For those who were wondering the price of a loadout card, I have done some maths.
Let’s assumed to begin with that we will buy everything (cases included) and that we will always receive lead cards. We will lose this assumption later on in this same message.
So here is what it looks like:
Cobalt = 6gold + 10 000 Cred = 1 186 000 Cred (864 cases)
gold = 4silver + 4 000 Cred = 196 000 Cred (144 cases)
Silver = 4bronze + 2 000 Cred = 48 000 Cred (36 cases)
bronre = 3iron + 1 000 Cred = 11 500 Cred (9 cases)
iron = 3*lead + 500 Cred = 3 500 Cred (3 cases)
lead = 1 000 Cred = 1 000 Cred (1 case)
If you aim at gold and cobalt and if you buy all cases, expect to have:
gold
0.144 cobalt loadout
0.576 gold loadout (1 chance over two to get a gold loadout from a case while digging for gold)
2.16 silver loadout
4.32 bronze loadout
21.6 iron loadout
115.2 lead loadout
colbat
0.864 cobalt loadout (almost 1 chance to get a cobalt loadout from a case while digging for cobalt)
3.456 gold loadout
12.96 silver loadout
25.92 bronze loadout
129.6 iron loadout
691.2 lead loadout
It is now tricky to figure out how much credits you will save by earning a higher tier loadout simply because it depends on your set. Every combination will lead to different saving, e.g: whether you gain the gold cards in the really last or really first cards.
A simple way to do that would be to create a Python script that mimic a player, i.e., algorithmically:
- Get a card
- if you can make a higher tier card with this card, do it, compute cost and goto 1, else goto 1
do this model say N times and check the stats.
If there is enough demands and if these thread go sticky, I can do this script and post the results. Let me know.
Note that I might edit this post accordingly.
First EDIT: copy paste of a message I wrote below.
http://i.imgur.com/UKn58YB.png
So I did the python script to compute the real price of a cobalt loadout card of the merc of your choice. I assumed that we will always buy the initial case (1000 cred). But I took into account the fact that you could earn higher tier loadouts than lead ones.
The results are in the image (the x-axis unit is the credit, my bad, and the y-axis is to have it normed, don’t look at the number).
Finally, the cost of a loadout card is in average about 500000 credits, half the price we would have if we could get only lead cases.
Meanwhile, by doing so, you will earn in average 0.316 spare cobalt card (of an unwanted merc).
For those who are interested, here is the python script (Edit, of course, indent doesn’t work here… I ll find a way later). Note that the code is absolutely not optimized and written in a way that people with no coding experience can read it.
[spoiler]
import random as ra
import numpy as np
import pylab as py
cost_list = []
second_cobalt_list = []
for i in range(1000):
cost = 0
second_cobalt = 0
cobalt = 0
gold = 0
silver = 0
bronze = 0
lead = 0
iron = 0
ra.seed(i)
while cobalt == 0:
new_loadout = ra.randint(1,1000)
cost = cost + 1000
if new_loadout == 1:
second_cobalt = second_cobalt + 1
elif new_loadout >= 2 and new_loadout <= 5:
gold = gold + 1
elif new_loadout >= 6 and new_loadout <= 21:
silver = silver + 1
elif new_loadout >= 22 and new_loadout <= 52:
bronze = bronze + 1
elif new_loadout >= 53 and new_loadout <= 203:
iron = iron + 1
else:
lead = lead + 1
if lead == 3:
lead = 0
iron = iron + 1
cost = cost + 500
if iron == 3:
iron = 0
bronze = bronze + 1
cost = cost + 1000
if bronze == 4:
bronze = 0
silver = silver + 1
cost = cost + 2000
if silver == 4:
silver = 0
gold = gold + 1
cost = cost + 4000
if gold == 6:
gold = 0
cobalt = cobalt + 1
cost = cost + 10000
cost_list.append(cost)
second_cobalt_list.append(second_cobalt)
print(‘In average, you will earn’,np.mean(second_cobalt_list),‘spare cobalt cards’)
n, bins, patches = py.hist(cost_list, bins=100, normed=1, histtype=‘stepfilled’)
py.xlabel(“Cobalt loadout cost”)
py.ylabel(“Probability”)
[/spoiler]