Post on 15-Dec-2015
transcript
game 1:
A box contains 1red marble and 3 black marbles.
Blindfolded, you select one marble.
If you select the red marble, you win $10.
If it cost you $8 to play this game, would you play?$1
What would be a FAIR PRICE ?
If you select the red marble, you win $10.
It will cost you $3 to play.
If you win, your prize is really $7.
the probability that you win
the amount that you win
the probability that you lose
the amount that you lose
( 1/4 ) ( $ 7 ) + ( 3/4 )( - $ 3 )
If you select the red marble, you win $10.
It will cost you $3 to play.
If you win, your prize is really $7.
( 1/4 ) ( $ 7 ) + ( 3/4 )( - $ 3 )
7/4 + - 9/4 = - 2/4 = - .50
the EXPECTED VALUE
If you select the red marble, you win $10.
It will cost you $3 to play.
If you win, your prize is really $7.
7/4 + - 9/4 = - 2/4 = - .50
the EXPECTED VALUE
If you play this game many times,you expect to LOSE an average of $.50 per game.
For example, if you play ten times, sometimesyou win $7 and sometimes you lose $3, but at the end you expect to be $5 poorer!
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
Die lands on 1You lose $2
( - 2 )
probability
( 1/6 )
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
Die lands on 2You lose $1
( - 2 )
probability
( 1/6 ) ( - 1 )( 1/6 )
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
Die lands on 3You breakeven
( - 2 )
probability
( 1/6 ) ( - 1 )( 1/6 ) ( 0 )( 1/6 )0
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
Die lands on 4You win $1
( - 2 )
probability
( 1/6 ) ( - 1 )( 1/6 ) ( 1 )( 1/6 )
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
Die lands on 5You win $2
( - 2 )
probability
( 1/6 ) ( - 1 )( 1/6 ) ( 1 )( 1/6 ) ( 2 )( 1/6 )
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
Die lands on 6You win $3
( - 2 )
probability
( 1/6 ) ( - 1 )( 1/6 ) ( 1 )( 1/6 ) ( 2 )( 1/6 ) ( 3 )( 1/6 )
game 2:
Roll a single die.
The number the die lands on is the amount you win.
It costs $3 to play this game.
( - 2 )( 1/6 ) ( - 1 )( 1/6 ) ( 1 )( 1/6 ) ( 2 )( 1/6 ) ( 3 )( 1/6 )+ + + +
-2/6 + -1/6 + 1/6 + 2/6 + 3/6 = 3/6
EXPECTED VALUEexpect to WIN an average of $.50 per game
EXPECTED VALUE - .50Game 1 is weighted againstthe player who expects to LOSE an average of $.50per game.
EXPECTED VALUE +.50Game 2 is weighted in favor ofthe player who expects to WIN an average of $.50per game.
A game is said to be FAIR if its EXPECTED VALUE is 0
Al flips coin number 4
If it lands heads up, Al pays Bob $8 and the game ends .If it lands heads up, Al pays Bob $8 and the game ends.
Al flips coin number 4
If it lands tails up, Al must flip a fifth coin .
Every time Al must flip the coin, his payoff to Bob doubles.If the first headsup occurred on the fifth flip, Al pays Bob $16.If the first headsup occurred on the sixth flip, Al pays Bob $32.
This game might never end!
What is a FAIR price for Bob to pay Al for the opportunity to play this game?
I want to play this gameHow much will it cost me?
If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
But if the game is allowed to continue until Al flips a coin that landsheads up, then Bob must pay Al an infinite amount of money to make the game fair!
If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
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If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
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The probability that the first coin lands heads up is 8/16
Al would pay Bob $1. Bob would LOSE $1.13
(8/16)(-1.13)
If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
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The probability that the second coin lands heads up is 4/16
Al would pay Bob $2. Bob would LOSE $.13
(8/16)(-1.13)
(4/16)(-.13)
If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
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Al would pay Bob $4. Bob would WIN $1.87
(8/16)(-1.13)
(4/16)(-.13)
(2/16)(1.87)
If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
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TThe probability that the fourth coin lands heads up is 1/16
Al would pay Bob $8. Bob would WIN $5.87
(8/16)(-1.13)
(4/16)(-.13)
(2/16)(1.87)
(1/16)(5.87)
If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
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(8/16)(-1.13)
(4/16)(-.13)
(2/16)(1.87)
(1/16)(5.87)
There is probability 1/16 that there is no payoff to Bob. Al returns the $2.13 Bobpaid to play.
If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13
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(8/16)(-1.13)
(4/16)(-.13)
(2/16)(1.87)
(1/16)(5.87)
Bob’s expected value is close to zero ~ almost fair!
= - .565
= - .0325
= .23375
= .3669
Now we will try to determine a fair price for Bob to pay Al to play this game, assuming that the game continues until Al flips a coin that lands heads up.
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Coin 1 lands heads up. Probability = 1/2. Payoff to Bob = $1. (1/2)(1)=.50
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Coin 1 lands heads up. Probability = 1/2. Payoff to Bob = $1. (1/2)(1)=.50
Coin 2 lands heads up. Probability = 1/4. Payoff to Bob = $2. (1/4)(2)=.50
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Coin 1 lands heads up. Probability = 1/2. Payoff to Bob = $1. (1/2)(1)=.50
Coin 2 lands heads up. Probability = 1/4. Payoff to Bob = $2. (1/4)(2)=.50
Coin 3 lands heads up. Probability = 1/8. Payoff to Bob = $4 (1/8)(4)=.50
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Coin 1 lands heads up. Probability = 1/2. Payoff to Bob = $1. (1/2)(1)=.50
Coin 2 lands heads up. Probability = 1/4. Payoff to Bob = $2. (1/4)(2)=.50
Coin 3 lands heads up. Probability = 1/8. Payoff to Bob = $4 (1/8)(4)=.50
Coin 4 lands heads up. Probability = 1/16. Payoff to Bob = $8 (1/16)(8)=.50
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Coin 1 lands heads up. Probability = 1/2. Payoff to Bob = $1. (1/2)(1)=.50
Coin 2 lands heads up. Probability = 1/4. Payoff to Bob = $2. (1/4)(2)=.50
Coin 3 lands heads up. Probability = 1/8. Payoff to Bob = $4 (1/8)(4)=.50
Coin 4 lands heads up. Probability = 1/16. Payoff to Bob = $8 (1/16)(8)=.50
Bob’s expectedwinnings =.50 + .50 + .50 + .50 + . . .
Infinitely many .50’s