### The Math Behind Danger — Half 2

#### Does the assault actually have a bonus within the recreation of world conquest?

In Half 1, we mentioned the relative possibilities for assault and protection in Danger, the sport of world conquest. On the finish of Half 1, we concluded that the assault has a 47.15% likelihood of successful the battle for the primary soldier and we questioned how the well-known conquerors had been capable of obtain their feats beneath these circumstances. We saved the dialogue of the second soldier for Half 2.

To refresh our recollections, in Danger, the assault rolls as much as 3 cube, whereas the protection rolls as much as 2 cube. The very best rolls of every are in contrast and the loser loses a soldier, with the protection successful within the case of a tie. Subsequent, the second highest rolls of every are in contrast, and as soon as once more, the loser loses a soldier, with the protection successful within the case of a tie as soon as once more.

Nicely, right here we’re. Let’s dive into it.

*(**Right here* y*ou can discover code wherein I verify the under chances.)*

After all, relating to the defender’s chances, we’re merely calculating the bottom roll, since he has solely two cube. Due to this fact, the possibilities are a mirror picture of the possibilities we noticed relating to the best roll. This time, there are 11 prospects yielding a 2nd highest roll of 1, 9 for two, 7 for 3 and so on. The chance may be calculated by dividing by 36, the entire variety of permutations attainable for the 2 cube of the protection.

Calculating the second highest roll among the many three cube of the attacker differs considerably from the calculations of Half 1. I’ll be trustworthy. I struggled with this slightly. Within the calculations that comply with, two issues should be borne in thoughts.

- We should take into account each what number of outcomes are attainable
**and**what number of methods wherein every final result can happen. For instance, an final result of (6, 2, 3) is after all a single final result, however it might happen in 6 methods, comparable to which die every worth happens on. It may be any of {(2, 3, 6), (2, 6, 3), (3, 2, 6), (3, 6, 2), (6, 2, 3), (6, 3, 2)}. This final result due to this fact corresponds to 1*6 = 6 permutations. For one more instance, an final result with**precisely**two ones is definitely a group of 5 outcomes, for the reason that remaining die can take any worth between 2 and 6.**And**it might happen in any of three methods, {(1, 1, x), (1, x, 1), (x, 1, 1)}, comparable to the three attainable areas for the remaining die, so this final result truly corresponds to five*3 = 15 permutations. - We should be cautious with doubles and triples. These should be thought-about individually since, whereas there are 6 methods to acquire an final result of (1, 2, 3), there are solely 3 methods to acquire a (1, 2, 2) and just one technique to get hold of a (2, 2, 2).

With the above issues in thoughts, we’re able to proceed.

Take into account the chance of getting a 2nd highest roll of 1. That is comparatively straight-forward. Clearly the bottom roll is a 1 as effectively. For now, we are going to disregard the case the place all 3 cube are 1. The very best die can then take any worth between 2and 6, and it might seem upon any of the three cube, since we’ve got not specified which of the three cube incorporates the best roll. This yields a complete of three*5=15 permutations. Including the case of a triple 1 yields a complete of 16 permutations. By a symmetrical argument, we are able to calculate that the identical variety of permutations yield a 2nd highest roll of 6.

Subsequent, what about getting a 2 because the second highest roll? For now, we are going to disregard the potential of a number of twos and assume that the best roll was increased than 2 and that the bottom roll was decrease than 2. The very best roll can take 4 values (3–6) and the bottom toll should be 1, for a complete of 4 outcomes, and these can happen at any of 6 permutations of cube areas(3 prospects for the placement of the best roll (die 1, die 2 or die 3) and the 2 remaining prospects for the placement of the bottom roll), for a complete of 4*6=24 permutations. We are going to now take into account double twos, however not triple twos. If there are precisely 2 twos, then the remaining die can take any of 5 values (excluding 2), and this remaining die could possibly be any of the three cube, for a further 5*3=15 permutations. Including the ultimate case of triple 2’s, we get hold of a complete of 24+15+1 = 40 permutations. A parallel argument yields the identical consequence for a second highest roll of 5.

Lastly, what about getting a 3 or a 4? Let’s begin with 3. As soon as once more disregarding the potential of a number of threes, the best roll can take any of three values (4, 5 or 6) and the decrease roll can take any of two values (1 or 2), for a complete of 6 outcomes. This may as soon as once more happen at any of 6 permutations of two cube, for a complete of 6*6 = 36 permutations. Within the case of precisely 2 threes, the opposite die may take any of 5 values (any moreover 3) and will happen at any of the three cube, for a further 5*3 = 15 permutations. Including the final risk of three threes yields a complete of 36+15+1=52 permutations. A parallel calculation yields 52 permutations for a second highest roll of 4 as effectively. These outcomes are summarized within the under visuals.

Notice that the possibilities of assault outcomes are precisely symmetrical. To be mathematically exact, P(x) = P(6-x). We are going to come again to this level.

We subsequent evaluate the possibilities of assault and protection instantly.

We will see that the assault has a big benefit right here. It’s more likely to acquire values of 4, 5 or 6, than protection is.

We at the moment are able to calculate the relative chances of victory for the second soldier. This section is essentially unchanged from the parallel calculations we did in Half 1. We have to depend the permutations wherein the protection achieves a 2nd highest roll of x, after which decide what number of of these permutations yield an outright victory for the protection, a tie, or a win for the assault.

For instance, since, as calculated above, there’s a 3/36 likelihood of the protection’s second highest roll being 5, and there are a complete of 6⁵= 7776 permutations, clearly (3/36) * 7776 = 648 of these permutations will yield a second highest protection roll of 5. To win, the assault then must get a second highest roll of 6, the chance of which is 16/216, as calculated above, so (16/216) * 648 = 48 of the 648 permutations which yielded a second highest protection roll of 5 will lead to a victory for assault. To attain a tie, assault should roll a 2nd highest roll of 5, the chance of which is 40/216, so (40/216) * 648 = 120 of these permutations will lead to ties, and the rest (648–120–48 = 480) will lead to outright protection wins.

We will additionally calculate the respective conditional chances, wherein we calculate the possibilities of victory for every workforce, **given** a selected protection roll.

Notice the precise symmetry of the above desk and charts. Thus, the possibilities of an assault victory given a protection roll of x is an identical to the chance of a protection victory given a protection roll of 6-x. Moreover, the chance of a tie given a protection roll of x is an identical to the chance of a tie given a protection roll of 6-x. That is after all due to the symmetry of the second highest roll of assault which was famous above.

As in Half 1, Desk 4 and Chart 4 give a deceptive impression. They indicate that assault and protection are on an equal footing, however these are conditional chances and due to this fact they ignore how more likely low rolls for protection are.

Complete chances, alternatively, will enable for this. Within the desk under, we see clearly that the assault will win from a protection roll of 1 considerably extra typically, in absolute phrases, than protection will win from a protection roll of 6, regardless of the conditional symmetry famous above.

Under is a chart of complete win chances by 2nd highest defensive roll.

The above chart lastly captures what most of us felt instinctively, whether or not from expertise of taking part in Danger or from mathematical intuition, that the assault has an enormous benefit. That benefit performs out particularly within the 2nd battle.

Lastly, we are able to calculate the joint chance of every attainable 2nd highest consequence for each assault and protection. A joint chance is just the chance of two or extra occasions co-occurring. Because the cube rolls of assault and protection are unbiased, the joint chance is just the product of the person chances of the respective rolls of assault and protection.

See the next graph for a cool visualization of the above chart. Notice the configuration of the axes, which has been chosen to permit for greatest viewing.

There may be one remaining issue we should always take into account, earlier than we wrap up, and that’s how doubtless it’s for the assault to win **each **troopers. Nonetheless, in comparison with calculating the 3667 permutations and 4724 permutations which resulted within the assault successful the primary and second troopers respectively, which I calculated in my head (I had no selection, since I used to be requested the query on a spiritual vacation, on which I’m forbidden to jot down, though admittedly I did have a selection about whether or not to brag about it), this calculation requires calculating and summing 36 separate permutation counts. It is because this calculation varies relying on **each **protection rolls concurrently, whereas the earlier calculations required contemplating the upper or decrease roll of the protection independently. I’ll due to this fact not bore myself otherwise you with calculating and summing all 36 permutation counts, and can depend on python to depend the permutations. I’ll suffice with offering the code.

from itertools import product

import numpy as np

import pandas as pd

def check_both(consequence, attack_dice=3):

"""

Given a results of a number of cube, and a lot of assault cube, decide whether or not the assault will win each troopers,

the protection will each troopers, or every will win a soldier.

:param consequence: array of cube outcomes, the place the assault cube are declared first

:param attack_dice: variety of cube which characterize the assault cube

:return:

"""

if np.partition(consequence[:attack_dice], -1)[-1] > np.partition(consequence[attack_dice:], -1)[-1] and

np.partition(consequence[:attack_dice], -2)[-2] > np.partition(consequence[attack_dice:], -2)[-2]:

return 'attack_both'

elif np.partition(consequence[:attack_dice], -1)[-1] > np.partition(consequence[attack_dice:], -1)[-1] or

np.partition(consequence[:attack_dice], -2)[-2] > np.partition(consequence[attack_dice:], -2)[-2]:

return 'one_one'

else:

return 'defense_both'

# produce array of all attainable permutations of 5 cube

arr = np.array(checklist(product([1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 6],

[1, 2, 3, 4, 5, 6]))).transpose()

# calculate and print counts of assorted consequence varieties

print(pd.Collection(np.apply_along_axis(check_both, axis=0, arr=arr)).value_counts())

Within the under graphics, a truce refers to a consequence the place each assault and protection lose one soldier, whereas assault refers back to the assault successful each troopers and likewise for protection.

I assumed for a second that I may lastly use a pie chart for the above graphic with out concern of criticism, however alas, the possibilities are too comparable. The wait continues.

We will see that though the protection has a small benefit within the first battle, as we calculated in Half 1, the assault however has the higher hand total, because of its important benefit within the second battle.

#### Conclusions

- The chance of a victory for the assault is 60.75%.
*This may be seen clearly in Desk 5.* - The chance of an outright victory for the protection is 22.58%.
*This may be seen clearly in Desk 5.* - The chance of a tie is, maybe a contact surprisingly, precisely 1/6, or 16.67%.
*This may be seen clearly in Desk 5.* - If the protection’s second highest roll is 3, his chance of successful, together with the chance of a tie, is strictly 50/50.
*This may be seen in Desk 4.* - The possibilities of assault’s 2nd highest roll are precisely symmetrical.
*This may be seen in Desk 2.* - The conditional chances are precisely symmetrical.
*This may be seen in Desk 4.* - The most probably total outcomes, at 7.36% every, are for the protection to roll a 1 and for the assault to roll both a 3 or a 4.
*This may be seen in Desk 6.* - The chance of the assault successful each battles is 37.17%.
*This may be seen in Desk 7.* - The chance of the protection successful each battles is 33.58%.
*This may be seen in Desk 7.* - The chance of each the assault and protection successful a battle is 29.26%.
*This may be seen in Desk 7.*

We at the moment are able to offer a definitive reply to our unique query. The assault **does** have an total benefit in a 3 to 2 cube battle.

I’ve to say I’ve loved this foray into the arithmetic and chance of Danger, and if in case you have too, effectively, that’s a bonus.

Till subsequent time.

The Math Behind Danger — Half 2 was initially printed in In the direction of Information Science on Medium, the place persons are persevering with the dialog by highlighting and responding to this story.