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Math Help - Dice Game

  1. #1
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    Dice Game

    You play the following game. You throw a six-sided fair die repeatedly. You may choose
    to stop after any throw, except that you must stop if you throw a 1. Your score is the
    number obtained on your last throw. Determine the strategy that you should adopt in order
    to maximize your expected score, explaining your reasoning carefully.
    First I deal with the obvious, If you get a 6 you stop and if you get a 1 your out of luck. So I guess all the question is really asking if wheater it is worth trying for a second roll if you get say a 2 , 3 ,4 or 5. I am having trouble defining a probability distribution for this.
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    Quote Originally Posted by bobak View Post
    First I deal with the obvious, If you get a 6 you stop and if you get a 1 your out of luck. So I guess all the question is really asking if wheater it is worth trying for a second roll if you get say a 2 , 3 ,4 or 5. I am having trouble defining a probability distribution for this.
    Seems to me that each time you roll, the probability of getting 1, 2, 3, 4, 5, 6 is 1/6 for each. So the expected number of spots each time you roll is 3.5.

    So to maximise my expected score I'd roll again if my score was currently less than 3.5 and stop if my score was currently greater than 3.5.

    Strategy:

    Roll again if you get 2 or 3. Stop if you get 4 or 5 (or 6!).
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    Quote Originally Posted by mr fantastic View Post
    Seems to me that each time you roll, the probability of getting 1, 2, 3, 4, 5, 6 is 1/6 for each. So the expected number of spots each time you roll is 3.5.

    So to maximise my expected score I'd roll again if my score was currently less than 3.5 and stop if my score was currently greater than 3.5.

    Strategy:

    Roll again if you get 2 or 3. Stop if you get 4 or 5 (or 6!).
    It's probably worth pointing out that if you decide to keep rolling until you get either a six or a one, the probability of getting each result is obviously 1/2. So the expected value using this strategy is ...... 3.5!
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    Quote Originally Posted by mr fantastic View Post
    Seems to me that each time you roll, the probability of getting 1, 2, 3, 4, 5, 6 is 1/6 for each. So the expected number of spots each time you roll is 3.5.

    So to maximise my expected score I'd roll again if my score was currently less than 3.5 and stop if my score was currently greater than 3.5.

    Strategy:

    Roll again if you get 2 or 3. Stop if you get 4 or 5 (or 6!).
    That seems a bit too easy, what about consecutive throws ?
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    Quote Originally Posted by bobak View Post
    That seems a bit too easy, what about consecutive throws ?
    The question asks for a strategy ...... That, I think, is the strategy.

    Let's say you roll the first time and get a 2. My strategy says roll again. When you do, the probabability of each number is obviously still equal to 1/6. Let's say you get a 3. My strategy is to roll again. On the third roll, the probabability of each number is still obviously equal to 1/6. Let's say you get a 5. My strategy is to stop.

    Obviously there's always a chance of getting a 1, in which case you're

    But it's all about having a strategy to maximise your expected score .....

    I haven't worked out exactly what that expected score will be, but it's seems clear (a dangerous statement) that it'll be greater than 3.5
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    Quote Originally Posted by mr fantastic View Post
    The question asks for a strategy ...... That, I think, is the strategy.

    Let's say you roll the first time and get a 2. My strategy says roll again. When you do, the probabability of each number is obviously still equal to 1/6. Let's say you get a 3. My strategy is to roll again. On the third roll, the probabability of each number is still obviously equal to 1/6. Let's say you get a 5. My strategy is to stop.

    Obviously there's always a chance of getting a 1, in which case you're

    But it's all about having a strategy to maximise your expected score .....

    I haven't worked out exactly what that expected score will be, but it's seems clear (a dangerous statement) that it'll be greater than 3.5
    The expected value will be 4. I'll post details later.
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    Quote Originally Posted by mr fantastic View Post
    The expected value will be 4. I'll post details later.
    If you keep rolling until getting a 1, 4, 5 or 6, then each of these has a probability of 1/4 of occuring (by symmetry).

    So the expected value is (1)(1/4) + (4)(1/4) + (5)(1/4) + (6)(1/4) = 16/4 = 4.

    As it happens, the strategy of rolling until getting a 1, 5 or 6 also has an expected value of 4 .....

    All other strategies have an expected value of less than 4.
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