play the game 18 times you expect to win 7 times and lose 11 times on average. So your winnings in dollars will be 3 × 7 − 2 × 11 = − 1. That is, you can expect to lose $1 if you play the game 18 times. On the average, you will lose 181 per game (about 6 cents). This can be found also using the equation
3 ×
− 2 ×
We call this number the expected value. More formally, let the outcomes of an experiment be a sequence of real numbers n 1 , n 2 , · · · , nk, and suppose that the outcomes occur with respective probabilities p 1 , p 2 , · · · , pk. Then the expected value of the experiment is
E = n 1 p 1 + n 2 p 2 + · · · + nkpk.
Example 36. Suppose that an insurance company has broken down yearly automobile claims for drivers from age 16 through 21 as shown in the following table.
Amount of claim Probability 0 0. 2,000 0. 4,000 0. 6,000 0. 8,000 0. 10,000 0.
How much should the company charge as its average premium in order to break even on costs for claims?
Solution. Finding the expected value
E = 0(0.80)+2, 000(0.10)+4, 000(0.05)+6, 000(0.03)+8, 000(0.01)+10, 000(0.01) = 760
Since average claim value is $760, the average automobile insurance premium should be set at $760 per year for the insurance company to break even
Example 36. An American roulette wheel has 38 compartments around its rim. Two of these are colored green and are numbered 0 and 00. The remaining compart- ments are numbered 1 through 36 and are alternately colored black and red. When the wheel is spun in one direction, a small ivory ball is rolled in the opposite direction around the rim. When the wheel and the ball slow down, the ball eventually falls in any one compartments with equal likelyhood if the wheel is fair. One way to play is to bet on whether the ball will fall in a red slot or a black slot. If you bet on red for example, you win the amount of the bet if the ball lands in a red slot; otherwise you lose. What is the expected win if you consistently bet $5 on red?
Problem 36. Suppose it costs $8 to roll a pair of dice. You get paid the sum of the numbers in dollars that appear on the dice. What is the expected value of this game?
Problem 36. An insurance company will insure your dorm room against theft for a semester. Suppose the value of your possessions is $800. The probability of your being robbed of $400 worth of goods during a semester is 1001 , and the probability of your being robbed of $800 worth of goods is 4001. Assume that these are the only possible kinds of robberies. How much should the insurance com- pany charge people like you to cover the money they pay out and to make an additional $20 profit per person on the average?
Problem 36. Consider a lottery game in which 7 out of 10 people lose, 1 out of 10 wins $50, and 2 out of 10 wins $35. If you played 10 times, about how much would you expect to win?
Problem 36. Suppose a lottery game allows you to select a 2-digit number. Each digit may be either 1, 2, 3, 4, or 5. If you pick the winning number, you win $10. Otherwise, you win nothing. What is the expected payoff?
Conditional Probability and Independent Events When the sample space of an experiment is affected by additional informa- tion, the new sample space is reduced in size. For example, suppose we toss a fair coin three times and consider the following events:
A : getting a tail on the first toss B : getting a tail on all three tosses
Since
S = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }
then P (A) = 48 = 12 and P (B) = 18. What if we were told that event A has occurred (that is, a tail occurred on the first toss), and we are now asked to find P (B). The sample space is now reduced to {T HH, T HT, T T H, T T T }. The probability that all three are tails given that the first toss is a tail is 14.
The notation we use for this situation is P (B|A), read ”the probability of B given A,” and we write P (B|A) = 14. Notice that
P (B|A) =
P (A ∩ B)
P (A)
This is true in general, and we have the following:
Given two events A and B belonging to the same sample S. The condi- tional probability P (B|A) denotes the probability that event B will occur given that event A has occurred. It is given by the formula
P (B|A) =
P (A ∩ B)
P (A)
Example 36. Consider the experiment of tossing a fair die. Denote by A and B the follow- ing events:
A = {Observing an even number of dots on the upper face of the die} , B = {Observing a number of dots less than or equal to 3 on the upper face of the die}.
Find the probability of the event A, given the event B.
Solution. Since A = { 2 , 4 , 6 } and B = { 1 , 2 , 3 } then A ∩ B = { 2 } and therefore
P (A|B) =
1 (^63) 6
If P (B|A) = P (B), i.e., the occurrence of the event A does not affect the probability of the event B, then we say that the two events A and B are independent. In this case the above formula gives
P (A ∩ B) = P (A) · P (B).
This formula is known as the ”multiplication rule of probabilities”. If two events are not independent, we say that they are dependent. In this case, P (B|A) 6 = P (B).
Problem 36. Two coins are tossed. What is the probability of obtaining a head on the first coin and a tail on the second coin?
Problem 36. Consider two boxes: Box 1 contains 2 white and 2 black balls, and box 2 contains 2 white balls and three black balls. What is the probability of drawing a black ball from each box?
Problem 36. A container holds three red balls and five blue balls. One ball is drawn and discarded. Then a second ball is drawn. (a) Whta is the probability that the second ball drawn is red if you drew a red ball the first time? (b) What is the probability of drawing a blue ball second if the first ball was red? (c) What is the probability of drawing a blue ball second if the first ball was blue?
Problem 36. Consider the following events.
A: rain tomorrow B: You carry an umbrella C: coin flipped tomorrow lands on heads
Which of two events are dependent and which are independent?
Problem 36. You roll a regular red die and a regular green die. Consider the following events.
A: a 4 on the red die B: a 3 on the green die C: a sum of 9 on the two dice
Tell whether each pair of events is independent or dependent. (a) A and B (b) B and C
