It is the typical "oh this changed so everything must be different" and "I did not do as well as I thought I would so it must now be harder".
Should make a general paranoid parrot meme for this category.
As far as I saw last night. Same old floor jumps, same old difficulty, and same old Nyzul.
Yeahhhhhhhhhh, I'm sorry, but an order lamp floor jumping from 90 seconds to a potential 5+ minutes is not "paranoid parrot meme"
The fact that you are insisting entirely proves the opposite. Lamps are necessarily going to fuck you unless you can still guess 1/1. Also nice job mentionning you can win without running through walls, we totally believe you (why even mention it ?). I've always wondered why the entire BG pseudo elitist cheaters try their hardest to justify the event, implying that they win 1/3 of the time ("without cheats"). Very funny.
I don't see what the issue is with people saying they go 1/3 legit. Seems completely plausible to me and I'd certainly buy that because if I went and grabbed 5 friends and had a worse win rate than 1/3-4 I'd be furious (unless we got fucked every run by Order floors) so 1/3 doesn't seem farfetched.
Order floors are going to happen but not every run (and sometimes they won't be terrible). Plus quite a few 100 wins could accommodate for 1 order floor anyway.
you will average like 2+ lamp floors per run, so the chance to get one is very high, can't do the math now, but since you need about 15-20 floors to win, everyone can guess. Also 3-lamps floor order are not the same as 5-lamps order.
Assume: 20% chance to land on a lamp floor (I don't want to find Kirschy's post with the exact percentage!)
Assume: All three lamp types have an equal probability of occurring
(0.2)*16=3.2 , so on a 16 floor run we should expect, on average, 3 lamp floors. On average, one should be code.
p(order)=0.066
p(no order)=1-0.066=0.934
p(no order after 16 floors)= 0.934^16 = 0.335389930618
p(no order after 17 floors)= 0.934^17 = 0.313254195
p(no order after 18 floors)= 0.934^18 = 0.292579418
p(no order after 19 floors)= 0.934^19 = 0.273269177
p(no order after 20 floors)= 0.934^20 = 0.255233411
Pretty sure it doesn't need to be any more complicated than that. Let me know if I'm wrong!
Number of Order Lamps in 25 runs
0, 1, 1, 1, 0, 1, 1, 1, 2, 0, 0, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 3, 0
Average clears: 19.8 (across those 25 runs, was worse prior to that but a different set-up)
Number of times an extra 3 mins per Order floor would have stopped us clearing: 3 (1 is marginal as I have it down as exactly 3 mins remaining, counting as a fail.)
Yep, you're right. I realized it and refreshed hoping to make a quick edit, but you caught me. I'm going to get rid of it anyway, so that no one goes away from the thread thinking it's correct.
I'm hardly the person to answer this question, seeing as I haven't used probability for very much at all in the last four years, but since I'm awake I thought I'd give it another shot.
I count 120 (16!/[14!*2!]) ways for two order floors to be arranged among 14 non-lamp floors. Each of these occurs with probability p(two order floors)=[(0.0666666666666)^2 * (1-0.06666666666666666)^14]. Hence the probability is 120*[(0.0666666666666)^2 * (1-0.06666666666666666)^14]=0.203008209425225.
I continued this for 0->16 order floors for a 16 floor run, and I will now paste the results from my spreadsheet here in a way that will likely be indiscernible by anyone.
# of order floors _ Arrangements _ P(arrangement) _ P(# of order floors)
0 _ 1 _ 0.331580075394535 _ 0.331580075394535
1 _ 16 _ 0.0236842910996096 _ 0.378948657593754
2 _ 120 _ 0.00169173507854354 _ 0.203008209425225
3 _ 560 _ 0.000120838219895967 _ 0.0676694031417418
4 _ 1820 _ 0.000008631301421140 _ 0.0157089685864758
5 _ 4368 _ 0.000000616521530081 _ 0.00269296604339585
6 _ 8008 _ 0.000000044037252148 _ 0.000352650315206599
7 _ 11440 _ 0.000000003145518010 _ 0.0000359847260414897
8 _ 12870 _ 0.000000000224679857 _ 0.00000289162977119113
9 _ 11440 _ 1.60486E-11 _ 0.000000183595541028009
10 _ 8008 _ 1.14633E-12 _ 0.00000000917977705140043
11 _ 4368 _ 8.18804E-14 _ 0.000000000357653651353263
12 _ 1820 _ 5.8486E-15 _ 1.06445E-11
13 _ 560 _ 5.8486E-15 _ 3.27522E-12
14 _ 120 _ 4.17757E-16 _ 5.01309E-14
15 _ 16 _ 2.13141E-18 _ 3.41026E-17
16 _ 1 _ 1.52244E-19 _ 1.52244E-19
Hopefully that's not completely wrong. I'd post numbers for three, etc. order floors on 15-20 jump runs, but it's almost totally unreadable pasted here.
The short answer is that on a 16 jump run you should expect two order floors only 20% of the time.
For what it's worth, these are the numbers BG wiki links to
77.53% Chance of Fighting Critters:
26.87% Chance of Defeat All
17.18% Chance of Defeat 1 Enemy
16.74% Chance of Defeat an Enemy Family
16.74% Chance of Defeat an Enemy Leader (I took boss encounters off of this measurement)
21.59% Chance of Lamps:
8.37% Chance of Single Lamp
4.85% Chance of Ordered Lamps
8.37% Chance of Same time Lamp
.88% Chance of Free floors
Not sure on sample size. And like he says he has removed 20/40/60/80/100 floor objectives which are always Leader.
Yeah, so it's about a 5% lamp rate. If you do get lamps, it can be 3, 4, or 5 lamps in the order (2, 3, and 4 average guesses required if I'm not mistaken). Assuming an even distribution of those, you're looking to lose about 3 guesses worth of time on average every time you have a lamp floor. I know that a lot of my wins would have been losses if it had taken 3 attempts instead of 1 to clear lamps.
At the same time though, the odds of getting no lamps are about 40%. So even if every set of lamps is an insta-loss, they've only reduced your win rate by 60%. Even if you do get lamps (and assuming an even distribution again), your odds of going 1/1 are about 7% on average. 1/2 is higher and may not kill the run depending how much extra time you would have had, etc. I'd say that they reduced your odds of winning by something like 30-50% with the lamp patch.
Pretty sure he wasn't talking to you lol
Nevermind
3 Lamps: 6 combinations
4 Lamps: 24 combinations
5 lamps: 120 combinations
If there's an even distribution of 3, 4, and 5 lamps, then you would see the average of your odds of guessing it on the first guess.
1/6 + 1/24 + 1/120 = 20/120 + 5/120 + 1/120 = 26/120
26/120/3 = 7.2%
Spoiler: show
Agggh, you guys reply so fast. I was cOming in to delete that.