Joe Kisenwether, co-designer of the Freaks and Fantasy Button Men expansions
Odds of Successful Tripping
Suppose you’re trying to trip a d10 with d6. How likely are you to get away with it? The most straightforward way to figure this out is making a table of the possible outcomes of the trip roll:
The columns indicate the result on the d10, the rows the result on the d6 and X’s appear in those boxes where the trip was successful. We can see that the trip works 21 out of 60 times, which simplifies to 7/20 = 35%.
The layout of the X’s in the table above should suggest something to you if you’ve read the article on “The Highest Lower Number”. Triangular numbers. They show up again in the formula for how likely a trip is to succeed. If you have a trip die with N sides and your opponent has a regular die with X sides, and his die is at least as large as yours, then the probability that you can successfully trip is: (The Nth triangular number)/(N*X). Since the formula for the Nth triangular number is N*(N+1)/2, we can simplify the previous formula to (N+1)/(2X). Considering the previous example, d6t vs. d10, we get a probability of (6+1)/(2*10) = 7/20 = 35%.
If the trip die N is larger than the target die X, then the formula is 1-(the (X-1)th triangular number)/(N*X). Which simplifies to (2N-X+1)/(2N). So a trip six trying to take a d4 would win (2*6-4+1)/(2*6) = 9/12 = 75% of the time.
One more time. The probability of an N sided trip die successfully tripping an X sided die is:
(N+1)/(2X) when X>=N
(2N-X+1)/(2N) when X<N
OK, Now lets do a practical example.
You’ve got a d10 showing a 10, he’s got a trip d6 showing a 4 and a d8 showing a 6 (10/10) vs. (4/6t , 6/8) You need to keep your 10 to win. Do you take the trip die, or the die showing the larger number?
If you take the trip die, you will have to re-roll the 10, and must deal with a 60% chance of rolling a 6 or less and losing. On the other hand, if you capture the d8, you will only have a 40% chance of rolling a 4 or less, but your opponent will then get a chance to trip you. The probability that you would then lose the game =
(Probability of rolling <=4) + (Probability of rolling >4)*(Probability of being tripped) = .4 + .6 *(7/20) = 61%
So in this case, the (slightly) wiser thing to do is take the trip die.
OK, for extra credit, here’s a tough one:
(6/6t , 8/12) vs. (5/6 , 20/20)
Your opponent will win if he keeps his 20. Do you:
a) Take the d6 with the trip die
b) Trip the d6
c) Take the d6 with the d12
d) Trip the d20
Article 9.2: Why are Shadow Dice powerful?
Shadow dice just make power attacks upward instead of downward. This doesn’t increase their field of attack (the set of numbers they could capture) at all, just changes it. A roll of a 4 on a regular six-sider corresponds to a roll of a 2 on a shadow six. In each case four of the six numbers in the die’s range can be hit. No net increase in attacking ability. On average, a randomly rolled N-sider will be able to attack (N+1)/2 different values. This doesn’t change when the die is shadow.
So why are they so effective? Well, the first reason is that it is USUALLY the dice showing the larger numbers that our opponent doesn’t want taken, and these are more likely to fall to a shadow die than a regular because the shadow’s field of attack consists of higher numbers. But that’s only “usually”. If our opponent has shadow dice as well, he wants to hold on to the shadow dice that show low numbers. The numbers that would not typically be in our shadow die’s range. This would seem to indicate that regular dice are more effective against shadow dice than other shadow dice are. And yet, I’d much rather face an all shadow character like Wastenott < (4) (8) (10) (20) (X) > with Starchylde < (6) 8 (10) 12 X > than with Niles < 6 10 10 12 X >. In fact, I would much rather face any character with Starchylde than with Niles. Why does making a few of the dice shadow make such a difference between characters that have almost the same die sizes? And if Shadow dice are so Powerful, why aren’t Wastenott and Peace < (10) (12) (20) (X) (X) > more potent than they are?
A pair of regular six-siders showing 2 and 4 could capture a 1,2,3,4, or 6. Five different values. The formula is simply MAX(die1,die2)+1. The plus one is the skill attack using both dice. On average, a pair of randomly rolled regular N-siders will be able to attack (4N^3 + 9N^2 – N) / (6N^2) different values. (I’m not going to go into detail as to how to derive these formulas. Just generate the first few values by hand then solve for the rational function that passes through them.)
A pair of shadow six-siders’s showing 2 and 4 could attack 2,3,4,5,6. Five different values. For two shadow dice, the formula is N-MIN(die1,die2) (plus 1 if die1+die2>N). For a pair of shadow N’s, the expected field of attack is (4N^3 + 6N^2 + 2N)/(6N^2). Calculating (or just referring to the chart below) we see that this means a pair of shadows actually has a WORSE average field of attack than a pair of regulars! Between 3% and 9% worse. What’s going on here?
Let’s look at what happens when you mix shadow and regular together. A regular die showing a 2 and a shadow showing a 4 could capture 1,2,4,5 or 6. Five different values. The formula is MIN[(regular die &endash; shadow die + N +1),N] ( plus 1 if die1+die2 >N) One regular N-sider and one shadow N-sider will have a field of attack of: (5N^3 + 6N^2 + N) / 6N^2. Considerably better than either type of unmixed pair. Between 6% and 19% better.
The question is why? The answer is that a regular die and a shadow die are less likely to have their fields of attack overlap one another. Less redundant attacks leads to a larger combined field of attack even though each individual die of the mismatched pair is no stronger than each individual die of matched one.
Conclusion: The strength of shadow dice comes largely from combining them with regular dice because when you do, your fields of attack are less likely to overlap.
OK, now what about Queer dice? They are shadow if showing an odd number and regular if showing an even. Individually, they are slightly stronger than either regular or shadow, having an average field of attack of (N+2)/2 rather than (N+1)/2.
A pair of queer six-siders showing 3 and 4 would be able to capture 3,4,5,6 (with the 3) 1,2,3,4 (with the 4) or 7 (with the skill attack). Seven different values. Taken in pairs, they have an average field of attack of (6N^3 + 5N^2 &endash; 6N) / (8N^2) when N is an even number and (6N^3 + 9N^2 + 2N – 1) / (8N^2) when N is an odd number. Not surprisingly, we find that their average field of attack lies between that of the matched pairs and the unmatched.
Here are a few things that I highly suspect, but have yet to “do the math” for:
Conjecture 1: The difference between mixed sets and all regular / all shadow sets will become more pronounced as more dice are used, with the optimal mix being 50% shadow, 50% regular.
Conjecture 2: Larger sets of Queer dice will behave more like mixed sets and less like unmixed ones i.e. queer dice work better when you have a bunch of them. (Which is good, since at present the only character who has them has nothing but Simon <4q 6q 12q 20q Xq> )
OK, one last question. Does any of this have any bearing on playing the game? Well, it can help with character selection, now that you know mixed sets are more powerful than unmixed. It can also come up during the game. If you have a choice between taking a shadow die and a regular, all other things being equal, you will want to take the one that leaves your opponents shadow/regular ratio furthest from balanced.
Die Size vs. Expected Size of Field of Attack
|N||2 Regular||2 Shadow||Mixed Pair||Queer Pair|