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help me with a math problem

proccoprocco Member, Mobile Tester Posts: 915
This will probably turn out to have an easy answer, but math isn't my strong suit:
Say the d20 is missing from my set of dice, but I want to replicate the range of 1-20. Easy enough...roll a d4, d8, and d10 together to get a range of 3-22, then subtract 2 to get the 1-20 range. My question is, is that actually equivalent to rolling a d20, or will the chance of each number coming up be weighted towards the numbers in the middle, and how do you figure that out?
Thanks!

Comments

  • FinneousPJFinneousPJ Member Posts: 6,455
    No, it's not equivalent to rolling a d20. In a d20 there is an even distribution of values. In your case, you can roll a 10 in many different ways, for example. Hence, it will be weighed towards the middle.
  • proccoprocco Member, Mobile Tester Posts: 915
    That's what I thought...how does one show that mathematically?
  • FinneousPJFinneousPJ Member Posts: 6,455
    What do you need to show? That the distribution is different? Or what the distributions are? It's easy to show that rolling a 1 in your scenario is way unlikelier than in d20. Rolling a 1 means rolling a 1 in each separate roll.

    P(1) = 1/4 * 1/8 * 1/10 = 1/320

    Whereas in d20

    P(1) = 1/20
  • semiticgoddesssemiticgoddess Member Posts: 14,903
    And Microsoft Excel has a random number generator built in. If you put
    =RANDBETWEEN(1,20)
    into a cell and click delete in an empty cell, it will give you a random number between 1 and 20.
  • FardragonFardragon Member Posts: 4,511
    procco said:

    This will probably turn out to have an easy answer, but math isn't my strong suit:
    Say the d20 is missing from my set of dice, but I want to replicate the range of 1-20. Easy enough...roll a d4, d8, and d10 together to get a range of 3-22, then subtract 2 to get the 1-20 range. My question is, is that actually equivalent to rolling a d20, or will the chance of each number coming up be weighted towards the numbers in the middle, and how do you figure that out?
    Thanks!

    No, it's not the same. A d20 has exactly the same probability of getting each number: 5%. Using your suggested method, the probability of getting a 1 or 20 is 1/(4x8x10), which works out at about 0.3%, while the probability of getting 10 or 11 is much higher (to early in the day for me to work out exactly what).
  • BillyYankBillyYank Member Posts: 2,768
    edited January 2017

    And Microsoft Excel has a random number generator built in. If you put
    =RANDBETWEEN(1,20)
    into a cell and click delete in an empty cell, it will give you a random number between 1 and 20.

    There's a button on the Formula tab called "Calculate Sheet" that will do the same as a delete. If you're going to do any extensive die rolling in Excel, I would recommend setting calculation options to "Manual". That way you won't accidently reroll the dice and lose a roll you wanted to keep. I use Excel for all my character creation die rolls. I like it because you can set up complicated rolls like "roll 4d6 and drop 1".

    Edit: There's a lot of free dice rolling apps for Android and, I assume, iOS. I played around with this one for a few minutes and it seems OK.
    https://play.google.com/store/apps/details?id=fr.sevenpixels.dice&hl=en


    Post edited by BillyYank on
  • proccoprocco Member, Mobile Tester Posts: 915
    Thanks, fellas. I didn't actually lose my d20, it was just something I was curious about for no good reason. I understand that the numbers towards the middle of the scale are going to have a higher chance of coming up since there are more ways of them to come up...I'm going to try to work out on my own how to figure out what the chance of each number is, then check back to see if anyone has a solution for that...
  • TeflonTeflon Member, Translator (NDA) Posts: 515
    Then throwing three d6s at once vs. d6 three times in a row is different? :p
    Anyway i think cannot replace d20 with two d10s, alas I cannot prove mathematically but just intuition.
  • AstroBryGuyAstroBryGuy Member Posts: 3,437
    Teflon said:

    Then throwing three d6s at once vs. d6 three times in a row is different? :p
    Anyway i think cannot replace d20 with two d10s, alas I cannot prove mathematically but just intuition.

    You can, but don't add them. Use them as percentile dice (one d10 for the tens place, one d10 for the ones place).

    01-05 = 1
    06-10 = 2
    11-15 = 3
    etc..
    96-00 = 20
  • FinneousPJFinneousPJ Member Posts: 6,455
    @procco It's simple. It's the number of combinations that result in a roll over the total number of combinations:

    P(1) = 1 / 320, because there is only one way to get 3 on 1d4+1d8+1d10
    P(2) = 3 / 320, because there are 3 ways to get 4 on 1d4+1d8+1d10

    and so on. Of course, finding out the number of combinations may not be so simple ;)
  • TeflonTeflon Member, Translator (NDA) Posts: 515
    :o : What school of magic are you?
    AstroBryGuy : Math.
  • BillyYankBillyYank Member Posts: 2,768
    The 1d4 + 1d8 + 1d10 - 2 method comes out as a standard bell curve:

    http://anydice.com/program/a8fc
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