Leader Factory!

Aaron Cleavin

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"What you have calculated is the p >=1 event occuring "

Isn't that exactly what you are needing to know ?
In 16 trials what is the probability of at least one event

Event being an Orig DR of 2
Not at all what we want to know is how many leaders are generated (on average) in a player turn, rather than the chance of one or more being generated.
 

Magpie

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Yes so the first thing you need to calculate is the probability of rolling 1 or more 2's out of 16 attempts.
That is the formula I posted earlier.
 

klasmalmstrom

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...put 16 248s in a hex with a CT AFV wreck, each player turn they will generate 16/36 leaders with no chance of burning the wreck ...
Since the wreck is treated as CE, Immobile, and without any MG, which give us a CC DRM of: -1 (CE/Abandoned) -1 (Immobile) -1 (no MG) = -3.

CCV of a half-squad is 3, so a CC DR of 4 (or less) should result in a Burning Wreck, no?

Seems like that is more likely to occur before the first 1,1 is rolled.
 

Magpie

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Since the wreck is treated as CE, Immobile, and without any MG, which give us a CC DRM of: -1 (CE/Abandoned) -1 (Immobile) -1 (no MG) = -3.

CCV of a half-squad is 3, so a CC DR of 4 (or less) should result in a Burning Wreck, no?

Seems like that is more likely to occur before the first 1,1 is rolled.
I think you add +1 for every squad equivalent overstacked 16 HS = +5 ???
 

jrv

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"What you have calculated is the p >=1 event occuring "

Isn't that exactly what you are needing to know ?
In 16 trials what is the probability of at least one event

Event being an Orig DR of 2
No, because you could create two (or more) leaders in a given CCPh. On average each CC attack will create one thirty-sixth of a leader, and on average sixteen CC attacks will create sixteen thirty-sixths of a leader, or roughly 0.44 leader per CCPh. Assuming you don't generate a burning wreck, and if you create a burning wreck, you have created a leader.

JR
 
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Magpie

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No, because you could create two (or more) leaders in a given CCPh. On average each CC attack will create one thirty-sixth of a leader, and on average sixteen CC attacks will create sixteen thirty-sixths of a leader, or roughly 0.44 leader per CCPh. Assuming you don't generate a blazing wreck.

JR
That's not how probability is calculated though.
The formula of the situation is the binomial probability function that I posted earlier
 

jrv

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That's not how probability is calculated though.
It is for the question of how many leaders are created on average. Your calculation, what is the likelihood is that at least one leader is created, is not accurate for determining how "productive" the factory is. If one-thousand halfsquads are used instead of sixteen, that will produce 1000*1/36 leaders per CCPh (ignoring the burning wreck issue). That number is much greater than one. The odds that at least one leader will be created will still be < one (albeit very near one). It has to be because it is a probability, not a calculation of the number leaders created.

JR
 
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Honza

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You mathematicians drive me nuts. You need to go and do a watercolour. ;)
 

Magpie

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To work out what the expected number of leaders created you have to work out the probabilities for the various number of leaders created.
Then from that you can work out what the most likely outcome is

The table is :

8623

So the most likely outcome is no leader at all, second most likely is 1 leader and outside chance of 2 and pretty much insignificant for the rest.
 

Aaron Cleavin

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Aha! - but isn't there an Unlikely Kill rule as well - kicks on when rolling a 1,1?
Overtstacking +5 Immobile, CE, No maneed useable MG -3

If you roll a 1,1 you may genrate an 8-1 leader

In this case

CCV = 3+1 = 4

DRM +5 -3 -1 (Ldr) == Net +1

Final DR = 3 which is not <= 1/2 of 4

This no burn is possible.

The 8-1 possibliity is why and overstacking of 5 is needed to avoid any chance of a burn


(OverStacking of 4 with a CX DRM or reduced CCV because of conscriptness also works but is less elegant/sustainable)
 

Aaron Cleavin

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To work out what the expected number of leaders created you have to work out the probabilities for the various number of leaders created.
Then from that you can work out what the most likely outcome is

The table is :

View attachment 8623

So the most likely outcome is no leader at all, second most likely is 1 leader and outside chance of 2 and pretty much insignificant for the rest.
We aren't talking about most likely outcome we are talking about expectancy which is 16/36
 

Aaron Cleavin

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To work out what the expected number of leaders created you have to work out the probabilities for the various number of leaders created.
Then from that you can work out what the most likely outcome is

The table is :

View attachment 8623

So the most likely outcome is no leader at all, second most likely is 1 leader and outside chance of 2 and pretty much insignificant for the rest.
Sum m*P and you will get 16/36
 

Aaron Cleavin

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That's not how probability is calculated though.
The formula of the situation is the binomial probability function that I posted earlier
It is exactly how expectancy is calculated in statistics, as per the formula I posted
 

jrv

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So the most likely outcome is no leader at all, second most likely is 1 leader and outside chance of 2 and pretty much insignificant for the rest.
Per your table the odds for one leader is 0.2912, the odds for two leaders is 0.0624, the odds for three leaders is 0.0083. Based on those three alone the expected number of leaders created per CCPh is 0.2912*1 + 0.0624*2 + 0.0083*3 = 0.2912 + 0.1248 + 0.0249 = 0.4409, which is getting close to the correct answer of 0.444444.... leaders per turn. Since I truncated the probabilities my estimate based on the first three is low even for that group.

JR
 
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Aaron Cleavin

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Per your table the odds for one leader is 0.2912, the odds for two leaders is 0.0624, the odds for three leaders is 0.0083. Based on those three alone the expected number of leaders created per CCPh is 0.2912*1 + 0.0624*2 + 0.0083*3 = 0.2912 + 0.1248 + 0.0249 = 0.4409, which is getting close to the correct answer of 0.444444.... leaders per turn. Since I truncated the probabilities my estimate based on the first three is low even for that group.

JR
Summing the first 5 terms we get 0.4444269675303 converging as you say to 4/9 == 16/36
 

klasmalmstrom

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Of course, while we are discussing this and figuring out the odds - the front that those 16 half-squads were supposed to cover is being overrun by the enemy. :D
 
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