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Thread: Dice Probability (2d6, 3d6, 4d6)

  1. #1
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    Dice Probability (2d6, 3d6, 4d6)

    I would like to find the probabilities of rolling a result using different numbers of six-sided dice. I am looking for the best way to figure out the probability for each number of dice. I have found information that explains finding the probability of rolling specific numbers or for the sum of all the dice rolled, but I am looking for the result of rolling two numbers or any numbers higher than the target numbers.

    It is easy enough for me to draw out the 36 possibilities of 2d6 to find these odds, but I would like to know what method can be used to calculate these odds for 2d6, 3d6, and 4d6. I have been banging my head against the wall with this problem for several days and I still fail to get a firm grasp of the proper concepts. I have not worked with probability calculations for many years and I would appreciate any guidance for this personal project of mine.

    Below is a table that I would like to complete. For example, what is the probability of rolling at least 3 AND at least 5 using 2d6? Initially I had wanted to calculate odds with re-rolls added in, but I am having a hard enough time with just the flat rolls.

    3+ & 3+ |___|___|___|
    4+ & 3+ |___|___|___|
    5+ & 3+ |___|___|___|
    6+ & 3+ |___|___|___|
    4+ & 4+ |___|___|___|
    5+ & 4+ |___|___|___|
    6+ & 4+ |___|___|___|
    5+ & 5+ |___|___|___|
    6+ & 5+ |___|___|___|
    6+ & 6+ |___|___|___|

    Thank you anyone who takes the time to help with my request.

  2. #2
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    Re: Dice Probability (2d6, 3d6, 4d6)

    So I decided to generate an Excel worksheet of all the possible outcomes and calculate the probability of each condition. Even though I have the data I need, I am still interested in learning how one would construct the formula to find the above data without creating a table the way I had.

    I am also curious if it is possible to calculate the probability of the same conditions by rerolling the highest or lowest die.

  3. #3
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    Re: Dice Probability (2d6, 3d6, 4d6)

    Ok in the previous post I am not very sure about the intend of the table, now I understand it better.

    For the sum of dice, it involve a technique called convolution. For random variables with bounded support like this, usually the closed-form solution cannot be easily generalized/calculated for large amount of random variables (dice). For small number, we can still do it, like in your case you just have 4.

    For the first table, things are easy for the first column as you know how to calculate for 1 dice. For 2 or more, things got more and more complicated. Not sure if the method is better than computer brute force. Shall try it later.

    E.g., the event 4+ & 3+, for 2d6, in probability context, is that

    \{D_1 \geq 4 \cap D_2 \geq 3\} \cup  \{D_1 \geq 3 \cap D_2 \geq 4\}

    where D_1, D_2 are the result of the 2 dice.

    The number of unions is calculated by the permutation. To calculate this you probably need to use inclusion-exclusion principle:

    \Pr\left\{\{D_1 \geq 4 \cap D_2 \geq 3\} \cup  \{D_1 \geq 3 \cap D_2 \geq 4\} \right\}

    = \Pr\{D_1 \geq 4 \cap D_2 \geq 3\} +  \Pr\{D_1 \geq 3 \cap D_2 \geq 4\}- \Pr\{D_1 \geq 4 \cap D_2 \geq 4\}

    (as \{D_1 \geq 4 \cap D_2 \geq 3 \cap D_1 \geq 3 \cap D_2 \geq 4\}= \{D_1 \geq 4 \cap D_2 \geq 4\})

    = \frac {1} {2} \times \frac {2} {3} +  \frac {2} {3} \times \frac {1} {2} - \frac {1} {2} \times \frac {1} {2}

    = \frac {1} {3} + \frac {1} {3} - \frac {1} {4}

    = \frac {5} {12} \approx 41.67\%

    For more dice, the terms will be more and will be more complicated.

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