View Full Version : Can someone help me out with dice odds?

Jeivar

2017-09-16, 08:54 AM

My group is playing FFG Star Wars, and I'm playing a Force adept. To activate a Force power I need to succeed on a Force die check, but only 5 of the 12 sides give a positive outcome. That means a 41,666 % chance of success. I am gradually clawing my way towards being able to roll a second die.

Math is a big weakness of mine. How does the probability here work? Is it just a plain 41,66 + 41,66?

Necroticplague

2017-09-16, 09:15 AM

My group is playing FFG Star Wars, and I'm playing a Force adept. To activate a Force power I need to succeed on a Force die check, but only 5 of the 12 sides give a positive outcome. That means a 41,666 % chance of success. I am gradually clawing my way towards being able to roll a second die.

Math is a big weakness of mine. How does the probability here work? Is it just a plain 41,66 + 41,66?

Roll a second die and do what with it? Add it to the first? Replace it if it's higher? What, exactly, are the positive outcomes on the die?

Either way, it's not simply adding the two independent probabilities together (by that logic, using three dice would give you a roughly 123% chance of success, which is clearly wrong).

For this kinds of stuff, Anydice (http://anydice.com/) is a pretty good site to use, since it can calculate pretty easily.

Assuming you mean 'roll two d12, take the highest, you want to roll a 8 or more', your chance of success is 65.97% (or, slightly less than two-thirds).

output [count {8,9,10,11,12} in [highest 1 of 2d12]]

however, if you use

output [highest 1 of 2d12]

and click 'at least' for data, you'll get the same result (by looking at the bar indicating 'at least 8')

If you want to do the math more manually, it's: chance the first die succeeds (41.6)+ chance first die fails (58.4)*chance second die suceeds (41.6, assuming independence)=65.89.

Now, the resultant numbers are slightly dissimilar, but it's small enough that I'm nearly completely sure this is due to rounding differences between computer's and mine numbers.

Hamste

2017-09-16, 09:18 AM

If you have a 5/12 chance to succeed and you roll twice your chance to succeed on either one is 1 -your chance to fail twice.

Your chance to fail twice is 7^12*7/12 or 49/144. (1-49/144) or 66% chance to succeed (you also have a 25/144 chance to succeed twice if that matters at all)

Faily

2017-09-16, 10:32 AM

Just wanted to note that Force Users can spend a Destiny point (turning a Light to Dark) and gain 1 Conflict to turn a Dark Side pip to Light Side pip on the Force-die.

I have also used this generator (http://game2.ca/eote/?montecarlo=100000) when playing Star Wars.

Jeivar

2017-09-16, 11:33 AM

Just wanted to note that Force Users can spend a Destiny point (turning a Light to Dark) and gain 1 Conflict to turn a Dark Side pip to Light Side pip on the Force-die.

I know, but it is hardly an optimal solution.

Roll a second die and do what with it? Add it to the first? Replace it if it's higher? What, exactly, are the positive outcomes on the die?

Roll a second die for a higher chance of scoring at least one success. But my question has been answered.

Thanks.

Lvl 2 Expert

2017-09-16, 02:23 PM

65.97%

65.89.

Now, the resultant numbers are slightly dissimilar, but it's small enough that I'm nearly completely sure this is due to rounding differences between computer's and mine numbers.

Yup. That's why it can be handy to leave numbers in fractions. 5/12+7/12*5/12=5/12+35/144=60/144+35/144=95/144 (lots of mental in between steps make it look more complicated than it is) = 0.6597 and a repeating 2 behind that.

But the method is the same, and the error is very small.

Knaight

2017-09-16, 03:05 PM

Math is a big weakness of mine. How does the probability here work? Is it just a plain 41,66 + 41,66?

The easy way to see that this can't be the formula is to consider the three die case - there's no way that 15/12 is the actual probability there. This is generally a good habit to be in with any equation, where you poke around a bit at the edges (drop values really high and really low) and see if it works there. In the case of the general equation of Ptot=P1+P2+P3...+Pn, going high (infinite events) produces an infinitely high probability when it is clearly supposed to approach 1.

*Also 41,666 rounds to 41,67 not 41,66.

If you want to do the math more manually, it's: chance the first die succeeds (41.6)+ chance first die fails (58.4)*chance second die suceeds (41.6, assuming independence)=65.89.

It's 65.9, not 65.89; you can't just pull a fourth significant figure out of nowhere when you've been working with three the whole time. There's also the matter of known error here - rounding to a tenth means that you have +/- 0.05 error on all terms. Given the terms F and S for first and second try probabilities, you're using the formula P=F+S-FS. The uncertainty there then works out to Utot=sqrt((UF-SUF)2+(US-FUS)2), which can then be treated as sqrt(UF2+SUF2+US2+FUS2), compliments of some error analysis specific equation simplifications. This works out to +/- 0.08, which checks out nicely.

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