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OneHoop
03-12-2013, 06:44 PM
Collected: %Drop: # of gold opens for >=50% chance at Collect 10:
0: 100%: 1
1: 90%: 1
2: 80%: 1
3: 70%: 1
4: 60%: 1
5: 50%: 1
6: 40%: 2
7: 100%: 1
8: 5.94%: 12
9. 2%: 35
56 opens * 15G/open = 840G. We got 5 days, so I'll assume the average CC addict can comfortably get 48 opens in with a clock hour between them, so we'll assume 8 resets for 960 total.

Collected: %Drop: # of gold opens for >=60% chance at Collect 10:
0: 100%: 1
1: 90%: 1
2: 80%: 1
3: 70%: 1
4: 60%: 1
5: 50%: 2
6: 40%: 2
7: 100%: 1
8: 5.94%: 15
9. 2%: 46
71 opens * 15G/open = 1065G, assume 23 resets for 1410G total.

Collected: %Drop: # of gold opens for >=70% chance at Collect 10:
0: 100%: 1
1: 90%: 1
2: 80%: 1
3: 70%: 1
4: 60%: 2
5: 50%: 2
6: 40%: 3
7: 100%: 1
8: 5.94%: 20
9. 2%: 60
92 opens * 15G/open = 1380G, assume 44 resets for 2040G total.

Collected: %Drop: # of gold opens for >=80% chance at Collect 10:
0: 100%: 1
1: 90%: 1
2: 80%: 1
3: 70%: 2
4: 60%: 2
5: 50%: 3
6: 40%: 4
7: 100%: 1
8: 5.94%: 27
9. 2%: 80
122 opens * 15G/open = 1830G, assume 74 resets for 2940G total.
Note: 5days*24opens/day=120 opens, so if you didn't sleep and started the event as soon as it was opened, you still would need gold resets at >=80% chance.

Collected: %Drop: # of gold opens for >=90% chance at Collect 10:
0: 100%: 1
1: 90%: 1
2: 80%: 2
3: 70%: 2
4: 60%: 3
5: 50%: 4
6: 40%: 5
7: 100%: 1
8: 5.94%: 38
9. 2%: 114
171 opens * 15G/open = 2565G, assume 123 resets for 4410G total.

Collected: %Drop: # of gold opens for >=98% chance at Collect 10:
0: 100%: 1
1: 90%: 2
2: 80%: 3
3: 70%: 4
4: 60%: 5
5: 50%: 6
6: 40%: 8
7: 100%: 1
8: 5.94%: 64
9. 2%: 194
288 opens * 15G/open = 4320G, assume 240 resets for 7920G total (more than 5 vaults!).


M@

OneHoop
03-12-2013, 06:58 PM
PS Since I assumed 48 free opens, just add 47*15=705G to the totals above for the "first" crowd.

mxz
03-12-2013, 07:15 PM
Cool analysis. I didn't check the math, but this is useful stuff.

OneHoop
03-13-2013, 06:45 AM
I did it by calculator, not spreadsheet so checking the math might be useful. If nothing else, it gives you an idea of how many boxes (kegs) to save for overtime.

M@

Killshot-CC
03-13-2013, 12:57 PM
Wait a minute. You mean I can't improve my odds if I wait 30 seconds, wave my phone around in the air 3 times, and tap very slowly? Let me know right quick because I might have to change my strategy.

(Good stuff with the tables. Thanks)

OneHoop
03-13-2013, 09:20 PM
Math:
1-(1-SUCCESS)^TRIES=LIKELIHOOD
(1-SUCCESS)^TRIES=(1-LIKELIHOOD)

TRIES= log(1-SUCCESS) / log(1-LIKELIHOOD)
[NOTE: can substitute natural log, ln, for log]

eg, if LIKELIHOOD=2%, SUCCESS=67%
TRIES = log(0.33)/log(0.98)
TRIES = 54.877
Ergo, need 55 successful opens from beer #9 to have a 67% chance of beer #10.
Check: 1-(1-2%)^54.877=1-0.98^54.877=1-0.33=0.67.
or 1-0.98^55=67.08%

I think it is Tommy Two Toes who posited that cash opens are about 33% successful, so you can roughly multiply that by 3.

M@

PS Mods: PastaSux Sucks

PawnXIIX
03-13-2013, 09:30 PM
I think it is Tommy Two Toes who posited that cash opens are about 33% successful, so you can roughly multiply that by 3.

That's honestly the only part that is speculation. For my calculations I use 35%, but I feel like that's really much lower than it is :/ Maybe more at 40%...

We'll never know D:

mxz
03-13-2013, 09:54 PM
Well, we could do a comprehensive study but it's probably not worth it. Even combing through the tracking threads should converge on a good approximation.

PawnXIIX
03-13-2013, 09:57 PM
Well, we could do a comprehensive study but it's probably not worth it. Even combing through the tracking threads should converge on a good approximation.

My simulations point towards a rate between 35% and 40%, but then again they're seeded a little differently than an open per hour over the course of a week. That is, if they do use just a simple random algorithm. I think they'd at this point be using some sort of predictable system so they can check the legitimacy of results.

mxz
03-13-2013, 10:02 PM
Why are you simulating it?

sum(special items,cash,RP, bikes/batons/other useless items)/# opens

The problem is we don't know how well/accurately people record their opens.

Tommy Two Toes
03-13-2013, 10:04 PM
I think it is Tommy Two Toes who posited that cash opens are about 33% successful, so you can roughly multiply that by 3.


That's honestly the only part that is speculation. For my calculations I use 35%, but I feel like that's really much lower than it is :/ Maybe more at 40%...

We'll never know D:

Agreed. I based my hypothesis on a really small sample. (Maybe 120 open attempts).

As the contest went on, I think it improved slightly

Last time I calculated, I had 62 attempts on each of 3 accounts for a sample size of 186 (still very small). 111 BLNTs for an open rate of about 40%.

bald zeemer
03-14-2013, 12:28 AM
So all the guys who got caned for 5-6k gold in their first-open attempts were getting normal open rates, it seems. I'm not sure how much solace that will bring, but it's interesting to know.

OneHoop
04-03-2013, 08:48 PM
Lots of people dropping gold for 30% off building upgrades!

We don't actually know the odds are the same this time, but lets assume they are and look at the special situation where you've managed to get 9 paper bonds for free and want to spend gold to get lucky number 10!

Assuming no resets, else double totGold:

N totGold %Chance
1 15 2.00%
2 30 3.96% (=100%-(100%-2%)^N)
3 45 5.88%
4 60 7.76%
5 75 9.61%
6 90 11.42%
7 105 13.19%
8 120 14.92%
9 135 16.62%
10 150 18.29%
11 165 19.93%
12 180 21.53%
13 195 23.10%
14 210 24.64%
15 225 26.14%
16 240 27.62%
17 255 29.07%
18 270 30.49%
19 285 31.88%
20 300 33.24%
21 315 34.57%
22 330 35.88%
23 345 37.17%
24 360 38.42%
25 375 39.65%
26 390 40.86%
27 405 42.04%
28 420 43.20%
29 435 44.34%
30 450 45.45%
31 465 46.54%
32 480 47.61%
33 495 48.66%
34 510 49.67%
35 525 50.69%
**1050 gold with resets before it gets to be probable***
36 540 51.68%
37 555 52.64%
38 570 52.59%
39 585 53.59%
40 590 54.52%
41 605 55.43%
42 620 56.32%
43 635 57.19%
44 650 58.05%
45 665 58.89%
46 680 59.71%
47 695 60.52%
48 710 61.31%
49 725 62.08%
50 740 62.84%

I'm going to stop here, because most people aren't being persuaded by math in this but also because it is already obvious that you are not increasing your probability of winning by 2% every time you open one. (eg, chances only go up by 0.76% between 49 opens and 50 opens) It's not additive--randomness doesn't care how many times you lost already. Although, you'd probably have a whole garage full of enhanced bikes if you got to that point! ;)

69 1035 75.19%
80 1200 80.14%
105 1575 88.01%
228 3420 99.00%

Good Luck---or should I say Gold?!?

Any computer scientists out there to chip in about how RNG's aren't really random and how that would affect this analysis?


M@

BigMoney
04-04-2013, 01:32 AM
Any computer scientists out there to chip in about how RNG's aren't really random and how that would affect this analysis?


M@

Not without knowing what type of PRG they use and how it's seeded.

Edit: and if they did use a PRG I can almost certainly say it wouldn't affect the probabilities, but that's based on the assumption that there are no other hidden factors at play (e.g. not saying this is at all the case, but for example something like top gold spenders are given better odds, etc). I feel like there are, but I don't know what.

Edit 2: if you knew how the PRG was seeded (typically computers use something like the computer's time converted into milliseconds), and how the seed affected the outcome, you could guarantee a successful "drop" each time. Though that's obviously not going to happen.

OneHoop
04-04-2013, 05:00 PM
Not without knowing what type of PRG they use and how it's seeded.
Edit 2: if you knew how the PRG was seeded (typically computers use something like the computer's time converted into milliseconds), and how the seed affected the outcome, you could guarantee a successful "drop" each time. Though that's obviously not going to happen.

Wait, so it really will help to "wait 30 seconds, wave my phone around in the air 3 times, and tap very slowly"???
::grin:: (you would have to be able to sync to the server computer's clock and account for network lag)

re: Edit 1: Occam's Razor almost always applies to computer programs--the easiest explanation is the most rational. This is because programmers will almost always reuse subroutines and variables when they can. I don't think that it is any more complex than they had to make it to maximize revenue.

M@

Tommy Two Toes
04-04-2013, 05:50 PM
Wait, so it really will help to "wait 30 seconds, wave my phone around in the air 3 times, and tap very slowly"???
::grin::

So that's what I'm doing wrong. I was waiting 3 seconds, then waving the phone 30 times. <SMH> So silly of me:D