Edge before or Edge after, when to use Edge dice to bump hits |
Edge before or Edge after, when to use Edge dice to bump hits |
Mar 12 2007, 04:11 AM
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#1
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Old Man of the North Group: Dumpshocked Posts: 9,674 Joined: 14-August 03 From: Just north of the Centre of the Universe Member No.: 5,463 |
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. |
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Mar 12 2007, 04:17 AM
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#2
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Moving Target Group: Members Posts: 668 Joined: 4-September 06 Member No.: 9,304 |
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. |
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Mar 12 2007, 05:18 AM
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#3
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Prime Runner Group: Banned Posts: 3,732 Joined: 1-September 05 From: Prague, Czech Republic Member No.: 7,665 |
Ack. Hellz yes it should. If you have the "Rule of Six" on your dice, your average number of hits per die converges to 2/5 of a hit. If you have a reroll up your sleave, your average number of hits per die is 5/9 of a hit. So the reroll is (unsurprisingly) superior to getting the rule of six. But if you get the Rule of Six you also get more dice. So let's compare like terms like good elementary school kids: That's 18/45 of a hit per die with "Ro6" and 25/45 of a hit per die with a reroll. So if 7 times your dice pool is more than 18 times your edge, you're better off rerolling. That means that the point where you are better off buying more dice is: Edge / Dicepool 1 / 2 2 / 5 3 / 7 4 / 10 5 / 12 6 / 15 7 / 17 Right. Calculus is your friend. -Frank |
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Mar 12 2007, 07:26 AM
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#4
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Great Dragon Group: Members Posts: 7,116 Joined: 26-February 02 Member No.: 1,449 |
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.
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Mar 12 2007, 01:35 PM
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#5
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Great Dragon Group: Members Posts: 7,089 Joined: 4-October 05 Member No.: 7,813 |
i can't help but notice that your figures are basically identical to his, with any values ending in .5 being rounded down. so i would say that his figures aren't really all that horrible of a way to do it :P |
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Mar 12 2007, 02:45 PM
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#6
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Running Target Group: Members Posts: 1,333 Joined: 19-August 06 From: Austin Member No.: 9,168 |
There's also the fact that you can succeed without the Edge, and save the point for later. It's all a big game, man! |
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Mar 12 2007, 02:55 PM
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#7
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Genuine Artificial Intelligence Group: Members Posts: 4,019 Joined: 12-June 03 Member No.: 4,715 |
It helps that when the OP did his lazy calculation of the expected number of hits on an exploding die he rounded it to exactly the correct value. :-) |
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Mar 12 2007, 05:18 PM
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#8
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Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
Sometimes you get lucky and rounding does the calculus for you. Other times it's transcindental and you're out of luck.
Where do you get this? Above you said that using rule of six it converges to 2/5 which is the same value I calculate. I don't get where that becomes 18/45. |
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Mar 12 2007, 06:14 PM
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#9
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Genuine Artificial Intelligence Group: Members Posts: 4,019 Joined: 12-June 03 Member No.: 4,715 |
Same value. He's just "showing his work" as it were. He's using an unreduced fraction to make it more evident where he is getting his numbers. |
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Mar 12 2007, 06:24 PM
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#10
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Moving Target Group: Members Posts: 200 Joined: 22-June 06 Member No.: 8,764 |
Would it be horrible of me to ask for someone to amplify on the calculus? I'm trying to follow along (if only for the sake of the mental exercise), but I'm kinda new to calc (never took it in college, now trying to muddle through on my own). If the answer is "no, off-topic," no worries. |
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Mar 12 2007, 07:02 PM
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#11
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Shadow Cartographer Group: Members Posts: 3,737 Joined: 2-June 06 From: Secret Tunnels under the UK (South West) Member No.: 8,636 |
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. |
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Mar 12 2007, 09:22 PM
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#12
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Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
How the 2/5 number is derived (this is not intended to be a mathematical proof): I first derive a formula to estimate a sort of expected value from a single die result. So I’ll introduce the following notation. P(n) as a probability of getting n on a roll of a single die, in this case I define probability as the average outcome. P(5) is the probability of rolling a 5, P(6) is the probability of rolling a 6. Then I use: P(Exploding Hit) = P(5) + P(6) + P(6)*P(Exploding Hit) What this means is the probability of a hit, is the sum of the probability of a 5, the probability of a 6 and then the last term is what makes this exploding, it says if you roll a 6 you get another die. Now this is an infinite sum, and it is the equivalent to:
Excuse the poor notation, it’s hard to do in text. That is the sum from n=1 to infinity over all that stuff. Now this is a geometric series, and it’s solution is well known, I looked it up in the CRC Standard Mathematical Tables just to be sure. The Result is:
P(5) = 1/6 //the odds of rolling a 5. P(6) = 1/6 //the odds of rolling a 6. Thus: P(Exploding Hit) = (1/6+1/6)/(1-1/6) = (2/6)/(5/6) = (2/6)*(6/5) = 2/5 That was the tough one to figure. I hope that assuaged your curiosity. |
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Mar 12 2007, 09:25 PM
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#13
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Moving Target Group: Members Posts: 200 Joined: 22-June 06 Member No.: 8,764 |
Demerzel, thanks.
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Mar 12 2007, 09:29 PM
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#14
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Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
You are welcome.
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Mar 12 2007, 09:40 PM
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#15
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Genuine Artificial Intelligence Group: Members Posts: 4,019 Joined: 12-June 03 Member No.: 4,715 |
You know, in this simple case you can skip the whole infinite sum business and go straight to basic algebra. Let me rename P(Exploding Hit) as simply E, the expected value. P(5) and P(6) are just 1/6, obviously. So substituting those: E=1/6+1/6+E/6 (this is just your equation rewritten) *algebra happens, solve for E* E=2/5 |
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Mar 12 2007, 11:37 PM
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#16
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Old Man of the North Group: Dumpshocked Posts: 9,674 Joined: 14-August 03 From: Just north of the Centre of the Universe Member No.: 5,463 |
I agree this is an important factor in deciding whether to use an Edge point or not. The original discussion arose in a situation in which a shot was being taken, and the player wanted to do as much damage as possible, so the decision to use Edge or not was moot. That decision is more likely to be important when the PC is facing a test with a threshold. At that point there is a specific target number of hits necessary, and a calculation of some sort could be made. That calculation would be muddied by factors such as how lucky the player feels, or how close to the final objective the team is (and presumably how little more opportunity there may be to need those Edge dice), which are hard to operationalize in an equation. Knowing the average number of hits is (1/3) the dice pool at least gives the player an idea of how many hits he can expect without using Edge. I too would be interested in seeing someone tackle this aspect of the decision. Not only for the obvious reason, but also because playing with numbers is fun. I see Demerzel dove into the calculus. Thanks. It was interesting to see that rounding off the algebraic approach got to the same point as the calculus. I don't feel so bad for not remembering the formulas. As an addendum, I didn't mention in the original post that if the Edge dice are saved till after the roll of the original dice pool, then comparing the two remaining options, OPTION 2 (with Edge dice only and Rule of 6s) vs. OPTION 3 (with rerolls of failures), the tipping point is at p = 1.74 e. So if the dice pool is more than 1.74 times the Edge dice, go with the rerolls. |
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Mar 13 2007, 02:00 PM
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#17
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Genuine Artificial Intelligence Group: Members Posts: 4,019 Joined: 12-June 03 Member No.: 4,715 |
Or p = 1.8 e, if you don't make rounding errors. ;) |
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Mar 13 2007, 02:06 PM
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#18
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Genuine Artificial Intelligence Group: Members Posts: 4,019 Joined: 12-June 03 Member No.: 4,715 |
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. |
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Mar 13 2007, 02:45 PM
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#19
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Old Man of the North Group: Dumpshocked Posts: 9,674 Joined: 14-August 03 From: Just north of the Centre of the Universe Member No.: 5,463 |
Good point, Moon-Hawk.
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Mar 13 2007, 04:08 PM
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#20
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Running Target Group: Members Posts: 1,333 Joined: 19-August 06 From: Austin Member No.: 9,168 |
That's really not the case, though - I had to beat it in to my head that 1/3 isn't really a good number to use, because I was failing far more often than I expected. I started using http://www.unseelie.org/cgi-bin/shadow.cgi to get a good estimate of successes. |
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Mar 13 2007, 04:41 PM
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#21
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Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
I put together some interesting charts some months ago. I should put them online somewhere and post a link...
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Mar 13 2007, 05:09 PM
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#22
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Genuine Artificial Intelligence Group: Members Posts: 4,019 Joined: 12-June 03 Member No.: 4,715 |
Well that's the problem. 1/3 is the right number to use when you're talking about expectation over a large number of trials. But just to reword what you'd essentially saying, the expected value is often not what you're looking for when you're doing an analysis of your chances. What you're really looking for are the odds of your doing so well or better, which a reasonably high degree of probability. For example, you may get 1/3 of your dice as hits as both the mean and mode, but what you really want to know is how many dice do you need to be reasonably sure (say 80%) that you'll succeed. The answer to this question is (approximately) four times as many dice as you need hits will leave you with approximately an 80% chance of getting that many hits or more. This is the root of the 4:1 trade-in mechanic for large dice pools and stressful situations. So 1/3 is exactly the right number to use when calculating expectation, it's just that expectation isn't always the value that you really want to know. |
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Mar 13 2007, 05:35 PM
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#23
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Moving Target Group: Members Posts: 200 Joined: 22-June 06 Member No.: 8,764 |
Great site, but I wish the link to the CGI source worked. I've been trying to work out the probabilities on my own (ha! :please: ), and haven't come up with anything that makes any kind of sense. Enjoying the thread anyway, though :) |
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Mar 13 2007, 06:17 PM
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#24
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Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
The probability actually starts at about 80% and goes up.
Unfortunately there's not a good rule of thumb for this kind of probability. I've made some tables that you can look up probabilities on and some plots of probability of success, I'll try and PDF 'em up and post a link soon. I keep them around the game table when I'm running a game. |
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Mar 13 2007, 08:34 PM
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#25
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Running Target Group: Members Posts: 1,333 Joined: 19-August 06 From: Austin Member No.: 9,168 |
Yes. That is precisely it, and you've actually just made it easier for me to get in my head. For some reason, probability and I just don't get along. I can do multivariate calculus just fine, but ask me to tell you the odds of getting a straight in Texas Hold 'em, and I blank out. Because you usually don't want to know whether you're 50% likely to succeed. You want to know when you're pretty likely to do it. |
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