ExcellencyMath

Second Edition provides a set of "excellency" charms that work in various ways. Each has advantages and disadvantages. This pages compares the First and Second Excellencies, describing the math and probabilities involved. Briefly, here is how they work:

• First Excellency: Buy dice for a roll. Most exalts spend 1m per die.
• Second Excellency: Buy successes for a roll. Most exalts spend 2m per success.

ExcellencyModelling gives an example of how to model Exalted dice, based (loosely) on the math below.

Success Per Die

The book claims that two dice, on average, generate one success. This implies one die generates 0.5 successes on average. This is easy to verify. If you were to roll a die 100 times, the ideal distribution on a fair d10 would be for each number to come up exactly 10 times. Count the successes this would generate:

60 rolls of six or less x 0 successes = 0 successes

30 rolls of 7, 8 or 9 x 1 success = 30 successes

10 rolls of 10 x 2 success = 20 successes

Total successes = 50 on 100 die, thus 0.5 successes per die.

So, First and Second Excellency would appear to be equivalent. Well, not exactly. It's true that, on average, they will generate equal numbers of successes. There are two big differences, however. The first is that the successes of the Second Excellency are applied later in the Oder of Modifiers than the dice from the First, making bonuses from the second immune to cancellation from, say, effects that reduce pools to zero. This doesn't happen often, but is a slight advantage for the Second. The advantage of the First is that it allows the possibility up to four times the effect of paying the same cost on the Second, though also the possibility of no effect at all. How does this break down?

First Excellency: Counting

With the Second Excellency, spending 2m buys a 100% chance of exactly one success. If a solar spends those two motes on the First instead, what happens. One way to figure this out is just to count. The solar buys two dice. They can come up 100 distinct ways. In the following table, the result of one die is on one axis, the result of the other die is on the other axis, and the fields of the table list how many successes occur for that combination of results.

|| ||1||2||3||4||5||6||7||8||9||10|| ||1||0||0||0||0||0||0||1||1||1||2|| ||2||0||0||0||0||0||0||1||1||1||2|| ||3||0||0||0||0||0||0||1||1||1||2|| ||4||0||0||0||0||0||0||1||1||1||2|| ||5||0||0||0||0||0||0||1||1||1||2|| ||6||0||0||0||0||0||0||1||1||1||2|| ||7||1||1||1||1||1||1||2||2||2||3|| ||8||1||1||1||1||1||1||2||2||2||3|| ||9||1||1||1||1||1||1||2||2||2||3|| ||10||2||2||2||2||2||2||3||3||3||4||

If you count these up, you get:

36 results of 0 successes

36 results of 1 success

21 results of 2 successes

6 results of 3 successes

1 result of 4 successes

First Excellency: Math

You can get the same result by doing math. For the purpose of this description, assume the two dice being rolled are different colors (say red and blue). You can figure out the probability of each success count just using math knowing that a single die has a 6/10 chance to roll zero successes, a 3/10 chance to roll exactly 1 success and a 1/10 chance to roll two successes. So:

To roll zero successes, both dice need to generate no successes:

P0 = 6/10 * 6/10 = 36/100 = 36%

To roll exactly one success, either the red die needs to roll 7,8 or 9 while the blue rolls six or under, or the blue die needs to roll 7,8 or 9 while the red rolls six or under:

P1 = (3/10 * 6/10) + (6/10 * 3/10) = (18+18)/100 = 36%

To roll exactly two successes, either both the red and blue die need to roll one success each, or one of them needs to roll two while the other rolls none:

P2 = (3/10 * 3/10) + (1/10 * 6/10) + (6/10 * 1/10) = (9+6+6)/100 = 21%

To roll exactly three successes, one die needs to roll two success and the other needs to roll one:

P3 = (3/10 * 1/10) + (1/10 * 3/10) = (3+3)/100 = 6%

To roll exactly four successes, both die need to generate two:

P4 = 1/10 * 1/10 = 1/100 = 1%

Note that these results match the count above.

So, it appears that the First Excellency has a 36% chance of being worse that the Second. It has an equal chance (36%) of being exactly as efficient as the Second, with a 63% (1 - P0) chance of being at least as efficient as the Second. It has a 28% chance of being twice as efficient, a 6% chance of being three times as efficient and a 1% chance of being four times as efficient. In all, there is a 35% chance of being more efficient that Second Excellency.

Second Excellency: Sanity Check

Do the results above really create an average outcome equal to that provided by the Second Excellency? Let's check. Say we rolled a pair of d10s 100 times. The percentages above suggest we would get the following results:

36 x 0 success = 0 successes

36 x 1 success = 36 successes

21 x 2 successes = 42 successes

6 x 3 successes = 18 successes

1 x 4 successes = 4 successes

Total successes = 100 successes on 100 rolls (or 100 successes on 200 dice, meaning 0.5 successes per die).

Rolling that pair 100 times would be like buying two dice with the First Excellency 100 times (200 motes for a solar). To buy 100 successes with the Second Excellency, a solar would need to spend exactly that many motes.

First Excellency: More Dice

So, how does this scale? That is, how to the probabilities work out with, say, four dice. Using the same method as above gets a bit ugly, so change our tactics slightly. (Really, we are doing the same thing, but this way will look cleaner). In this method, we figure out how many possible ways the dice can give a certain number of successes, then divide that by the total number of ways the dice can come up total.

In this description, C(x,y) means taking the combinations of x things taken y at a time where order does not matter. The technique use in what follows can be a little confusing, but the basic idea can be illustrated by an example. The technique breaks the problem into situations where you figure out how many combinations of dice could result in a certain pattern. Suppose, for example, that you need to figure out how many ways you can get two of the four dice to have one success each, and the others to have none. What you do first is figure out how many ways you can select two distinct dice from the pool. Say you have a blue die (b), a green die (g), a red die (r) and a yellow (y). The number of distinct pairs you can select from this pool is C(4,2), which is read "4 choose 2". This value is six (read the link above on how to calculate this), with the six distinct pairs being bg, br, by, gr, gy, and ry. One you have selected a pair, each dice in that pair can generate exactly one success three ways. Since there are two dice, that means that there are 3^2 possible ways these dice can be rolled where each one shows exactly one success (which is the pattern you are looking for in this example). The remaining two dice must both come up with zero success. Each of these die can do this in six ways, so there are 6^2 possible ways these two dice can generate zero successes on each die. So, to figure out how many possible ways the pattern can be matched, you multiply the number of ways you can choose the success pair by the successful pair possibilities and then by the possibilities of the remaining dice: C(4,2) * 3^2 * 6^2 = 6 * 9 * 36 = 1,944. Note that this is not the only way to generate two successes, just the the way where two dice have one success each and the other two have none.

With four motes on the Second Excellency, a solar has a 100% chance of getting two successes. Spending four motes on the First Excellency, the solar can expect:

Total possible dice results (paying attention to order): each die can come up 10 ways, and there are four die:

10^4 = 10,000 possibilities

To roll zero successes, all dice need to generate no successes. Each die can generate no successes six different ways, so all of these combine:

6^4 = 1296 possibilities; 1296/10000 = 12.96%

To roll exactly one success, only a single die can roll a single success. There are four ways to choose the first die, and that die can generate a single success three ways. The remaining die can generate zero successes six ways each. So

C(4,1) * 3 * 6^3 = 2592 possibilities; 2592/10000 = 25.92%

To roll exactly two successes, either two dice roll one success or one die rolls two. In the first case there are six ways to choose the dice of the successful pair, and nine ways these two dice can get one success each, with the remaining die generating zero successes six ways each. In the second case, there are four ways to choose the successful die, and only one way that die can generate two successes, with the remaining die generating zero successes six ways each.

C(4,2) * 3^2 * 6^2 = 1944 possibilities

C(4,1) * 1 * 6^3 = 864 possibilities

1944 + 864 = 2808 possibilities; 2808/10000 = 28.08%

To roll exactly three successes, one die needs to roll two success and another needs to roll one, or three die need to roll one success. There are four ways to choose the first die, and that die can generate a two successes one way, with three ways to choose the second die and three ways it can generate a single success. In the second, there are 4 ways to choose the successful three dice and 27 ways that trio can generate one success each. In both cases, remaining die each have six ways to generate no successes.

C(4,1) * 1 * C(3,1) * 3 * 6^2 = 1296 possibilities

C(4,3) * 3^3 * 6 = 648 possibilities

1296 + 648 = 1944 possibilities; 1944/10000 = 19.44%

To roll exactly four successes, either two dice roll 10s, one die is a 10 and two others single successes, or four dice roll one success each:

C(4,2) * 1 * 6^2 = 216 possibilities

C(4,1) * 1 * C(3,2) * 3^3 * 6 = 648 possibilities

C(4,4.) * 3^4 = 81 possibilities

216 + 648 + 81 = 945 possibilities; 945/10000 = 9.45%

To roll five successes, success combinations are either (2,2,1,0) or (1,1,1,2):

C(4,2) * 1 * C(2,1) * 3 * 6 = 216 possibilities

C(4,3) * 3^3 * 1 = 108 possibilities

216 + 108 = 324 possibilities; 324/10000 = 3.24%

To roll six successes, success combinations are either (2,2,2,0) or (2,2,1,1):

C(4,3) * 1 * 6 = 24 possibilities

C(4,2) * 1 * 3^2 = 54 possibilities

24 + 54 = 78 possibilities; 78/10000 = 0.78%

To roll seven successes, success combination is (2,2,2,1):

C(4,3) * 1 * 3 = 12 possibilities; 12/10000 = 0.12%

Only one way to roll eight successes

1 possibility; 1/10000 = 0.0001%

So, with four dice, you are only 13% likely to get no benefit, but you are more likely (38.88%) to do worse than Second Excellency. You are 61.22% likely to do at least as well as Second Excellency. You have a 33.04% chance of doing better than Second Excellency. Essentially, as you add more dice, the more likely you are to have an average result.

And, just for purity:

1296 x 0 successes = 0 successes

2592 x 1 successes = 2592 successes

2808 x 2 successes = 5616 successes

1944 x 3 successes = 5832 successes

945 x 4 successes = 3780 successes

324 x 5 successes = 1620 successes

78 x 6 successes = 468 successes

12 x 7 successes = 84 successes

1 x 8 successes = 8 successes

Total scccesses = 20,000 on 10,000 rolls (or 20,000 on 40,000 dice, averaging 0.5 successes per die)

First Excellency: Odd Dice

What if you have three motes to spend (or, more likely, that your added dice limit is an odd number)? Since you can't "fractionally buy" the Second Excellency, if you have the motes, First is usually the way to go. Sticking just to three motes, suppose you are in a situation where you can activate both First and Second Excellencies on the same test. If you spend 2m on Second and 1m on first, from the math above, the following are the possibilities:

2 motes gives a 100% of one success, plus 1m that buys:

60% chance of failure (total 1 success)

30% chance of a single success (total 2 successes)

10% chance of two successes (total 3 successes)

If you spent these three motes on the First, you'd get the following:

Total possible dice results (paying attention to order): each die can come up 10 ways, and there are three die:

10^3 = 1,000 possibilities

To roll zero successes, all dice need to generate no successes. Each die can generate no successes six different ways, so all of these combine:

6^3 = 216 possibilities; 216/1000 = 21.6%

To roll exactly one success, only a single die can roll a single success (1,0,0):

C(3,1) * 3 * 6^2 = 324 possibilities; 324/1000 = 32.4%

To roll exactly two successes, either (2,0,0) or (1,1,0):

C(3,2) * 3^2 * 6 = 162 possibilities

C(3,1) * 1 * 6^2 = 108 possibilities

162 + 108 = 270 possibilities; 270/1000 = 27.0%

To roll exactly three successes, either (2,1,0) or (1,1,1):

C(3,1) * 1 * C(2,1) * 3 * 6 = 108 possibilities

3^3 = 27 possibilities

108 + 27 = 135 possibilities; 135/1000 = 13.5%

To roll exactly four successes, (2,2,0) or (2,1,1):

C(3,2) * 1 * 6 = 18 possibilities

C(3,1) * 1 * 3^2 = 27 possibilities

18 + 27 = 45 possibilities; 45/1000 = 4.5%

To roll five successes, success combination is (2,2,1):

C(3,2) * 1 * 3 = 9 possibilities; 9/1000 = 0.9%

Only one way to roll six successes

1 possibility; 1/1000 = 0.001%

Again for purity, on 1000 rolls:

216 x 0 successes = 0 successes

324 x 1 successes = 324 successes

270 x 2 successes = 540 successes

135 x 3 successes = 405 successes

45 x 4 successes = 180 successes

9 x 5 successes = 45 successes

1 x 6 successes = 6 successes

Total scccesses = 1,500 on 1,000 rolls (or 1,500 on 3,000 dice, averaging 0.5 successes per die)

You can see, though, that 3 motes on the First gives a chance at at least three successes that is almost twice as high as 2m on Second and 1m on First.

Variations

Just for kicks, it may be useful to change the rules a bit and see what happens.

Flat dice

Consider a case where 10s do not count as two successes, but only one, like damage dice. This makes the math much easier. When rolling the same pair as above, but not counting 10s as two successes:

To roll zero successes, both dice need to generate no successes:

P0 = 6/10 * 6/10 = 36/100 = 36%

To roll exactly one success, either the red die needs to roll 7,8, 9 or 10 while the blue rolls six or under, or the blue die needs to roll 7,8, 9 or 10 while the red rolls six or under:

P1 = (4/10 * 6/10) + (6/10 * 4/10) = (24+24)/100 = 48%

To roll exactly two successes, both need to roll 7 or higher:

P2 = 4/10 * 4/10 = 16/100 = 16%

You'd expect an average of 0.4 successes per die, which is what you get. For example, an average for 100 rolled pairs:

36 x 0 success = 0 successes

48 x 1 success = 48 successes

16 x 2 successes = 32 successes

Total successes = 80 successes on 100 rolls (or 80 successes on 200 dice, meaning 0.4 successes per die).

Lower target

Suppose 10s worked like normal, but the target number was six, not seven. First then becomes a better deal:

P0 = 5/10 * 5/10 = 36/100 = 25%

P1 = (4/10 * 5/10) + (4/10 * 5/10) = (20+20)/100 = 40%

P2 = (410 * 4/10) + (1/10 * 5/10) + (5/10 * 1/10) = (16+5+5)/100 = 26%

P3 = (4/10 * 1/10) + (1/10 * 4/10) = (4+4)/100 = 8%

P4 = 1/10 * 1/10 = 1/100 = 1%

Average should go up by 10%, which is what happens:

25 x 0 success = 0 successes

40x 1 success = 40 successes

26 x 2 successes = 52 successes

8 x 3 successes = 24 successes

1 x 4 successes = 4 successes

Total successes = 120 successes on 100 rolls (or 120 successes on 200 dice, meaning 0.6 successes per die).

Flat dice and a lower target

Suppose we lower the target to six and only count 10s as one success. This means that 1-5 is failure, 6-10 is success, so works just like flipping coins. If you flip two coins and call heads "successes", you have four equal probability outcomes, both tails, one head and one tail, one tail and one head, two heads. Doing this ten siders looks like this:

P0 = 5/10 * 5/10 = 25/100 = 25%

P1 = (5/10 * 5/10) + (5/10 * 5/10) = (25+25)/100 = 50%

P2 = 5/10 * 5/10 = 25/100 = 25%

You'd expect an average of 0.5 successes per die, which is what you get. For example, an average for 100 rolled pairs:

25 x 0 success = 0 successes

50x 1 success = 50 successes

25 x 2 successes = 50 successes

Total successes = 100 successes on 100 rolls (or 100 successes on 200 dice, meaning 0.5 successes per die).

So, this works out to have the same average, but different payoffs than standard First Excellency.

Third Excellency Variations

From what comes above, it should be clear that the average roll for a set of dice is one success for every two dice. This is a good metric for determining if you should use the Third Excellency to reroll. If your original roll exceeds the average, on average a reroll will not improve the result. Likewise if the original roll doesn't meet the average, a reroll will improve it on average.

Some use a house rule that allows the Third Excellency to reroll only those dice that failed, keeping any that generated successes. So, on average, a roll of x dice will generate x/2 successes. Using the Third Excellency with this house rule, you'd roll the x/2 failures, generating x/4 more successes on average. So, with this house rule, motes spent on the Third Excellency will turn X dice into 3x/4 successes, or x/4 more successes than the roll alone.

Compare this to the Second Excellency, where every two motes adds a success. In this case, you get the x/2 successes from the roll, plus m/2 successes, where m is the number of motes spent. Using the Third Excellency house rule, m is a constant of 4. If you spent that much essence on the Second Excellency, you'd get x/2 + 2 successes on average. Since the extra successes provided by the Third Excellency house rule are x/4, if the size of the dice pool (x) is greater than eight, the Third Excellency would provide more successes per mote, on average.

Comments

What kind of math could we put here reflecting the fact that each individual die on an individual roll only has a 4/10 chance of producing successes? (ie, treat success as a binary/boolean state) I'm sure things would look different if 10s were a single success, and 6s succeeded also. - IanPrice

It might also change things to put up four or more dice, since then you'd have instances of below average success besides zero. - IanPrice

Which Excellency is better when used to increase DV in terms of motes spent? nikink (1st Ex means you throw away excess successes, and could possibly roll no successes, thus the 2nd is better when raising DV. Alternatively 1st means you could spend half as many motes to get as many successes as using 2nd.)

On average, there is no difference. So, it depends on your definition of "better". The 2nd is more reliable, in that it always generates the average result. 1st can do both worse and better, but depend on luck. Read the paragraph above that starts with "So, it appears that the First Excellency has a 36% chance of being worse that the Second". - Wordman

Yep, but with the 1st you may end up rolling excess successes which are discarded, as well as the chance of rolling less. So the 1st can never do better, but may do worse. So if you are spending your maximum motes, 2nd is always better; 1st is only good if you cannot/will not spend your maximum, and thus need to risk getting more successes than (die pool/2). nikink (This is stemming from a debate with another person in my gaming group, he might chime in here with his view which seems to be 1st is uniformly sub-optimal for raising DVs due to the preceding idea.)

Then there's the third, which, for say, 4 motes, gives you +3 DV in many cases. That's 1.33 motes per DV, guaranteed. It's a fabulous deal - if you'll only ever need a +3DV to save your butt. It's equal in cost if you need a +2DV to save your butt, and only a penalty if you need either 1DV boost (stunt more!) or worse, 4+. In the case of 4+, however, you're already pushing most 2nd excellencies - you'd need 8 Att+Ability to hit it. Which, you likely have, I'll admit. But really, the most you'll ever get is a 4-5 DV need that you could hit. Thus, the question is raised - What % of the time do you need that 4-5 boost, vs. needing a 3 or less? That'll determine the usefulness of the 3rd excellency. Especially since once you've used an excellency, you're locked into it. (As a side note, it's frigging amazing for DB's, as they're capped at Ability+Specialty of 8, and can use it reflexively for 3 motes - always a bargain!) -- GreenLantern

I see what you are saying with the 1st vs. 2nd for DV. I totally spaced the "roll dice for DV from the 1st" rule, because I consider it so stupid. What's the point of using unrolled values for defense if you are just going to roll during defense anyway? It adds additional complexity all in order to, as you say, create something demonstrably inferior to to just using a "two dice gives +1 DV" rule (according to the errata, they did slightly address this with stunt dice for DV, where the "don't roll" version is the official one). Dumb. When I get a minute, I'll add a DV section. - Wordman