Critical Math

Never tell me the odds, give me an hour and a lot of paper and I'll figure them out.

In this downtime of little new TV and a need to distract once in a while, I proposed we watch all of S1 Critical Role (which I really only half ass watched until Percy died and came back). So on Thursday at late Saturday nights we watch old-school Critical Role. Sometimes we sneak in another episode but basically we’ve been grinding through.

The other day we got to Omens and had to pause while we argued about math.

We’d hit a point where Matthew Mercer was rolling for an absent Ashley Johnson. She was the cleric and he rolled her healing spell which restores people for 4D8+5 hit points. Matt got a total of 20.

“Oh,” says my wife, disappointed. “Below average.”

I mused aloud, “I know the average of 2D6 is 7, but I forget the math.”

This led to an old argument about the law of averages and statistics, which is to say there is a one in eight chance of rolling an eight (8) on any given roll of a 1D8. No matter how many times people say the dice are good or bad, you have the same chance every time. Flipping a coin 100 times will give you roughly 50 heads and 50 tails. Over time, the wiggliness of Xeno’s Paradox will keep the numbers from a true 50-50,but the point remains you will always have a 50% chance of tossing a heads or a tails.

Related to this, though, are the odds of two heads in a row. The problem with the math is a combination of fairness and a concept known as “the Gambler’s Fallacy.” It’s also called “the Monte Carlo Fallacy” because of the time it happened at Monte Carlo. See if the odds of a heads is 1 in 2, then the odds of heads twice in a row is 1 in 4. Three in a row is 1 in 8, because you double it on down. Four in a row is 1/16, and five is 1/32, and this goes on.

And this confuses people because even if the odds of flipping a coin to heads 20 times in a row is 1 in 1,048,576, this does not change that the odds of flipping heads for #21 is both one in 2 and 1 in 2,097,152. To bake your noodle a little more, if you’ve flipped a coin 20 times for heads, the odds of flipping the 21st to be tails is … also 2,097,152.

This is because we’re looking at two different things here. The probability of a heads or a tails will always and forever be 50%. Period. The odds of rolling a Natural 20 on 1D20 will always be 1/20th (5% for those wondering). Those will never change. This is Bayes’ theorem in action.

We all get screwed up about this because we believe that fairness (i.e. the permanent 50%-ness of the toss) means that previous failures (or successes) will change the probability of the next toss. It won’t.

Let’s say we’re rolling 1D20 and it’s a fair roller. Our goal is to get a 20 and the probability of that is 5%.

But what about the odds of rolling a 20 at least once in 20 rolls? Well that’s 64%.

There’s a 95% chance of not rolling a 20 (or any other given number) but again that doesn’t change. Every roll you have a 95% chance of not rolling a 20 and a 5% chance of rolling a 1 or a 20 (unless you’re Wil Wheaton or Taliesin Jaffe). But this makes us ask another question. If there’s a 64% chance of a critical success (i.e. a 20) in 20 rolls of 1D20, does that change if my first roll is a non-20?

Kind of. It’s 62%.

Yes, it went down. Why? Because you have fewer chances to win! So the odds of a critical drop every time you roll a non-critical. And frankly this explains half of Wil Wheaton — he’s rolled so many non-critical, the odds are astronomical for him to actually roll one. On the other hand, his capacity for rolling 1s makes him the Joe DiMaggio of Failure for D&D. Sorry, Wil.

This is exactly the same kind of argument my father made back in 2016 (see Hot Hands And Playoffs from 2016). And I came to the same conclusion. Wheaton and Jaffe are what we call outliers.

An Aside: For anyone who’s confused, Wil Wheaton (yes Wesley Crusher from Star Trek, also a great writer, a gamer, and a generally amazing human) plays D&D. He has an incredibly capacity to fail his rolls, to the point that a term was coined: Wheatoning (see also the math on Wheatoning). Opposite this is Taliesin Jaffe (an actor, director, voice actor, an all around cool guy) who has a bizarre talent at successes. At least, he did as Percy in Series 1 of Critical Role, with a dice called “The Golden Snitch.” This die was later stolen. Long story. Taliesin still rolls crazy high, so most Critical Role fans see them as the opposite ends of the spectrum.

Whew. Okay that was a lot of math to tell you that while people want to think “If I roll a non-20 10 times in a row, the odds go UP for my next roll to be a 20!” and the truth is that the odds go down. There is no universal mathematical law of fairness, just of averages.

Back to the Critical Role episode in question. So Mercer rolled a 20 for 4D8+5. That’s pretty normal for Ashley Johnson (she has a tendency to roll below average, but not to the point that you think she’s cursed like poor Wheaton). At the time, I couldn’t remember the math, but I remembered the experiment, which was that the majority of the faces on a d6 add up to 7.

You may be thinking “Mika, there are only three: 6 + 1, 5 + 2, 4 + 3.” and that’s true but it’s actually 6 because the inverses are true. But ask yourself how many possibilities are there with 2D6? There are 11. (2 through 12, you can’t roll a 1 on two dice). And 3 of the 11 are a seven so that’s 27% odds of a 7?

No. It’s 16%. Again we’re talking about frequency versus probability. There are 11 possible outcomes, but there are 36 possible variants. You always have six chances to get a 7, one to get a 2, and one to get a 12. This means that to get the probability (i.e. the odds) of rolling a 7, you have to take the frequency of getting a 7 (again, that’s 6) and divide it by the total number of frequencies (36). And 6/36 is 16.67%.

Here’s a table to help:

Dice RollFrequencyProbability

It’s annoying to think that you always have two sets of probabilities, but if it helps we’re talking about likelihood versus probability. The likelihood of rolling a 7 on 2D6 is 16.8% (rounding up). The probability is 1 in 11. With one die, it’s always a flat sameness (likelihood of rolling a 20 is 1 in 20, so is the probability).

We’ve gone around a bit. I know. Here’s the fun stuff. Before arguing math stuff, my wife says “Oh just add up the opposite sides of a die.”

In my defence I never looked.

A quick check on this and we learned some really hilarious factoids:

  1. She’s right (12+1 = 13 and 2d12 is in fact average 13).
  2. This works because the math is “highest number on the die plus the lowest number equals the average of two dice of the same face.”
  3. On standard dice the 1 and the high number are opposite (6 and 1, 8 and 1, etc).
  4. All the opposite sides add up to the average.
  5. On a D10 this breaks because the 0 is opposite a 9, which is technically highest and lowest, but they add up to 9 (or 19, D10s are weird). The average is 11.
  6. One of my D8s is misprinted and has 8 opposite 7. It was the first one we checked..

Of course the first die we checked happened to be the misprint. The purple die in the photo below is buck wild and all wrong.

Two eight-sided dice. The off-white one (properly) shows a 5 below the 8, the purple shows a 3.

This is the only die in that set (in fact, of all our regularly used dice) that has been misprinted. And while technically the change in weight (different numbers are different sizes and have a different amount of paint on each bit) means the die rolls wrong, it’s not statically significant for me to worry much. Dice have a standard set so that numbers can be trusted. Also it makes math easier for people. The average of 2D20 is *drumroll* 21.

Let’s go back to Ashley Johnson. Mercer rolled a 20 on 4D8+5. Is that below average? Oh yes it is.

The average of 4D8 is 18 because 2D8 = 8 + 1 – 9 times 2 is 18. Add 5 and the new average becomes 23. a 20 is 3 points under. But at least you can’t roll under a 9!

Bonus Section

The wonderful thing about Math is you can see parallels. Once you know 2D6 is 7, you can start to jump. 4D16 is 14 and so on. But… what does ‘6 + 1’ mean. How would you get there?

Me I look for patterns. 6 is the maximum of one die. One is one-half the number of dice rolled. Then it stands to reason that 4D6 is going to be 6 plus something with 4, right? 6 plus 8 is 14, so is that our pattern? No because going back one means “6 plus 2 is 8” and that isn’t seven. So how did we get the one? One is one half the number of dice…

DiceAverageMax + DiceMax + 1/2 Dice
2D676 + 2 = 86 + 1 = 7
4D6146 + 8 = 146 + 2 = 8
6D6216 + 6 = 186 + 3 = 9

None of those are right but … Do you see the patterns?

First the average starts at 7 and you add 7 for every 2 dice you add.

We need to add something to a six, but it has to factor in with the number of dice. So how do we get to that one again? We know that the only way to make a one out of the first example (2D6) is to divide the dice by 2. For 2D6 we’re adding 2, which means what plus 2 gets to 14? A 12. And what is 12 divided by 6? A 2…

Hang on then. We have a possible pattern!

( maximum score of one die * half the dice rolled ) + ( half number of dice rolled )

DiceAverage( Max * 1/2 Dice ) + ( 1/2 Dice)
2D67( 6 * 1 ) + ( 2 / 2 ) = 6 + 1 = 7
4D614( 6 * 2 ) + ( 4 / 2 ) = 12 + 2 = 14
6D621( 6 * 3 ) + ( 6 / 2 ) = 18 + 3 = 21
8D628( 6 * 4 ) + ( 8 / 2 ) = 24 + 4 = 28
10D635( 6 * 5 ) + ( 10 / 2 ) = 30 + 5 = 35

Bingo baby! We have our pattern! But does this work for odd numbers? It must for the formula to be correct…

1D67( 6 * .5 ) + ( 1 / 2 )
3D610.5( 6 * 1.5 ) + ( 3 / 2 )
5D617.5( 6 * 2.5 ) + ( 5 / 2 )
7D624.5( 6 * 3.5 ) + ( 7 / 2 )
9D631.5( 6 * 4.5 ) + ( 9 / 2 )

Yep, the math keeps working!

And this, friends, is how you reverse engineer a formula.

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