Probability / Math question:I have 6 dice, labelled 0 to 31. Two of the dice are guaranteed to roll 31 each time (chosen randomly) and the remaining four can be any number 0 through 31.What are the odds that I can roll three of the dice with a 31?Four dice?Five dice?All six?How about the probability of getting a more specific combo, like dice number 1, 3, 5 and 6 being 31 on a four or higher roll?Thanks for any help.
>>1557818>Two of the dice are guaranteed to roll 31 each time (chosen randomly) and the remaining four can be any number 0 through 31.This means you can disregard those 2 dice and just rephrase the questions as such:>What are the odds that I can roll one of the dice with a 31?>Two dice?>Three dice?>All four?
>>1557829Doing the easiest (rephrased) one first:Chance of all four getting 31 is: (1/31)^4 = 0.00000108>>1557818If anyone is curious about the image in OP here's a video about it: https://www.youtube.com/watch?v=i4EFkspO5p4That's a redraw of one of the earliest digital artworks.
>>1557829>This means you can disregard those 2 dice and just rephrase the questions as such:The last part of the question is why I mention that, if it makes a difference.>0.00000108What's that as a percentage? Because hearing it like that sounds WAY lower than I was hoping.Getting four or maybe even five on a very good day was my main hope. Currently I'm seeing something like maybe 10% for a three result, my end goal is to get a variety of four 4x31 result rolls or maybe even a five or two.I've the possibility of guaranteeing three of the rolls as being 31 instead of just two, but it would be so much more of a pain in the ass to do that I deemed it not worth doing.
