Acescent goldstone

acescent (adj.): becoming, or tending to be, sour.

goldstone (n.): a type of glass (gemstone) made with copper or copper salts in the presence of a reducing flame.

Again, it's been a while since I've posted. But luckily (or perhaps not...), I should be posting more often now =).

Anyway, I was going to post about Brawl - mainly to convince you that, contrary to what you might expect, Peach does not suck, and in fact is one of the best characters in Brawl. But then, when you think about it, such discussions about "who the best character in Brawl is" don't really make sense. Especially not at this early stage, when there isn't even enough information from tournaments to get a rough sense of what characters belong to which tiers. You should just use whatever characters you're the best with.

But do try Peach for a bit before you condemn her to the bottom of your preferred list =).

Onto other stuff. Some people have recently rediscovered the board game Risk, or something. In any case, they seem to like playing it a lot now. Which is, in a sense, very strange, considering that if you really feel like playing a strategy game, there are many electronic alternatives. Most of them also have much deeper gameplay, with a much smaller luck factor. In fact, there are probably tons of other board games with a much smaller luck factor than Risk.

But bashing Risk is not the point of this post. Attacking in Risk has a very interesting mechanism. The way the people I play with attack in Risk (which actually is not the official way) is:

1. If the attacker has 1 army, he rolls one die, if he has 2 armies, he rolls two dice, if he has 3 or more armies, he rolls three dice.
2. If the defender has 1 army, he rolls one die. Otherwise, he has a choice of rolling one or two dice.
3. If the attacker's largest die is greater than the defender's largest die, the defender loses one army. Otherwise, the attacker loses one army. If the defender rolled two dice, if the attacker's second largest die is greater than the defender's other die, the defender loses another army. Otherwise, the attacker loses another army.

Now, the question is, when should the defender roll one die as opposed to two dice? Clearly, if the attacker rolled two sixes, the defender is probably better off rolling only one die - that way he will probably only lose one guy, as opposed to probably losing 2. Similarly, if the attacker rolled three ones, then the defender might as well roll two dice if he can, since he automatically kills two of the attacker's armies. But of course, there are more hazy scenarios, like if the the attacker rolls two 4s. Should the defender roll one die or two dice?

Firstly, it's clear that only the attacker's two largest rolls matter - the smallest die doesn't affect anything. So let these two rolls be r and s, with r >= s. Now we can consider both cases:

Defender rolls one die

In this case, only the attacker's largest roll, r, matters. The defender can roll any number from 1 to 6 with equal probability. For numbers 1 through r-1, the attacker wins, for the other 7-r numbers, the defender wins.

We are considering the change in relative strength between the attacker and the defender, i.e. (the number of guys the defender has) - (the number of guys the attacker has). If the defender wins, his relative strength increases by 1, and if the attacker wins, the defender's relative strength decreases by 1. So the expected change in relative strength is:

(expected change)
= ((7-r)-(r-1))/6
= (4-r)/3

Defender rolls two dice

In this case, there are three possible outcomes - the defender loses 2, the defender and attacker both lose 1, and the attacker loses 2. Note that if the defender and attacker both lose 1, this does not change the relative strength, so we only need to consider the other two outcomes.

Let the defender's rolls be a, and b, with a >= b. Looking at it like this, there are 21 possible rolls for the defender, since rolls like (4, 3) and (3, 4) are considered the same. (21 since 1+2+3+4+5+6 = 21). Then, he loses when both a(r-(s/2))(s-1)

cases where the defender loses 2 guys.

How many cases are there where the attacker loses 2 guys? We must have a>=r and b>=s. This works out similar to the above case... it turns out that there are:

(7-r)(4+(r/2)-s)

cases where the attacker loses 2 guys. So, the expected change in strength is:

(expected change)
= (2(7-r)(4+(r/2)-s)-2(r-(s/2))(s-1))/21
= ((7-r)(8+r-2s)-(2r-s)(s-1))/21
= (56 + r - 15s + s^2 - r^2)/21


Now we can finally resolve the initial question - it is better to roll two dice whenever:

(56 + r - 15s + s^2 - r^2)/21 >= (4-r)/3

or

28+8r-15s+s^2-r^2 >=0

Technically, if it equals 0, it doesn't matter whether you roll two dice or one. Now, while that formula isn't that large at all, it is still pretty random. But note there are only 36 (and really only 21) possibilities for (r,s) pairs. And if we check whether to use two dice or one for each pair, we get the following very nice pattern:

1 2 3 4 5 6

-+-----------
12 2 2 2 2 2
22 2 2 2 2 2
32 2 2 2 2 2
42 2 2 N 1 1
52 2 2 1 1 1
62 2 2 1 1 1

That is, it's better to roll two dice whenever your opponent's smallest roll is smaller than 4.

Now with this, it's not too bad to calculate chances of winning. For example, assuming you're smart, the expected change in relative strength if the attacker rolls r, s, is:

2.0000 1.9047 1.7142 1.4285 1.0476 0.5714
1.9047 1.3333 1.1428 0.8571 0.4761 0.0000
1.7142 1.1428 0.6666 0.3809 0.0000 -0.476
1.4285 0.8571 0.3809 0.0000 -0.333 -0.666
1.0476 0.4761 0.0000 -0.333 -0.333 -0.666
0.5714 0.0000 -0.476 -0.666 -0.666 -0.666

Looping over all possible 56 different rolls of three dice for the attacker, the expected change in relative strength each time is 5/56 - with favor towards the defender.

Figuring out the probability that the attacker will win if he attacks with n armies and the defender defends with m armies isn't much harder either - you can get a recursive formula fairly easily. But maybe I'll post that another time.

Until then, ciao.

-squidout.