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| Author | Luck = 1 and 14 hits without befallen luck in a row :) |
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Just curious if anyone had worse luck than me
https://www.lordswm.com/warlog.php?lt=-1&warid=23617609
Seems like a record of BAD luck :)
From this table
http://eng.witchhammer.ru/viewpage.php?page_id=37
you can see that maximum 9 hits can be without befallen luck if your luck is 1 (100% probability that 10th strike will be lucky if your 9 strikes were unlucky).
But each stack has it's own Luck counter:
The probability of morale/luck triggers = F^(1+[moves befallen so far]-[moves not befallen so far]*F/(1-F)), whence F is luck/10 or morale/10, each stack having its own independent counters for morale and luck; F∈[0;0.5]
So such BAD luck is possible, i.e. 14 hits without befallen luck in a row for whole army (but for a single stack such bad luck is impossible). | | Topic moved from "General game forum" to "Queries and help". | .9^14 = 0.2287
The chance to not get luck 14 times in a row is 23%. I would say that's a pretty chance and your luck isn't that bad.
*Yes, technically I should have used the witchhammer table for each of your stacks and then multiplied those probabilities together. But my above calculation is much simplier and easier. :) | for Pang:
Any logical reason to move this topic?
This wasn't a question, nor did I need help.
I just wanted to share my bad luck with others.
So, General forum was the perfect place for it.
for Pantheon:
You are wrong here.
If 22% is the probability of such bad luck then you would have it in about every 5 battles, but you don't remember if you ever seen such a bad luck, right? | yea, my bad... i click too fast just now while moving a few topic.
and i realized that i move this topic to wrong section; it suppose go to OGF :\ | I know I overestimated the probability. The basic formula I used doesn't take enough account the balancing aspect where every 10 counts (from the same counter) your luck triggers.
But the 22% probability is accurate for 1 count each on 14 different counters. | | Topic moved from "Queries and help" to "Off-game forum". | it suppose go to OGF :\
Ah, now I understand, you just wanted to move it, no matter where :)
You have a great power!
I'm trembled! :) | I review the battle again and realized that there were even 15 unlucky hits, because Bowmen hit 2 stacks of Behemoths, which is 2 hits, not 1 hit.
Luck can befall against one stack or two, or none.
For example, when I did (double + double + triple) damage with Sprites here (on 3 stacks):
https://www.lordswm.com/forum_messages.php?tid=1839414
(that means it was 3 different hits)
Battle against Behemoths and probabilities of no Luck for that turn:
1) Bowmen, 0.9
2) Bowmen, 0.8708450335
3) Griffins, 0.9
4) 25 Swordsmen, 0.9
5) 1 Swordsmen, 0.9
6) Bowmen, 0.8331899463
7) Griffins, 0.8708450335
8) 25 Swordsmen, 0.8708450335
9) Bowmen, 0.7845565310
10) 25 Swordsmen (back), 0.8331899463
11) Bowmen (part 1), 0.7217440598
12) Bowmen (part 2), 0.6406186336
13) Bowmen, 0.5358411166
14) Bowmen (back), 0.4005157497
15) Bowmen, 0.2257363173
Multiplying all those probabilities gives the probability of such bad luck:
0.0052862149
It means that such bad luck occurs once in about 200 battles on average.
Actually I thought that this probability will be smaller than 1/200, but it is still small.
By the way, Bowmen had bad luck 9 times in a row!
It means that the next time would be 100% lucky!
So, my Bowmen were as unlucky as they could be :) | #9
Okay, there's 2 problems with that post.
First I counted 14 attacks and Second you are using the formula incorrectly.
In the end I only counted 4 bowmen attacks; not 5.
As for the formula it should be multiplied this way:
Bow 8: 0.4005
Griffins 2: 0.8708
25 Sword 3: 0.8332
1 Sword 1: 0.9000
=0.2615 or 26% chance | I'm not sure which attack you don't count here:
11) Bowmen (68 shooting, part 1)
12) Bowmen (68 shooting, part 2)
13) Bowmen (68 melee)
14) Bowmen (32 melee, back)
15) Bowmen (7 melee)
I counted 11 and 12 as two different attacks (I explained why in the beginning of 9th post)
Your number 0.2615 is correct IF you are calculating the probability that each of these 4 hits will be unlucky:
5) 1 Swordsmen, 0.9
7) Griffins, 0.8708450335
10) 25 Swordsmen (back), 0.8331899463
14) Bowmen (back), 0.4005157497
I.e. you have taken into account only these 4 events, not all 15 events (for these 4 events other events changed only probabilities).
The only thing I'm not sure is how the probability of luck changes if you hit 2 stacks at once.
Case 1 (as before):
11) Bowmen (part 1), 0.7217440598
12) Bowmen (part 2), 0.6406186336
13) Bowmen, 0.5358411166
14) Bowmen (back), 0.4005157497
15) Bowmen, 0.2257363173
Case 2:
11) Bowmen (part 1), 0.7217440598
12) Bowmen (part 2), 0.7217440598
13) Bowmen, 0.6406186336
14) Bowmen (back), 0.5358411166
15) Bowmen, 0.4005157497
I am sure that one of these cases are correct, but not sure which.
I previously guessed that the case 1 is correct.
Ok, I can give the probability that each 15 hits will be unlucky in both cases:
Case 1: 0.005286214895
Case 2: 0.01837247312 | Sorry, I read too quickly and kind of glazed over the volley shot being treated as 2 separate counts. I'm not sure how that affects the luck formula. In fact I don't even remember ever seeing luck happen for volley. But for simplicity let's treat that volley as 1 attack.
Your number 0.2615 is correct IF you are calculating the probability that each of these 4 hits will be unlucky:
Actually if I were using 4 hits, then it would be 78.46% chance to not get any luck trigger. The witchhammer table (1 morale; 0 trigger) shows the trigger probability for each turn. For example turn 9 has 77.43% chance of a trigger if the previous 8 turns had no trigger. 1-77.43% = 22.57% that turn 9 will not trigger. This 22.57% doesn't mean for the event of turn 9 that luck will not trigger. This # represents the probability of 9 consecutive no triggers.
This is why I tallied up the # of no triggers for each stack (separate counters) and then multiplied them together. You treated each and every event as separate when the table obviously takes into account the previous events. | For example turn 9 has 77.43% chance of a trigger if the previous 8 turns had no trigger. 1-77.43% = 22.57% that turn 9 will not trigger.
Correct!
This 22.57% doesn't mean for the event of turn 9 that luck will not trigger.
Not sure what you mean here.
This # represents the probability of 9 consecutive no triggers.
Wrong, 22.57% is not the probability of 9 consecutive no triggers!
22.57% is the probability of no luck in the 9th turn when 8 the last turns had no luck trigger (this is the probability of no luck in the 9th turn, not all 9 turns).
The formula in is this:
Probability of Luck now (in this one turn) when n times no luck before = F^(1-n*F/(1-F))
F = 0.1 (when Luck =1)
F = 0.2 (when Luck =2)
...
It is interesting that from this formula we get probability 1 when F=0.1 and n=9:
F^(1-n*F/(1-F)) = 0.1^(1-9*0.1/(1-0.1)) = 0.1^(1-9*0.1/(0.9)) = 0.1^0 = 1
(i.e. luck indeed will happen when 9 times were no luck before)
Let's take simple example of just Bowmen (Luck = 1, i.e. F = 0.1):
Probability of "no luck" in 1st turn = 1-F^(1-0*F/(1-F)) = 1-F = 0.9
Probability of "no luck" in 2nd turn when 1st turn was unlucky = 1-F^(1-1*F/(1-F)) = 1-0.1291549665 = 0.8708450335
Probability of "no luck" in 3rd turn when 1st and 2nd turns were unlucky = 1-F^(1-2*F/(1-F)) = 1-0.1668100537 = 0.8331899463
And so on.
Now, what is the probability that Bowmen will have no luck the first 2 turns?
Answer:
P(NoLuck1 and NoLuck2) = P(NoLuck1)*P(NoLuck2 | NoLuck1) = 0.9*0.8708450335 = 0.7837605302
Probability P(NoLuck2 | NoLuck1) is conditional probability here (it is probability of the event NoLuck2 when we know for sure that event NoLuck1 already happened).
Similarly, the probability that there will be no luck in the first 3 turns:
P(NoLuck1 and NoLuck2 and NoLuck3) = P(NoLuck1)*P(NoLuck2 | NoLuck1)*P(NoLuck3 | NoLuck1 and NoLuck2) = 0.9*0.8708450335*0.8331899463 = 0.6530213941 | | im confused... | | Don't worry, we all are. Those two are in a world of their own now... | | good luck to them i hope they can find their way home lol | | well i have had even worse than this in a battle without any luck even though havin an amulet of luck | im confused...
The analysis above gives simple answer:
- 15 hits with no befallen luck in a row (when you have Luck 1) happens approximately once in about 200 battles on average (i.e. only 0.5% chance of such bad luck)
To be precise:
- if you have 4 stacks
1st stack attacks 9 times,
2nd stack attacks 2 times,
3rd stack attacks 3 times,
4th stack attacks 1 time
- and you have Luck 1
- then there is 0.5% chance that in all those 15 attacks luck won't befall. | | On the face of it, robai = win i would feel unlucky to not see a lucky trigger, although I noticed that one luck was never enough ). This belongs in OGF because ofc it's nothing to do with the game... But anyway since when did the mods around here moderate and not just enjoy pressing their buttons ) | More examples:
1)
If you have you have 3 stacks
1st stack attacks 5 times,
2nd stack attacks 5 times,
3rd stack attacks 5 times
- and you have Luck 1
- then there is 4.86% chance that in all those 15 attacks luck won't befall.
Proof:
(0.90*0.87*0.83*0.78*0.72)^5 = 0.0486
2)
If you have you have 5 stacks
1st stack attacks 3 times,
2nd stack attacks 3 times,
3rd stack attacks 3 times,
4th stack attacks 3 times,
5th stack attacks 3 times
- and you have Luck 1
- then there is 11.6% chance that in all those 15 attacks luck won't befall.
Proof:
(0.90*0.87*0.83)^5 = 0.116
3)
If you have you have 15 stacks and each of them attacks only once
- and you have Luck 1
- then there is 20.6% chance that in all those 15 attacks luck won't befall.
Proof:
(0.90)^15 = 0.206 |
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