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Math Help - Dice

  1. #1
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    Dice

    If you roll five six-sided dice, then put aside any one set of same numbered dice/die that show the/a most occurring number and re-roll any remaining dice/die, what is the chance that within those two rolls (or one roll if all the dice showed the same number on the first roll) all the dice will show the same number?

    To clarify:
    If you roll 1, 1, 1, 4 & 6 on the first roll, you will re-roll the dice that show 4 & 6 on the second roll.
    If you roll 1, 2, 2, 5 & 5 on the first roll, you will re-roll either the dice that show 1 & 2 or the dice that show 1 & 5 on the second roll as 2 and 5 are both most occurring numbers.
    If you roll 3, 3, 4, 4 & 4 on the first roll, you will re-roll the dice that show 3 on the second roll as 4 is more commonly shown.
    If you roll 5, 5, 5, 5 & 5 on the first roll, you will not re-roll any of the dice as they already all show the same number.
    If you roll 1, 2, 3, 4 & 6 on the first roll, you will re-roll all except one of the dice on the second roll as all of the numbers present are most occurring numbers (even though it would make no difference to the chance of all the dice showing the same number if you were to just re-roll all of them).
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  2. #2
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    I'm not 100% sure about it, but I think its 1:3888.

    What I did was treat each die as an independent roll, not all relating to each other, because when drawing a random number from a set (like a set of dice) it doesn't matter if you draw (roll) all numbers at once or if you do them each separately. So you have (1/6)(1/6)(1/6)(1/6)(1/6), to get 1:7776. Multiply by two, since you said to roll twice, and you get 1:3888.

    Again, I'm not completely sure about it, but that's my best guess.
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  3. #3
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    Answer: 0.0126; 1.26% or about 1 out of 79.

    I made a Excel spread sheet to solve this witch I added as an attachment.

    Basically you were on the right track in your post Obsidantion, you noted that there is only a few types of outcomes of the first roll. You can divide it into 5 different categories:

    1. All 5 dice show the same number
    2. 4 dice show the same number
    3. 3 dice show the same number
    4. 2 dice show the same number
    5. and finally no dice show the same number


    So we need to know how many out of 7776 possible outcomes goes in each category. This is what I came up with:
    None: 720
    Pairs: 5400
    Trips: 1500
    Quads: 150
    All: 6
    Here I have just counted all the occurrences from a list of all possible outcomes. Of course it is possible to calculate this one at a time, but the math is a bit complicated...

    I will give one example:
    //
    For three of a kind or Trips as I have called them there is basically 6 different kind of result that interest us: 111; 222; 333; 444; 555 and 666.

    Looking on one of them, 555 it can occur in lots of different ways, 5X55X; XX555; and 5X5X5 to name a few. Here the X is the dice showing anything other then 5. The number we are seeking here is the number of combinations to pick 3 out of 5 dice in no specific order, the number is 10.

    The dice showing anything but 5 has each 5 different possible values. A total of 25.

    So the total number of Trips is:
    6*10*25 = 1,500
    //

    When calculating the number of Pairs, make sure to exclude any Pair where the other dice is a Trips! The number of combinations that does not have a pair or better is calculated by how much is left when summing up the other categories.

    The chance to complete the five of a kind is calculated for each case as:
    (1/6)^n where n is the number of dice rolled.

    So to complete the calculations the probability for each case is multiplied by the chance to complete the five of a kind and the results are summed up!
    Attached Files Attached Files
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  4. #4
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    Quote Originally Posted by Sweed View Post
    Answer: 0.0126; 1.26% or about 1 out of 79.
    This is what I have (using Excel as well ) so I think you're right. Congrats.
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  5. #5
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    Quote Originally Posted by Obsidantion View Post
    This is what I have (using Excel as well ) so I think you're right. Congrats.
    Yea, but It will be much more fun to do the calculation without trying all combinations... It will be more math that way! At least rnow we know the answer so it's a bit easier now.
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