Ok gang, here is some number crunching for those who get a rise out of that sort of thing.

In a recent game, a fellow player and I discussed what is the best way to use Edge dice to get more hits on a Test. My buddy contended that using Edge right away gave a better result (more hits). I felt that saving the Edge point to get re-rolls was often better. Neither one of us convinced the other to change his mind, so I decided to do a little algebra to figure out how they stack up on average.

From the SR4 manual, there are three ways in which Edge can be used to bump up the number of hits on a test.

OPTION 1: declare use of Edge before rolling, include Edge dice in the dice pool, and use the Rule of 6s.

OPTION 2: declare use of Edge after rolling, roll Edge dice, and apply Rule of 6s for Edge dice only.

OPTION 3: declare after rolling, re-roll failures, and do not apply Rule of 6s.

So here’s the math:- (Skip to the end if you don’t want to check out the algebra)

5s and 6s are hits, so on average (1/3) of the dice rolls are a hit. For re-rolling 6s, (1/6) of the dice get to be re-rolled.

Let p= dice pool dice
e= Edge dice
n= total number of hits.

Then

For No Edge Dice: n = (1/3) p

For OPTION 1: (Since re-rolls could theoretically go on forever, with diminishing likelihood, calculus should be used here, but I haven’t done that stuff in decades, so I will just round the equation off at three terms, which will be close enough)

n = (1/3) (p + e) + (1/3) ((1/6) (p + e)) + (1/3) ((1/6) (1/6) (p + e))
= (36/108) (p + e) + (6/108) (p + e) + (1/108) (p + e)
= (43/108) (p + e)
= 0.4 (p + e)
= 0.4 p + 0.4 e

For OPTION 2: (Ditto for rounding)

n = (1/3) (p + e) + (1/3) (1/6) e + (1/3) (1/6) (1/6) e
= (1/3) p + (43/108) e
= 0.33 p + 0.4 e

For OPTION 3: (2/3 of the original roll are failures, and are re- rolled)

n = (1/3) p + (1/3) (2/3) p
= (5/9) p
= 0.56 p

SO:

OPTION 1 is always marginally better than OPTION 2, but I imagine a player may want to risk OPTION 2 in order to try for enough successes before committing to using a point of Edge. The larger the original dice pool, p, the more likely a typical threshold of hits are to happen without use of Edge, but the average number of hits is also smaller with OPTION 2.

The fun comparison, and the one we discussed, is between OPTION 1 (declare Edge before and use Rule of 6s) and OPTION 3 (declare Edge after and re-roll failures).

The turning point between the two is when the two equations are set equal to each other:

0.4 p + 0.4 e = 0.56 p

This gives p = 2.5 e.

So if the original dice pool is more than 2.5 times as large as the character’s Edge Attribute, the player profits by using the edge point to re-roll failures. If the dice pool is less than 2.5 times the Edge, then it is more profitable to use the point of Edge before rolling and using the Rule of 6s.

Ouch
My head hurts

Way to logical for me.

I use edge when I don't think the number of successes will be enough, or when I know so, like damage resistance checks.

Another advantage to declaring Edge afterwards is that you can only apply Edge once to any dice test, so if you add Edge first, you are out of luck if you roll a Glitch or a Critical Glitch.

Sometimes you get lucky and rounding does the calculus for you. Other times it's transcindental and you're out of luck.

All very nice, folks, but this is half the story. The big factor in comparing the benefit of before and after edge use is knowing whether or not you need to. The above calculations do not account for this aspect of the game and so don't model it very well.

The most valid criteria for assessing the best approach would not be which gets the most hits on a single test, but which gets the most hits over a series of n tests. Because on some occasions, the player using "after spending" will achieve sufficient hits without the use of edge, it will be available on more of the tests than the player using "prior spending."

The advantage of prior spending will increase as the probability of success decreases. To illustrate that: if I've got a fifty-fifty chance of succeeding using my normal pool, then edge use will be wasted approximately half the time. If the chance of succeeding were a difficult 1 in 5, then my edge would only be wasted 20% of the time (assuming edge made enough difference to turn it into a success, but we'll allow that as it's no guarantee in either pre- or post- case).

It would be fun to work out the tipping point with this principle accounted for, but I'll have to come back to it as I don't have time for now, unless someone else wants to take a stab at it. The general rule though would be the more edge you have, the more likely you are to use it for easier tests, and therefore the more likely you are to waste it in the long term. But then with more edge, the advantage of a prior spend is greater. It's tricky, but in a very edge-scarce environment (infrequent refresh or small pools) the post spend will be rewarded more. In an edge-saturated environment, the pre-spend will be rewarded more. Assuming that the number of trials n is fixed and not a player choice, then the tipping point for any given difficulty level of test is going to be a ratio of n:e (where e is edge pool including refreshes).

I might take a stab at this for a range of different probabilities unless someone beats me to it, but I don't have time right now. Looks fun, though.

-K.

Demerzel, thanks.

You are welcome.

Although, really, if you're talking about rolling after the fact, you don't have to base your decision on p, the number of dice in the original pool.
Say p is divided into s and f, the number of successes and failures where s+f=p.
Once you've rolled, s and f are known quantities.
So really your decision to add edge dice e or reroll failures is based on the relationship between f and e, not p.
You're looking for the point where 1/3*f = 0.4e
Which is when 5/6 * f = e, or when f = 1.2 e. So when the number of failures is higher than 1.2 times your edge, you should reroll failures, but if it's less you should add edge.

The original calculations are only for the situation where you have already decided that you WILL use edge, but you're trying to decide choose which strategy will yield the most expected successes overall.

Good point, Moon-Hawk.

I put together some interesting charts some months ago. I should put them online somewhere and post a link...

The probability actually starts at about 80% and goes up.

The math seems to say that re-rolling is better than than rolling one more dice.

I almost always go with spending my edge after the initial roll simply because I'd hate to waste an edge point on something that I have a chance of getting without it. I mean, if its a desperate situation I will use it before hand if I have no shot at getting the target number without it, but that doesn't happen often. Generally the concept of "well...4 hits on 7 dice could happen...sometimes my dice like me..." is just too much gamer instinct to ignore. I guess maybe thats why I never got to calc or trig

here's a little spreadsheet calculator to tell you what your hits are likely to be with any given combination of edge use:

http://www.smarmybastard.com/srun/SRED.xls

@yesman:
There are a few flaws there.

When you say "what your hits are likely to be". In the case of no edge and a 15 dice pool, the answer that yields of 5 is only going to be correct 21.4% of the time. Now it is the most likely result, but it is not really likely.

You also treat edge dice as only ever exploding once. In fact they could explode infinitely. Even if you accounted for that (Using the 2/5 multiplier instead of 1/3 + 1/18 per die) you'd still be having issues from the previous flaw.

Not to dig up a topic people were done talking about, but.. am I misunderstanding one of the available Edge rules? Particularly...

"You may re-roll all of the dice on a single test that did not score a hit."

It would appear that I'm the only person that does NOT read this to mean:

"You may re-roll all of the dice that did not score a hit on a single test."

Do you see the syntax difference I'm pointing out? The first version is as it is printed in the book. I've always understood that to mean that you can do a re-roll on a test if you totally got screwed and got zero hits on the test. Am I wrong in this interpretation? If you're to read it with as pure a meaning as can be derived from the original text, the portion of the sentence that says "that did not score a hit" is referring to the test, not the dice.

It's sort of the.. "Throw the cow over the fence some hay" thing. Are we throwing cows over the fence? Or are we throwing hay to the cow located on the other side of the fence. English teachers and common understanding of english syntax would say the latter is true. If that same logic is applied to this rule, then the method of Edge spending that people seem to have calculated as superior is just flat out against the rules.

Probability is good and all, but the real question is how have you been rolling that gaming session. If you hot, then definitely uses it before rolling. If you are having an off session, then use it after unless your original dice pool is small. Granted this is not as scientific as previous post, but sometime the dice are with you and other times you are going to die from a baby throwing its pacifier at you.