probabilities of winning a DRN
 

I found it fun to do this, and thought others might find it useful in planning strats; so here it is.


The formula for the average of an open ended roll
M= n(n+1) / 2(n-1)

The formula for the variance of an open ended roll
V= n(n+1)(n^2+7n-2) / 12(n-1)^2

n= the number of sides in both these formulas, in this case 6.

also, multiple dice are additive in the formulas, so doing them will return the results of a single d6, simply double them for a 2d6.

they solve out to:
M= 8.4
V= 3.2

now, the DRN will actually cancel out in all rolls, so all that matters is the difference in the constants. Thus, in an attack roll where the attacker has an attack 13 and the defender a defense of 14, there is a difference of 1 between the attacker and defender. divide this difference by the variance to get the z score. the z score can then be compared to a z table to see what the probability that the defender will block the attack is. below though i've listed all results for a difference up to 6. the percent chance is basically the percent that the unit with the lower number can roll a high enough DRN to win the roll. so a unit that adds up to being 6 less than the opponent on whatever number (attack - defense, fear - moral, missile - missile dodge, ect), will only have a 3% chance to win a roll.

X-Xbar:
6......3%
5......5%
4......11%
3......17%
2......26%
1......38%
0......50%

Lastly, note that this table assumes normality, which is not the case. The distribution oscillates in a pattern, but which is difficult to represent in a formula. The oscillation is due to the open ended roll, thus there is a higher chance to roll both a 6 and an 8 than there is a 7. There is effectively a 'hump' in the distribution at 6 then, meaning that the estimates above for a difference of 6 are slightly underestimated. Additionally, the distribution isn't normal, but log normal, and asymptotically approaches infinity. A log normal distribution requires recomputing everything with a log link, which I may do at some point in time to give more exact estimates. Basically, all estimates are slightly underestimated due to the long positive tail that the distribution should have. This long positive tail has a special effect in some checks where a 'win' for some actions is particularly meaningful. Such as, an attack that rolls unusually high most likely gets a guaranteed hit, as it is unlikely a defender will also roll an unusually high roll on the same check. A successful attack results in additional things that a successful defend does not. This makes the quality of log normal distributions to allow higher rolls particularly meaningful for most rolls in a way that can't be captured in even a log normal distribution. Rather, a Binomial or perhaps Poisson distribution would have to be used to model the effect of a number of success over a time period.

Other than all that though, it is a decent table. I post it again so it doesn't get lost in all that text http://forum.shrapnelgames.com/images/smilies/happy.gif

X-Xbar:
6......3%
5......5%
4......11%
3......17%
2......26%
1......38%
0......50%

Re: probabilities of winning a DRN
 

Is this different from the table on the first page of the manual?

Re: probabilities of winning a DRN
 

lol... RTFM.

Re: probabilities of winning a DRN
 

Yes, it is different because it assumes normality and uses statistical methods. Use the manual table for values within the table's range, but otherwise Omniziron's formula might be a useful approximation if you want to extrapolate to +/- 20 or so.

-Max

Re: probabilities of winning a DRN
 

i never look in that end of the manual http://forum.shrapnelgames.com/images/smilies/happy.gif

but hey! it appears they are even with the log distribution. nice

Re: probabilities of winning a DRN
 

it doesnt' account for the wave caused by the open ended roll though, the probability for 8 should actually be higher than for 7

Re: probabilities of winning a DRN
 

Values outside of the table's range really aren't that necessary to extrapolate, now are they Max? There's not a whole lot of difference between a 1% chance and a 1.1% chance of success, is there?

I just feel kinda bad for Omni after that whole thing. But it IS funny! lol

Re: probabilities of winning a DRN
 

hmmm... and I tossed and turned all last nite in bed thinking about it too. I couldn't wait to get up this morning and look up mean and variance formulas for open ended rolls.

course, i did notice the wave, which the manual doesn't mention.


the wave! the wave! it matters. maybe not really, very rare for anything to get to a difference of 8.

Re: probabilities of winning a DRN
 

hmmm...

and actually the oscillation is very significant right at 7 and 8, the 6 isn't affected by it. the probability for 7 is dropped, while the probability for 8 is raised. where normally 7 is 8% and 8 is 6%, 7 is closer to 5% and 8 is 11%!

A crafty player may find ways to exploit that!

The oscillation continues and increases in frequency but drops precipitously in magnitude. the reason is simple, every number that would usually be attained by rolling a 6 on the dice is now having a way in which that number could usually be rolled shifted upward. A depression in the bell curve is created where the number losing a way to be rolled is, while the number above essentially gains that extra opportunity to be rolled.


think about this when kitting up your mage with extra penetration. maybe can hit that sweet spot, and you might actually hurt your chances to soul slay that SC by pumping penetration up one more!

Re: probabilities of winning a DRN
 

Rerolls are actually made with a dice that's 'numbered' 0-5. IIRC, probability of rolling anything under 6 with one open-ended dice is equal, and probability of getting anything above 5 and below 12 is equal, etc.

So something like:
5/6 chance of result 1-5.
1/6 chance of first 6; 1/6*5/6 chance of a result 6-10
1/36 chance of second 6; 1/6*1/6*5/6 chance of a result 11-15


Then I tried thinking about two dice, and gave up and decided that the manual table is good enough.

Re: probabilities of winning a DRN
 

You know, I couldn't find this damned table for some time (after being told to go look at it), because it's on page *5*! Page 5 is for installing instructions, and "what this first menu means" kind of stuff. http://forum.shrapnelgames.com/image...ies/tongue.gif

Anyway, it explains right there that a 6 is not just rerolled and added. You subtract one, and reroll. So there is no statistical variation in odds between 5, 6, 7 etc. I roll of 5 is 5. A roll of 6 is 5 + (d6). So your odds of success are not affected by the DRN mechanics.


But Endo, it's not a 0-5. It's 1-5, with each roll of 6 simply being another 1-5 on top of that. 0 would just suck. http://forum.shrapnelgames.com/image...ies/tongue.gif

Re: probabilities of winning a DRN
 

ah, that would solve that problem then.

that KO and JK, they thought of everything.

Re: probabilities of winning a DRN
 

That turns out not to be the case. It is always true that a success by X+1 also succeeds by X.

-Max

Re: probabilities of winning a DRN
 

well, with jim morrisons post, the issue is solved, however it would be true, you can count it up yourself and see.

the rows show how many possible ways to roll each number on 2d6 open ended

2
33
444
5555
66666
7777
88888

normally, 7 would be the mean. here it dips because it is no longer possible to roll a 7 with 6+1, it would now be a 6,?,1; and the area of the curve normally under seven gets pushed towards the end. count it yourself and see.

Re: probabilities of winning a DRN
 

But the area for 8 gets redistributed too (2+6). I guess that probably makes 6 the mode.

-Max

Re: probabilities of winning a DRN
 

The ripple is self correcting. Each number between 7-11 contains a 6. So yes, there is a slight deviation in favor of 6 - but not really. The fact that each value above 6 has a slightly lower chance of actually occurring is irrelevant, because it is ONLY replaced by a higher number.

Look at it this way, we'll expand that DRN rate table. Here are the normal probabilities, up to 12.


2
33
444
5555
66666
777777 **
88888 **
9999 **
10 10 10 **
11 11 **
12 *


Each asterisk, represents a roll that contains a 6. We see that out of 36 nominal rolls of 2d6, we get at least one 6, 11 times. This is absolutely not a reduced chance of success. Every time you roll a 6, you have a 17% chance of actually getting the number that you rolled (if you reroll a 1), but you have an 83% of rolling a higher number. While the 2d6 is capped at 12, all of these higher rolls that have a "reduced statistical chance" of occurring, are only shifted upwards. So you may have a reduced chance of hitting a natural 7, but that is nicely offset by a greatly increased chance of hitting a natural 30 - which would have otherwise been impossible. http://forum.shrapnelgames.com/image...ies/tongue.gif

Re: probabilities of winning a DRN
 

This thread is good and all, but I have a question.

Why was 3 scared of 7?

Re: probabilities of winning a DRN
 

789!