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Math Help - Dice rerolling probabilities

  1. #1
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    Dice rerolling probabilities

    Goodday,

    I'm writing an application and need to have it calculate the chances of rolling, for example a 4, 5 or 6 on a dice with the ability of a reroll. The current math I have is;

    3/6 + (3/6 * 3/6) = 0.75 * 100 = 75%.

    The problem I have is when I'm trying to have it calculate the chances of rolling a 3, 4, 5 or 6 with the reroll;

    4/6 + (4/6 * 4/6) = 1.1111 * 100 = 111.11%

    Which is ofcourse incorrect as you could still roll two 2's. What would be the correct calculation that I would need to use?


    Thank you,
    With regards,
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  2. #2
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    Quote Originally Posted by DaredevilWasAlreadyTaken View Post
    Goodday,

    I'm writing an application and need to have it calculate the chances of rolling, for example a 4, 5 or 6 on a dice with the ability of a reroll. The current math I have is;

    3/6 + (3/6 * 3/6) = 0.75 * 100 = 75%.

    The problem I have is when I'm trying to have it calculate the chances of rolling a 3, 4, 5 or 6 with the reroll;

    4/6 + (4/6 * 4/6) = 1.1111 * 100 = 111.11%

    Which is ofcourse incorrect as you could still roll two 2's. What would be the correct calculation that I would need to use?


    Thank you,
    With regards,
    To be honest, I'm having trouble tying to work out what your exact question is.

    Rolls are independent. So what happens in the second roll is not affected by what happened in the first roll.
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  3. #3
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    Quote Originally Posted by DaredevilWasAlreadyTaken View Post
    Goodday,

    I'm writing an application and need to have it calculate the chances of rolling, for example a 4, 5 or 6 on a dice with the ability of a reroll. The current math I have is;

    3/6 + (3/6 * 3/6) = 0.75 * 100 = 75%.

    The problem I have is when I'm trying to have it calculate the chances of rolling a 3, 4, 5 or 6 with the reroll;

    4/6 + (4/6 * 4/6) = 1.1111 * 100 = 111.11%

    Which is ofcourse incorrect as you could still roll two 2's. What would be the correct calculation that I would need to use?


    Thank you,
    With regards,

    Think of the chance of not getting those numbers at all and take it from 1:

    First scenario: 1 - (1/2 * 1/2) = 3/4 = 75%

    Second scenario: 1 - (2/6 * 2/6) = 1 - 4/36 = 32/36 = 8/9 = 88.9% (3sf)
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  4. #4
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    Please, could you clarify the problem? Say you want a 5 or a 6.
    Do you mean that you will continue the roll the die until success regardless of the number? Or do you have an upper limit on the number of rolls?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    To be honest, I'm having trouble tying to work out what your exact question is.

    Rolls are independent. So what happens in the second roll is not affected by what happened in the first roll.
    Sorry for the confusion, I'm not a native English speaker so I'll try to explain this as good as I can.

    I have a 1 die and need to roll a 5 or a 6, but whenever I fail to do so I may reroll that die 1 time to have another shot at the 5 or 6, how would I put this into a mathmatical formula?

    Does that make more sense?


    Thank you,
    With regards,
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by DaredevilWasAlreadyTaken View Post
    Sorry for the confusion, I'm not a native English speaker so I'll try to explain this as good as I can.

    I have a 1 die and need to roll a 5 or a 6, but whenever I fail to do so I may reroll that die 1 time to have another shot at the 5 or 6, how would I put this into a mathmatical formula?

    Does that make more sense?


    Thank you,
    With regards,
    Your chance, as I put in numerical form in my last post will be 1-(the chance of not getting a 5 or 6 on any roll). In numbers:

    P(5,6) = 1 - (P'(5,6))^{n}. No doubt I have the wrong notation but P(5,6) means the chance of getting a 5 or a 6 and P'(5,6) means the chance of not getting a 5 or a 6. n is the number of rolls you can have. In your case:

    P(5,6) = 1 - (\frac{2}{3})^2 = \frac{5}{9}
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