Hey there. My copy of Burning Wheel just arrived a few days ago, and I've been rather enjoying poking around in it. Now all I need are some players...

Anyway, I spotted the nice table of probabilities on the website, but I can't help but ask: How does open-ending a roll affect the probability? Sure, it goes up, but could anyone provide actual numbers? I'd do the calculations myself, but don't really know enough about probability, and don't have handy access to a computer with programming tools to test empirically (or the patience to do so by hand).

Presumably, the benefit of open-ending a roll increases with the number of dice in the roll. Specifically, I'm looking at elven skill-songs right now, so the relevant question of the moment is: skill-songs take two skill points to open. at what number of dice does the open-endedness of a skill-song give a greater benefit to one's chances than merely taking that extra die and putting it in the normal version of the skill?

An entirely crunchy question, I know, but I'd at least like to know whether I'm paying that extra skill point for mechanical advantage, or just color.

Hey Albert, I'm the guy who made those tables, and I don't have the math skills that are required to calculate those numbers. I would love to see the formulas though that would solve those probabilities. I still have the spreadsheet that was used to create the PDF's on the website, so if I can be convinced how to figure out how to solve these figures, I would submit those new tables.

Hm. It did occur to me that what I was asking might be a bit more complicated than I originally thought.

On the other hand, it also occurred to me that I could ssh into the machines back at Stanford and hack some code. Let's see how long it takes me to get some empirical test data on a few million tests! :-)

Here are some hard numbers. Sorry about the utter lack of formatting; I need to be out of here fifteen minutes ago, and might not have net access until after the weekend.

A couple of comments:

- This is on a million tests for each combination of ob and skill. If anyone would like to check my code, try here

- I don't know why the ob1 probabilities are a couple of percentage points low. When I tested the program on non-open ended rolls, the probabilities were all within about .2 percent of your calculated values. This suggests that other probabilities might be similarly low, but even if so...

- Pumping another straight skill die into a closed-ended skill is almost universally higher probability, excepting the 1ish percent chance of making a successful roll of an obstacle two levels higher than the skill. So it looks like that extra skill die to open song-skills is going into color and the awe effect. (And, of course, the fact that most life-paths and song-skills don't give you the option of choosing the non-song version) Given that character color seems to have its own currency in BW chargen, this is not inconsistent.

And without further ado:

ob1 sk1: 0.483137
ob2 sk1: 0.067209
ob3 sk1: 0.009487
ob4 sk1: 0.001827
ob5 sk1: 0.0
ob6 sk1: 0.0
ob7 sk1: 0.0
ob8 sk1: 0.0
ob9 sk1: 0.0
ob10 sk1: 0.0
ob1 sk2: 0.739882
ob2 sk2: 0.305475
ob3 sk2: 0.074019
ob4 sk2: 0.019946
ob5 sk2: 0.001361
ob6 sk2: 3.69E-4
ob7 sk2: 0.0
ob8 sk2: 0.0
ob9 sk2: 0.0
ob10 sk2: 0.0
ob1 sk3: 0.864391
ob2 sk3: 0.547956
ob3 sk3: 0.220111
ob4 sk3: 0.064569
ob5 sk3: 0.02168
ob6 sk3: 0.005487
ob7 sk3: 3.36E-4
ob8 sk3: 0.0
ob9 sk3: 0.0
ob10 sk3: 0.0
ob1 sk4: 0.928551
ob2 sk4: 0.71216
ob3 sk4: 0.392437
ob4 sk4: 0.175011
ob5 sk4: 0.059995
ob6 sk4: 0.015929
ob7 sk4: 0.004402
ob8 sk4: 0.001492
ob9 sk4: 0.0
ob10 sk4: 0.0
ob1 sk5: 0.96422
ob2 sk5: 0.821399
ob3 sk5: 0.581222
ob4 sk5: 0.306162
ob5 sk5: 0.134244
ob6 sk5: 0.052968
ob7 sk5: 0.015147
ob8 sk5: 0.005107
ob9 sk5: 0.001951
ob10 sk5: 5.28E-4
ob1 sk6: 0.98051
ob2 sk6: 0.900541
ob3 sk6: 0.717767
ob4 sk6: 0.468727
ob5 sk6: 0.246408
ob6 sk6: 0.118944
ob7 sk6: 0.041568
ob8 sk6: 0.01556
ob9 sk6: 0.00499
ob10 sk6: 0.001471
ob1 sk7: 0.991543
ob2 sk7: 0.937498
ob3 sk7: 0.807688
ob4 sk7: 0.593227
ob5 sk7: 0.381997
ob6 sk7: 0.196904
ob7 sk7: 0.095697
ob8 sk7: 0.036041
ob9 sk7: 0.015661
ob10 sk7: 0.005809
ob1 sk8: 0.997641
ob2 sk8: 0.965327
ob3 sk8: 0.876113
ob4 sk8: 0.71226
ob5 sk8: 0.515882
ob6 sk8: 0.313581
ob7 sk8: 0.166351
ob8 sk8: 0.072895
ob9 sk8: 0.033211
ob10 sk8: 0.012469
ob1 sk9: 0.998798
ob2 sk9: 0.981422
ob3 sk9: 0.92293
ob4 sk9: 0.813491
ob5 sk9: 0.621168
ob6 sk9: 0.428165
ob7 sk9: 0.265587
ob8 sk9: 0.128829
ob9 sk9: 0.062304
ob10 sk9: 0.027976
ob1 sk10: 0.999712
ob2 sk10: 0.983855
ob3 sk10: 0.950565
ob4 sk10: 0.864149
ob5 sk10: 0.721704
ob6 sk10: 0.538775
ob7 sk10: 0.365328
ob8 sk10: 0.207698
ob9 sk10: 0.112336
ob10 sk10: 0.05055

Wow. Give yourself a Deeds point!

I know that this is an acceptible way of coming up with the odds, but I like to base the probabilities on stistical formulas. I know that I could come up with the right numbers if I work at it for a while, but it might take me quite a while as well as it probably being some time before I could get around to doing it. If anyone is too impatient to wait around for me, you're more than welcome to use the spreadsheet that I've created to build from. Just send me an email.

Think of it in this way: Every die you roll for an open ended test counts as 1 and 1/6of a die. Statistically speaking, if you roll 6 dice open ended it's like rolling 7 dice non open-ended.

Note however that this is a mnemonic. It doesn't cover the extremes (it is possible to roll a single die and get 20 6's in a row).

Thomas

Well, I'm not enough of a probability theory badass to have a general solution, but this is what I came up with.

For a single die, the distribution is this:
P(0) = 1/2
P(1) = = 1/2 - (1/6) * (1/2) = 5/12
P(i) = (5/12)*(1/6)^(i-1) for i > 1

For two die:
P(0) = (1/2)^2 = 1/4
P(1) = 2*(1/2)*(5/12) = 5/12
P(i) = (5/12)*(1/6)^(i-1)+(i-1)*(5/12)^2*(1/6)^(i-2) for i > 1

For three die:
P(0) = (1/2)^3 = 1/8
P(1) = 3*(1/2)^2*(5/12) = 5/16
P(i) = (1/2)^2*(5/12)*(1/6)^(i-1)+i*(5/12)^2*(1/6)^(i-2)+(1/2)*i*(i-1)*(5/12)^3*(1/6)^(i-3) for i > 1

Whew. As you can see, it gets pretty ugly for more than two dice, and I wouldn't want to go any further than three.

If you just want the expectation and variance, that's a lot easier.

For just one die:
E(X) = (5/12)*SUM(i*(1/6)^(i-1)) over i:0 -> infinity
E(X) = 0.6

Variance is E(X^2) - E(X)^2 = 0.84 - 0.36 = 0.48

Since each die is independant, to get the expectation n die, you just multiply by n, so:

E(X) = 0.6 * n
V(X) = 0.48 * n