Interesting use-case of probability in Defense (and attack!)

Saturation attack is a well-known military tactic where the attack strategy is to overwhelm the opponent’s defense system with bombardments (more quantitatively and relatively less qualitatively), thus saturating their ability to respond. This results in either less available time or less available fire power or both due to saturation and hence failure in neutralizing the incoming targets.

For targeted attack, similar tactic is used to hit the target with less accurate but multiple such missiles. Idea is to throw more missiles than required (in an ideal case) to increase the certainty of hitting the target. Sort of like throwing multiple dices so that at least one dice’s outcome will be, say, 6!

Take three cases for example: One with highly accurate but only single missile, second with less accurate but two missiles and third one with even lesser accurate but three missiles. We will assume that each missile can destroy the target and hence no individual missile can be ignored. We will calculate probability of hitting the target in each case. ?

Before that note that no missile is ever 100% accurate or will remain 100% accurate throughout its lifecycle (courtesy uncertainties arising from changes with time e.g. environmental conditions, technological breakthroughs and advancements and many more). Also, the cost of accuracy shoots up as we reach the perfect 100% mark. And so, the cost of not hitting the target also goes through the roof! Overall, it’s the perfect (demo) missile that nobody wants to buy in bulk. It’s unique but only an expensive exhibit. Enough about tradeoffs, let’s get to the math!

Case 1:

Success rate = 90% (our precious!), # trials = 1

P(hit) = 0.9, i.e. 90% ?

Case 2:

Success rate = 70%, # trials = 2

A: first missile hits the target with P(A) = 0.7

B: second missile hits the target with P(B) = 0.7

Since A and B are not mutually exclusive events (A can happen, so can B, and so can A AND B in which case both the missiles hit the target at once). ?

P(hit) = P(A OR B) = P(A) + P(B) – P(A AND B)

= P(A) + P(B) – (P(B|A) * P(A)) = P(A) + P(B) – P(A) * P(B)

P(hit) = 0.7 + 0.7 – 0.7 * 0.7 = 0.91 i.e. 91% well we have almost the same probability of hitting the target as that of the case 1!

Case 3:

Success rate = 50%, # trials = 3

A: first missile hits the target with P(A) = 0.5

B: second missile hits the target with P(B) = 0.5

C: third missile hits the target with P(C) = 0.5

Like case 2, we have 3 events which aren’t necessarily mutually exclusive. For 3 events, we need to add probability of all the three events happening at the same time. ???

P(hit) = P(A OR B OR C) = P(A) + P(B) + P(C) – P(A AND B) – P(B AND C) – P(C AND A) + P(A AND B AND C)

= P(A) + P(B) + P(C) – P(A) * P(B) – P(B) * P(C) – P(C) * P(A) + P(A) * P(B) * P(C)

P(hit) = 0.5 + 0.5 + 0.5 – 0.5 * 0.5 – 0.5 * 0.5 – 0.5 * 0.5 + 0.5 * 0.5 * 0.5

= 1.5 – 0.75 + 0.125 = 0.875 i.e. 87.5% this isn’t too off from case 1 and case 2!

An individually launched missile with a 70% success rate will get to its target only 70% of the time. But with two such missiles the chances are 91%. Similarly, for three missiles with individual success rate of 50%, together they have 87.5% chance of hitting the target.

But then one might ask, how to decide between the success rate and number of missiles! Well, that becomes an optimization problem. Each missile has a cost associated with it. As success rate varies, so does the cost. Of course, there are many more parameters to consider. But a simplified objective function (optimization problems are modeled as objective functions) can be stated as below:

Objective Function = cost of each missile (M) * no of same missiles used (N)

N needs to be derived or estimated using the constraint: P(hit) = 90% (for example)

Objective function is evaluated for each available missile or combination of missiles! And the one with the minimum value indicates a case with minimum cost and hence optimized use of resources. The objective functions are also called "cost functions" since the problem is, generally, converted such that cost becomes the function’s output which is to be minimized to arrive at an optimal solution.?It also goes by the aliases of "loss function", "error function", "reward function" and so on. You guessed it right, this alias depends on the domain and the use-case.

I hope you enjoyed reading this article especially the application example of probability theory in the defense domain. Our precious! Do consider hitting the like button. This serves as your feedback to me for this post and inspires me to continue writing about similar topics from technology and engineering. Please feel free to share your suggestions and feedbacks in the comments section.

Thank you and I'll see you soon!

-Sanjay