Interlude 3: Another mathematical conundrum

Here’s another one, apropos of nothing.

I’m planning my birthday party (6×6 @ #6, Bar Äijä’s, watch this space for the report), and it will have a role-playey 6D6 theme. I’m planning a set of drinks, each with a number assigned to it. You roll 6D6, and depending on the outcome of the roll, that’s what you drink.

I’m thinking, 6 1s will mean you have to drink a Prairie God[1], while 6 6s will mean you have to drink one of everything else.

[1] Naga liqueur, fireball cinnamon and Blair’s Sudden Death.

Here’s the thing.

There’s only one way to roll a natural six with 6D6 (6 1s), and only one way to roll a natural thirty-six (6 6s), always assuming legit rolls, no dropping out of dice or adding any or substituting a D20 or any shenanigans like that.

But, for example, there’s more than one way to roll a thirty. You can roll 6 5s, or you can roll 4 5s and a 4 and a 6 … and so on. More than one way, that’s what I mean. So you’ve got a better-than-one-in-thirty-six chance of rolling one of those middle numbers.

I imagine it’s one of those bell-curve things, with a 0 chance of rolling anything under 6 or over 36, a 1 in 36 chance of rolling 6 or 36, and a climbing chance of the numbers in between. Where I can scatter the safer, more standard drinks for guests to roll.

Save me, mathy-types! My poor artsy brain just can’t handle the confusion of thirty-six numbery things.

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9 Responses to Interlude 3: Another mathematical conundrum

  1. Hokay. Well your suspicions are indeed correct, of course. While there is one way to roll 1-1-1-1-1-1 and one way to roll 6-6-6-6-6-6, aside from those you need to use a permutation calculation. NOT a combination one, by the way. The combination calculation is for when you have several options but the order does NOT matter. However in this case, 6-5-4-3-2-1 needs to be counted separately from 1-2-3-4-5-6 (a permutation of 6-5-4-3-2-1), so you use the permutations formula.

    Since there are 6 die, this becomes terribly complicated. But basically, the more of a middle-number like 18 you are considering, the far higher are the chances. You can see this if you consider 6-6-6-6-6-5 (total of 35). You can see, right now, there are 6 versions of that option, right? So, a little more common than rolling 36. The equation is (6!)*(6-5)!/(6-1)! (just roll with it) The exclamation point is factorial, which means 6! = 6*5*4*3*2*1. The equation above basically amounts to 6!/5!, which is really just 6, if you cancel out the other numbers that are the same

    If instead you want to consider a total of 34, you have even more freedom. Any 2 die can be 5’s. This becomes much trickier to calculate because you are specifying the die options, but you have more than 6. 5-5-6-6-6-6, 5-6-5-6-6-6, 5-6-6-5-6-6, 5-6-6-6-5-6, 5-6-6-6-6-5, 6-5-5-6-6-6, 6-5-6-5-6-6, 6-5-6-6-5-6, and so on. I’d put out the equations but it gets complicated very fast.

    Anyway basically the closer your total to the middle, yes, the more chances there are to get it.

    Now in fact the mid point, which you didn’t specify but I just implied is 18, is in fact NOT 18. It is (6+36)/2 or 21. So keep in mind 21 is the most common roll.


    • stchucky says:

      Awesome, I knew I could depend on both of you. So the 19-25 range is a good one to put my standard, safe drinks on (or maybe a surprise shot of hard stuff on that 21 sweet spot), and there’s basically no chance of a natural 6 or a natural 36.

      Funny stuff. If it was a 36-sided die, logic suggests there’d be a 1:36 chance of rolling any one number. But move the numbers onto six separate dice and suddenly the odds are astronomical. What?

  2. dreameling says:

    Can you come up with a single equation that gives you the number of possible permutations (x) for each die roll total (y)? 🙂 Then you could just count the chances from that…

    There are a total of 6^6 = 46656 permutations, so the chances of rolling a 1 or a 36 is 1/46656 = really friggin low…

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