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A measurement of fairness

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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 ?

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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 )

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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

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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, sometimes you win $7 and sometimes you lose $3, but at the end you expect to be $5 poorer!

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game 2: Roll a single die. The number the die lands on is the amount you win. It costs $3 to play this game.

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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 1 You lose $2 ( - 2 ) probability ( 1/6 )

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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 2 You lose $1 ( - 2 ) probability ( 1/6 )( - 1 )( 1/6 )

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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 3 You breakeven ( - 2 ) probability ( 1/6 ) ( - 1 ) ( 1/6 ) ( 0 )( 1/6 ) 0

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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 4 You win $1 ( - 2 ) probability ( 1/6 ) ( - 1 ) ( 1/6 ) ( 1 )( 1/6 )

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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 5 You win $2 ( - 2 ) probability ( 1/6 ) ( - 1 ) ( 1/6 ) ( 1 )( 1/6 )( 2 )( 1/6 )

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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 6 You win $3 ( - 2 ) probability ( 1/6 ) ( - 1 ) ( 1/6 ) ( 1 )( 1/6 )( 2 )( 1/6 )( 3 )( 1/6 )

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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 VALUE expect to WIN an average of $.50 per game

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EXPECTED VALUE -.50 Game 1 is weighted against the player who expects to LOSE an average of $.50 per game. EXPECTED VALUE +.50 Game 2 is weighted in favor of the player who expects to WIN an average of $.50 per game. A game is said to be FAIR if its EXPECTED VALUE is 0

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Al flips a coin If it lands heads up, Al pays Bob $1 and the game ends.

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Al flips a coin If it lands heads up, Al pays Bob $1 and the game ends.

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Al flips a coin If it lands tails up, Al must flip a second coin.

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Al flips coin number 2 If it lands heads up, Al pays Bob $2 and the game ends.

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Al flips coin number 2 If it lands heads up, Al pays Bob $2 and the game ends.

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Al flips coin number 2 If it lands tails up, Al must flip a third coin.

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Al flips coin number 3 If it lands heads up, Al pays Bob $4 and the game ends.

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Al flips coin number 3

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If it lands tails up, Al must flip a fourth coin.

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Al flips coin number 4 If it lands heads up, Al pays Bob $8 and the game ends.

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Al flips coin number 4 If it lands heads up, Al pays Bob $8 and the game ends.

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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!

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What is a FAIR price for Bob to pay Al for the opportunity to play this game? I want to play this game How 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 lands heads up, then Bob must pay Al an infinite amount of money to make the game fair!

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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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If both agree that Al will flip a maximum of 4 coins, then it would be fair if Bob paid Al $2.13 H T H T H T H T H T H T H T H T H T H T H T H T H T H T H T

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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 H T H T H T H T H T H T H T H T H T H T H T H T H T H T H T 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)

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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 H T H T H T H T H T H T H T H T H T H T H T H T H T H T H T 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)

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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 H T H T H T H T H T H T H T H T H T H T H T H T H T H T H T The probability that the third coin lands heads up is 2/16 Al would pay Bob $4. Bob would WIN $1.87 (8/16)(-1.13) (4/16)(-.13) (2/16)(1.87)

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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 H T H T H T H T H T H T H T H T H T H T H T H T H T H T H T The 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)

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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 H T H T H T H T H T H T H T H T H T H T H T H T H T H T H T (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 Bob paid to play.

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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 H T H T H T H T H T H T H T H T H T H T H T H T H T H T H T (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

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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 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 expected winnings =.50 +.50 +.50 +.50 +... Infinitely many.50’s

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Produced by MEI on behalf of OCR © OCR 2013 Introduction to Quantitative methods Probability and risk © OCR 2014.

Produced by MEI on behalf of OCR © OCR 2013 Introduction to Quantitative methods Probability and risk © OCR 2014.

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