Leader Factory!

Magpie

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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

Sure but what that gives you is the probability that at least 1 and no more than 3 leaders are created.

It doesn't tell you the most likely (expected) number of leaders that would be created per turn, which is 0
 

jrv

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Sure but what that gives you is the probability that at least 1 and no more than 3 leaders are created.

It doesn't tell you the most likely (expected) number of leaders that would be created per turn, which is 0
Your table gives all that information. Line one is the probability that zero leaders are created in any one run of sixteen CC DRs, 63.7 percent of the time. There's no point in adding it to the expected number of leaders created because .6372*0 = 0. Zero is the most likely number. It is also the median number. However it is not the expected number. The expected number is the sum of the number created times the chance of that number created. The expected number is 0.4444..., or sixteen divided by thirty-six. If in one-hundred player turns sixteen halfsquads attacked as described above, one would expect that forty-four leaders would be created.

To find the odds that at least one and no more than three leaders are created you sum lines 2-4, 0.2912… + 0.0624… + 0.0083…. This value is not particularly interesting to determining how "productive" (on average) the leader factory is because while it gives the probability, the "productivity" is probability times the value, i.e. the expected value. The value is, roughly, the percent of turns that at least one leader will be created, or put another way, the odds that one or more leaders will be created this turn. But you can get an exact answer by subtracting the probability that zero leaders are created this turn from one, i.e. 1 - 0.6372 = 0.3628 (approximately).

Similarly an event with a probability of one-tenth that produces ten leaders will on average produce one leader per turn. On most turns (nine out of ten on average) it will produce zero leaders and on the tenth turn (on average) it will create ten. The event's median and most common values are zero, but its expected value is one.

JR
 
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Juan SantaX

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Uh... That must be red skin man magic, or maybe witchcraft... We need a spanish Inquisition revival...
 

Philippe D.

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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

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.
This is where you're wrong. If you want to compute an expectation (average), you need to take into account the possibility of multiple leaders being created, and count these multiple times.

But you don't need to make all these calculations. Expectation is linear, so if one attempts gives on average 1/36 leader creation (and that is easy to check, as you generate exactly one with probability exactly 1/36), then N attempts give on average N/36 creations (and please note that independence is not needed for this - it works even if they are not independent, though in the present case they are). So just computing the probability that one attempt yields a leader, lets you compute the expected number of creations for an arbitrary number of attempts.

Looking for the most likely outcome has no effect on the expectation. It may be relevant to your tactical decision, but the most likely outcome doesn't enter the calculations.

(I know it seems counter-intuitive. My students have difficulty with this too.)
 

Philippe D.

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Sure but what that gives you is the probability that at least 1 and no more than 3 leaders are created.

It doesn't tell you the most likely (expected) number of leaders that would be created per turn, which is 0
OK, so I see where the confusion comes from. The word "expected" is interpreted as meaning whatever is most likely to happen. (The confusion is the same in french, with "espérance").

The "expectation" (or "expected value") has nothing to do with what is most likely. In many situations, it is not even among the possible results (in the leader creation attempt, the expectation is 1/36, but we all agree you cannot create a fractional number of leaders). You have to see it as an average if you do a large number of attempts (it's actually a limit, so "large" means "let the number of attempts tend to infinity, and compute the number of successes, divided by the number of attempts). And it's a theorem.
 

Aaron Cleavin

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OK, so I see where the confusion comes from. The word "expected" is interpreted as meaning whatever is most likely to happen. (The confusion is the same in french, with "espérance").

The "expectation" (or "expected value") has nothing to do with what is most likely. In many situations, it is not even among the possible results (in the leader creation attempt, the expectation is 1/36, but we all agree you cannot create a fractional number of leaders). You have to see it as an average if you do a large number of attempts (it's actually a limit, so "large" means "let the number of attempts tend to infinity, and compute the number of successes, divided by the number of attempts). And it's a theorem.
A proven theorum :)
 

Sparafucil3

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This is a lot of back and forth for something I think most people would walk away from the game over. A little sleaze here and there is to be expected, but torture the rules like this? I don't think so. JMO, YMMV. -- jim
 

jrv

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It really isn't
Per wikipedia, "the expected value of a discrete random variable is the probability-weighted average of all possible values. In other words, each possible value the random variable can assume is multiplied by its probability of occurring, and the resulting products are summed to produce the expected value." In this case the values of the random variable are 0..16, the possible number of leaders created in each trial. The probabilities are given in your table. The expected value is 𝚺pᵢvᵢ, where the vᵢ are the number of leaders created (column labeled "success times created") and pᵢ are the probabilities (column labeled "probability P") in your table.

The most likely value (the "mode") for this random variable is zero. If you had to bet on a single number with the same odds for all outcomes, bet zero. For analyzing the worth of the "sleaze" proposed, the mode isn't of much interest. The player is going to poll the random variable multiple times. Usually (more than half the time) the result is zero leaders, but sometimes the result is between one and sixteen (inclusive) leaders. What should interest a perpetrator of this sleaze is the expected number of leaders he will create per player (or game) turn, not the most likely number. In the long term he will create the expected number per turn times the number of turns he operates this sleaze. With sixteen halfsquads the expected number of leaders per player turn is 0.444..., or slightly less than one leader per game turn.

Perhaps this might guide you further: https://math.stackexchange.com/questions/1186474/what-is-the-difference-between-expected-value-and-most-likely-value

JR
 

Aaron Cleavin

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This is a lot of back and forth for something I think most people would walk away from the game over. A little sleaze here and there is to be expected, but torture the rules like this? I don't think so. JMO, YMMV. -- jim
The idea came after a wonderful game of ASL with Ken Young a true gentleman, and virtuoso player. He very legitimately CC'd his own wreck to attempt
to create some smoke cover. He was as embarrassed as I was astonished when a field promotion occurred form this; which led my twisted mind on a
torturous journey

Which is why I think the rule should be adjusted. Think of it as a reductio et absurdium gendanke experiment.
 

jrv

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This is a lot of back and forth for something I think most people would walk away from the game over. A little sleaze here and there is to be expected, but torture the rules like this? I don't think so. JMO, YMMV. -- jim
If you want to take eight squads out of the line to operate this "factory" in a game against me, have at it. I think I can kill an extra leader's worth of stuff per game turn because of such an employment of resources.

JR
 

jrv

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Which is why I think the rule should be adjusted. Think of it as a reductio et absurdium gendanke experiment.
I don't see it as a grave threat to the balance of ordinary CGs. Employing eight squads in such a manner is an expensive way to create leaders. Let 'em play it if that's what they want to do.

JR
 

Aaron Cleavin

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If you want to take eight squads out of the line to operate this "factory" in a game against me, have at it. I think I can kill an extra leader's worth of stuff per game turn because of such an employment of resources.

JR
RG CG 3 or the big new RO CG are the only cases where I think it might work. Even then somewhat marginal and dependent on creating a wreck in the right sheltered spot (would normally need to be a tank of one's own)
 

Paul M. Weir

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As an ex-programmer and a player of a game with a monstrous set of rules, I regard myself as quite logic ruled. Even so, I still have gut feelings born of experience and my gut feelings, with a splash of logic, tell me this: If you have such an advantage that you can deploy such a force for leader creation you have already won the game or if you don't have such an advantage you should be using that force to win the game.

So I am completely with jrv on this.

The only possible exception I could see is in a monster scenario or CG where you need to always keep a few squads on a flank. In that case there might be some point in giving them something to occupy their idle hands whilst on guard duty, in this case banging rifle butts on a dead tank. Otherwise it would be a criminal waste of a significant force. Just because you can do something doesn't mean you should, except for giggles, maybe.
 

Mister T

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Room for a new counter for wrecks to avoid milking the cow several times:
"Successfully extracted leader" 😜
 

Paul M. Weir

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Like clown cars it's possible to pack a lot of leaders inside a wreck. Perhaps there should be a DRM for vehicle size.

JR
The biggest clown would be the player who tried that trick.
 
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