Leader Factory!

Aaron Cleavin

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Just found a way to wreck a long CG , 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 2/3 chance of 8-1 and 1/3 of 8-0. The leader balance per Amierican, British, German, not quite as good for others.
 

Vinnie

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Why no chance of burning it? They will be rolling with a -3 so will burn on a 4 or less.

Of course, they're overstacked! Duh!
 

Aaron Cleavin

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Best Fuel is I think 15 237's Leader mix not quite as rich but still very good and taking a 2nd line Coy out much better than an elite one
 

jfardette

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Never thought of that. A little silly, if my opponent tried that I think I would suggest we try another game. I did have an opponent change his TCA and knock a squad off, which then battle hardened and generated a hero. Wasn’t intentional, but the rules do seem to remind us it’s a game and not a simulation from time to time.
 

klasmalmstrom

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...I did have an opponent change his TCA and knock a squad off, which then battle hardened and generated a hero. Wasn’t intentional, but the rules do seem to remind us it’s a game and not a simulation from time to time.
No Heat of Battle on a Bail Out MC though - PRC are exempt.
 

Magpie

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Your odds of generating a single leader each turn, ie. out of 16 CC rolls, is about 0.29 so that's quite an investment for a less than 1/3 chance of a leader.
 

jfardette

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Can you still take casualties for crew small arms if there is no crew? Not to sidetrack your thread Aaron, I think you are on the right track.
 

Magpie

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Can you still take casualties for crew small arms if there is no crew? Not to sidetrack your thread Aaron, I think you are on the right track.
No. It has to be "non-Abandoned "
 

Aaron Cleavin

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Your odds of generating a single leader each turn, ie. out of 16 CC rolls, is about 0.29 so that's quite an investment for a less than 1/3 chance of a leader.
That's per player turn so with 16 HS say you get 32/36 leaders per turn, or almost a whole Leader.
BTW 16/36 = 0.44 not 0.29 (Possibly 0.29 of exactly one leader, and sometimes you get >1)
 

Magpie

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The probability of at least 1 occurrence though is
p = 1 - (1 - .027)^16 -> that's about 0.35

Where do you get 16/36 from ?
 

Aaron Cleavin

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The probability of at least 1 occurrence though is
p = 1 - (1 - .027)^16 -> that's about 0.35

Where do you get 16/36 from ?
16/36 is the Expectancy. for independent events E(nX) = n*E(X)


Let {\displaystyle X}
X
be a random variable with a countable set of finite outcomes {\displaystyle x_{1}}
x_{1}
, {\displaystyle x_{2}}
x_{2}
, ..., occurring with probabilities {\displaystyle p_{1}}
p_{1}
, {\displaystyle p_{2}}
p_{2}
, ..., respectively, such that the infinite sum {\displaystyle \textstyle \sum {i=1}^{\infty }|x{i}|\,p_{i}}
{\displaystyle \textstyle \sum _{i=1}^{\infty }|x_{i}|\,p_{i}}
converges. The expected value of {\displaystyle X}
X
is defined as the series

{\displaystyle \operatorname {E} [X]=\sum {i=1}^{\infty }x{i}\,p_{i}.}
{\displaystyle \operatorname {E} [X]=\sum _{i=1}^{\infty }x_{i}\,p_{i}.}



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

Magpie

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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
 
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