In the September 4, 2006 issue of Space News, Thomas Christie, the former chief weapons tester at the Pentagon, was quoted as saying the Ground Based Mid-course Defense System “likely would have less than a 20% chance of shooting down an incoming missile from North Korea”. When asked his take on the system’s effectiveness, the president of the Missile Defense Advocacy Alliance, Mr. Riki Ellison, responded “that could be from firing just one missile, and, there are about 9 or 10 interceptors that could take multiple shots at the incoming target, thereby increasing the chances of a hit.”

If one interceptor has a 20% chance of hitting a target, what exactly are the chances of a successful defense against an incoming missile with 10 interceptors?

Basic probability theory can help answer that question. Let’s start simple. Suppose an antiballistic missile (ABM) interceptor has exactly a 20% chance of hitting an incoming missile. That means it has an 80% chance of missing. Not good.

Suppose two interceptors are shot, and for simplicity, let’s assume that each interceptor launched is an independent event (the action of one interceptor does not effect the other). In this case, there are several possible situations that can occur. The first ABM could hit the target, and the second could miss. Likewise, the first ABM could miss and the second one could hit. Or, they both could hit the target, or, both miss it.

Let’s abbreviate these four possibilities {HM, MH, HH, MM} using H for a hit and M for a miss. This list of all the possible outcomes for two interceptors is called the *sample space*. An *element* of the sample space is called an *event*. Now assuming each interceptor is independent, and one interceptor does not effect the other, the probability of any compound event of two interceptors can be computed by multiplying the probability of each of the single events.

For example, if both ABMs hit the target, we can multiply the probability of each one hitting to get the probability of the *compound event* of them both hitting:

P(HH) = P(H)P(H)

= (0.20)(0.20)

= 0.04

We have a 4% chance of BOTH missiles hitting the target.

The probability that both *miss* the target is computed as

P(MM) = P(M)P(M)

= (0.80)(0.80)

= 0.64

giving a 64% chance of both ABM’s missing the target.

Now, what if one hits, and the other misses:

P(HM) = P(H)P(M)

= (0.20)(0.80)

= 0.16

or a 16% chance of one hitting and one missing. It’s the same computation and the same result for the first ABM missing and the second one hitting:

P(MH) = (0.80)(0.20)

= 0.16

Now it is a basic tenent of probability that the sum of all probabilities of events in the sample space must add to 1. We can confirm this by noting that

0.04 + 0.64 + 0.16 + 0.16 = 1

Let’s now ask what is the probability that AT LEAST one hit will occur? There are three possibilites in the sample space where at least one hit can occur: HM or MH or HH. To find the probability of at least one hit occuring, we add the probabilities for each of these compound events. (In these simple cases, OR measn ADD in probability.)

P(at least one hit) = P( HM or MH or HH)

= 0.16 + 0.16 + 0.04

= 0.36

So there is a 36% chance that at least one of the two ABMs will it the target missile. Since all probabilities in a sample space must add to 1, and the probability that both interceptors miss is P(MM) = 0.64, we have that:

P(at least one hit) + P(all miss) = 1

0.36 + 0.64 = 1

Subtracting the P(all miss) from both sides of the equation, we have

P(at least one hit) = 1 – P(all miss)

Since there is only ONE WAY that any number of interceptors can ALL MISS the target, this simple equation gives an easier way to compute the probability of at least one hit for any number of ABMs.

With three ABMs you have a sample set of eight different possible outcomes.

{HMM, MHM, MMH, HHM, HMH, MHH, HHH, MMM}

To computer the probabiltiy that at least one of the three interceptors will hit the target, we have

P(at least one hit) = 1 – P(no hit at all)

or

P(HMM or MHM or MMH or HHM or HMH or MHH or HHH) = 1 – P(MMM)

It’s easy to compute the probability that all three interceptors miss:

P(MMM) = (0.80)(0.80)(0.80)

= 0.512

= 51.2%

Then, substituting this into our equation, we get the probability that at least one of the interceptors will hit as

P(at least one hit) = 1 – P(all miss)

= 1 – P(MMM)

= 1 – 0.512

= 0.488

= 48.8%

At least with three ABMs we are getting closer to a 50% chance of at least one of them taking out the target.

So what if we had 10 interceptors. The number of total outcomes increases to 2^10. What is the probability that at least one of the ten interceptors will hit the target?

P(at least one hit) = 1 – P(all miss)

= 1 – P(MMMMMMMMMM)

= 1 – (0.8)^10

= 1 – 0.107374

= 0.892626

> 89%

It’s close to a 90% chance that at least one of the ten antiballistic missiles will hit an incoming missile, assuming independent events.

Now, inreality, the events are not independent. In other words, if the first missile missed, one assumes there would be some information gleaned that would add to the accuracy of the second one. However, enemy efforts to disguise the incoming missile containing the warhead with a number of decoys further complicate the computation, as well as the probability of a hit.

Nevertheless, Mr. Ellison is correct; 10 interceptors increase the chances of a successful hit. As we have shown, the probability increases from 20% to 89%. He also states that “while the system’s capability might not be 100%, I think it would have a pretty good shot at intercepting the North Korean missile.” Given the price tag in the trillions through 2015 for Ballistic Missile Defense, it had better be “a pretty good shot”.

Hey Robin,

Thanks for the post. I understand it but not well enough to use it. My math brain was destroyed in grade 11 by a really obnoxious teacher. I’d love to revive it, but it would probably take a lot more patience than you have to spare. Anyway, if you have any simple examples of how I might be able to formulate probabilities in day to day living I’d love to hear them.

Here’s a collection of probabilities I came across and saved recently.

Odds of being struck by lightning (though not necessarily dying) in a given year: 1 in 400,000 (Source: National Weather Service)

Odds of dying in a car accident in a given year: 1 in 18,400 (Source: National Safety Council [NSC])

Odds of getting killed somehow while walking around outside: 1 in 49,000 (NSC)

Odds of drowning: 1 in 88,000 (NSC)

Odds of choking to death: 1 in 97,000 (NSC)

Odds of dying in an air (or space) accident: 1 in 392,000 (NSC)

Odds of freezing to death: 1 in 469,000 (NSC)

Odds of death from falling off the bed or a chair: 1 in 347,000 (NSC)

Odds of choking to death on your own vomit: 1 in 740,000 (NSC)

Odds of getting killed by fireworks: 1 in 26,440,000 (NSC)

Odds of death due to overly hot tap water: 1 in 11,100,000 (NSC)

Odds of death due to burning pajamas: 1 in 97,000,000 (NSC, and no that’s not a joke)

Odds that you’ll kill yourself: 1 in 9,200 (NSC)

JH

Hey John, Sorry to hear about your traumatic collision with math. It’s a cryin shame that so many people, including myself, had such abusive experiences in grade school. It doesn’t have to be this way!

I mean just think about going to class and learning the odds of dying in your burning pajamas! The kids would go crazy!

Odd are probabilities put in another form. For instance, toss a coin and the odds are 1:1 that you’ll get heads (or tails). The probability of getting heads (or tails) is P(H) = 1/2.

Notice that when you add the numbers in the odds form, 1:1, you get 2, which happens to be the denominator of the probability fraction.

So you can always turn odds into a probability by adding the numbers in the odds, putting that in the denominator for the probability form. In the numerator, you put the other odds number.

So the odds of dying in your pajamas is 1:97,000,000.

The probability of dying in your pajamas is P(PJ) = 1/97,000,001

You can compute probabilities from the basic ratio formula. Let E be some event. Then, the probability of E occuring is

P(E) = # ways E can occur / total # outcomes.

With this formula, you just have to count the differents ways things can happen.

For instance, roll a die. The total possible outcomes are {1,2,3,4,5,6}.

What is the probability of rolling an even number?

Well, there are three ways to roll an even number, 2, 4, 6 out of the six total possible outcomes. So the probability of rolling an even number is

P(even) = 3/6 = 1/2 or 50%.

What’s the probability that you’re going to ride your bike on the bike path today? Look at the past 100 days. Out of those days, how many days did you ride your bike? Suppose you rode your bike 85 out of the past 100 days. Then you could say that there was an 85/100 or 85% chance that you’re going to ride your bike today.

With this method, the more days you have measured, the better the probability. LIke, it would be better to make a ratio of the past 1000 days for a more accurate picture. Or, you could get even more detailed and only count days that have the same weather conditions as today and see how many days you rode your bike on days with the same weather conditions, etc…

anyway, that ratio is a pretty good way to use probability in a simple, but revealing way.

And sleep naked!!