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Math Help - Pricing Dice Game

  1. #1
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    Pricing Dice Game

    Stage 1

    In this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you 4 pounds.

    A fair price to play this game should be 3.50 as then the player or myself should not win any money if this game is played infinty times.

    Stage 2

    How much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.

    I priced this option at 3.57, however I been told that is wrong. Can anyone explain why?

    Thanks in advance

    Calypso
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by calypso View Post
    Stage 1

    In this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you 4 pounds.

    A fair price to play this game should be 3.50 as then the player or myself should not win any money if this game is played infinty times.

    Stage 2

    How much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.

    I priced this option at 3.57, however I been told that is wrong. Can anyone explain why?

    Thanks in advance

    Calypso
    Choose a  t such that if the first roll a\le t roll again otherwise accept the prize of   a.

    Now for any given  t you can work out the value of this game and hence a fair price. The fair price for the game is the maximum fair price.

    (I make  t=3 the value of  t that maximises the return and a fair price is then  4.25

    CB
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  3. #3
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    Sorry for the late reply I have just got back from holiday.

    The text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is

    ( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25

    I can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:

    So I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout

    So therefore the possible outcomes of the game are

    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
    5
    6

    Therefore fair price should equal 78/21 = 3.71?

    Thanks again

    Calypso
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  4. #4
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    Quote Originally Posted by calypso View Post
    I am still confuced as why I cant get this answer with my method:

    So I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout

    So therefore the possible outcomes of the game are

    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
    5
    6
    The reason this goes wrong is that those outcomes are not all equally probable. Each of the final three outcomes (4, 5, 6) occurs with a probability of 1/6. But the initial outcomes 1, 2, 3 are subdivided into six subcases, each of which occurs with a probability of only 1/36. Multiplying each outcome by its probability, you get \tfrac3{36}(1+2+3+4+5+6) + \tfrac16(4+5+6) = \tfrac{17}4 = 4.25.
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  5. #5
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    Quote Originally Posted by calypso View Post
    Sorry for the late reply I have just got back from holiday.

    The text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is

    ( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25

    I can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:
    I don't like their decision rule, not that it is wrong, but because it refers to something not in the sample space (that is "anything greater than 3.50" when it could have said anything greater than 3 or even anything greater than or equal to 4).

    CB
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  6. #6
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    Great, thanks everyone for your help

    Calypso
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