ORIGINAL: Bletchley_Geek
Some math on this.
We have a T-34 factory of capacity 50, with damage of 47% at Turn X. Let's see how many tanks it might be producing. Let p be the probability of the factory producing its capacity.
At Turn X: p0 = 0.53
At Turn X+1 : p1 = 0.56
At Turn X+2 : p2 = 0.59
Now we have three independent random events (i.e. the factory produces its full cap of T-34s).
Okay, yeah, I was right before. You produce your WHOLE capacity, or nothing. This will help anyone who wants to understand these equations understand what Bletchley has done.
Probability of producing 150 T-34's after 3 weeks = p0 * p1 * p2 = 0.1751, i.e. in seventeen out of every 100 games with the same situation.
This is the straight-forward multiplication of the original 3 probabilities.
The probability of 150 tanks (which is the factory's 50-tank capacity multiplied by the 3 turns we're looking at) REQUIRES those probabilities to be achieved. These are independent, exclusive probabilities on each turn. Thus, they are multiplied. If any of those probabilities had failed, it is no longer possible to get 150 tanks, and instead you're in one of the situations where you produce 100 tanks (which I look at next), 50 tanks, or 0 tanks.
Probability of producing 100 T-34's after 3 weeks
To understand the equation below, you have to understand what you're looking at in independent events.
To get 100 tanks, ANY 2 of your 3 probabilities have to be true.
So...
= p0*(1-p1)*p2 + p0*p1*(1-p2) + (1-p0)*p1*p2 = 0.4145, i.e. in 41 out of every 100 games.
You're looking at 3 independent equations. These are:
The probability that on T1, the factory fails to produce, which is this equation.
= (p0*(1-p1)*p2)
Explained. Bletchley is showing the scenario where the factory produces on T0 and T2, but NOT T1.
I added the parentheses to the beginning and end of that equation because that is only ONE scenario where you reach 100 tanks produced. For me, these are really important in understanding probability equations.
Note that the equation(1-p1) is the compliment of the p1 probability (specifically it is 1.0 minus .56, which is .44)- in other words, it is the chance that, on T1, the factory FAILED to produce. All probabilities are between 0 and 1 (think of this as a percentage between 0 and 100 percent). So you have a 100 percent chance of the factory either producing 50 tanks, or not producing 50 tanks - those are your only possible probability outcomes for T1.
The probability that the factory produces 50 tanks on T0 and also T1, but NOT T2, which is this equation:
(p0*p1*(1-p2))
Again, I add the extra parentheses to delineate exclusive scenarios. They are
exclusive because the only way to get 100 tanks is to have 2 successful production turns and 1 failure.
(Note you have a similar equation for the compliment of T2 failing to produce any tanks).
And finally the probability that the factory produces 50 tanks on T1 and T2 but not T0, which is this equation:
((1-p0)*p1*p2)
Now to note one more point on the equation:
p0*(1-p1)*p2 + p0*p1*(1-p2) + (1-p0)*p1*p2 = 0.4145
Let's abbreviate these raw numbers with what they mean:
The chance of producing 100 tanks equals:
The chance that you produce 50 on T0 and T2
plus
The chance that you produce 50 on T0 and T1
plus
The chance that you produce 50 on T1 and T2.
Each of those equations is ONE possible way to produce 100 tanks, so we add them together to fine the total probability.
The same thing is happening with the other two sets of equations, only in those, we are measuring only the chance of producing 50 tanks (in other words, two failures and one success over those 3 turns), or 0 tanks (all three turns result in failures to produce).
Probability of producing 50 T-34's after 3 weeks = p0*(1-p1)*(1-p2) + (1-p0)*p1*(1-p2) + (1-p0)*(1-p1)*p2 = 0.3255, i.e. in 32 out of every 100 games.
Probability of producing 0 T-34's after 3 weeks = (1-p0 * 1-p1 * 1-p2) = 0.08478, i. e. in 8 out of every 100 games.
You can see that this slightly beneficial to the factory owner, since the probability of getting the full production in those three weeks is about the double of getting none, being the norm to get just 50 T-34s in three weeks.
But this omits one critical fact, and that omission may cause people to mistakenly believe the system favors Soviet production.
What's missing is the fact that at 51% damage or more, a factory produces nothing (100% chance).
If factories with more than 50 percent damage could produce using the same equations as 50 percent or less, you would see that the chance of producing 0 tanks is the same as producing 150 tanks (right, Blethcley?). But the number of calculations would be far, far higher.
As is, the chance of producing 0 tanks in a damaged factory is actually much higher than the chance of producing 150, because at 51% damage or more, you produce nothing.