December 10, 2002

One Hundred Interesting Mathematical Calculations, Number 1
#### One Hundred Interesting Mathematical Calculations, Number 1

**Returning From World War II Bombing Missions: **Suppose that you are a bomber pilot flying a B-17 in the European Theater of Operations during World War II. You know that 4% of bombers get shot down on average on each mission. You want to calculate the chance that you would successfully fly all the missions of your tour of duty--to make things simple, let's say 50 (a larger number than was actually asked of air crews)--without getting shot down.

You start by thinking that there is a 96% chance you return from your first mission--0.96. Then, if you successfully return from your first mission, there is a 96 percent chance you will return from your second mission. So the chance you will successfully return from your first and your second missions is 0.96*0.96.

Continuing on, you see that the chance you will return from all fifty of your missions is 0.96*0.96*...*0.96, where there are 50 0.96s--where you multiply 0.96 by itself 50 times. And we know that we can write 0.96 multiplied by itself 50 times as (0.96)^{50}.

Now we have two choices:

- We can find a calculator with a "y
^{x}" button, and press in "0.96", "y^{x}", "50", "=" - We can do more math...

Let's do more math.

Let's start by using one of the grand useful approximations of math. For any number x that is close to 1, x is approximately equal to e^{(x-1)}, where the transcendental number e=2.718281828 is one of the master constants of all mathematics.

Since 0.96 is close to 1, it is approximately the case that: (0.96)^{50 }= (e^{[0.96-1]})^{50 }=^{ }(e^{[-0.04]})^{50}.

And now we can use the principle that a number with an exponent raised to a further exponent is just equal to the number raised to the product of the exponents: (e^{[-0.04]})^{50 }= e^{[-2]}.

What is e^{[-2]}? It is just one divided by e^{[2]}. What is e^{[2]}? It is, approximately, 2.72*2.72. That is easy to solve: 2.72*2.72 = 7.39. And 1/7.39 = 0.135.

A bomber pilot assigned to fly fifty missions would have a 13.5 percent chance of coming back from mission 50 without ever having been shot down.

Is it any wonder that bomber pilots did not want to fly anywhere near fifty missions?

Posted by DeLong at December 10, 2002 10:19 PM | TrackbackEmail this entry

Thanks for posting the math puzzle; this

is a neat trick I'd like to wish I could

teach my undergrad students. :-) But

it's probably at least as worthwhile to

harp on the fact that e^-1 = .368 as it is

note that e=2.718. So in this case, it

was easy to see that the answer would be

about e^-2, so without even reaching for

a napkin, I could say that the chance of

survival was better than (1/3)^2 = .11 and

worse than (.4)^2 = .16, and probably

close to the average of those, or .135.

Direct hit! :-)

Posted by: Jonathan W. King on December 10, 2002 11:54 PMA nice calculation.

If you want 500 wonderful pages like that, rush out and get TW Körner's The Pleasures of Counting. Coincidentally, he has lots of examples from the analysis of anti-sub operations in WWII.

Posted by: Lance Knobel on December 11, 2002 02:01 AMI don't get it: the event of getting shot down on flight 50 should be indenpendent of having survived through 49 (although obviously one needs to be alive to try one's chance - and this assumption might be false in reality if stress weakens the pilot or his plane over time).

Hence, I can see how ex ante a pilot would want to limit the number of flight he would commit himself to, but not necessarily why he would not accept the 50th flight e.g. (or 15th for that matter.) Of course, I am assuming that willingness to take a given risk is independent of previous risks taken, which is probably inconsistent with human psychology.

But then again, this wasn't the main point of Professor DeLong's post...

Posted by: Jean-Philippe Stijns on December 11, 2002 10:01 AMJean-Philippe:

The point is that missions were assigned in blocks; that is, a bomber pilot arrived in e.g. England knowing that he would be assigned x missions before being eligible for rotation out of regular combat. The question is, what is x?

If I recall correctly, x was 25. And for a lot of the war that was very high, because toward the end German resistance was feeble, but earlier it was very effective.

The model assumes that all pilots are the same, i.e. that they never learn anything from their previous missions.

I think actual casualty rates were much higher for "rookie" pilots, and went down very quickly as they built up experience.

Posted by: Kimon Berlin on December 11, 2002 12:57 PMIsn't that trick essentially another version of the taylor expansion? e^x is approximately 1+x, or for y=x-1, e^y =1+x-1 = x. The trick isn't a trick at all--it is an approximation--and it is useful to know where it came from.

That is the key to so much useful math--the taylor expansion.

Of course, you also ought to link to eric Weisensteins world of mathematics.

http://mathworld.wolfram.com/

Posted by: Brennan on December 11, 2002 03:45 PMI have to disagree with Jean-Philippe.

Disregarding any pschological and/or physiological influences that may build up over time, we are still not dealing with a simple independent events.

The odds of flipping a coin a getting tails is .5.

The fact that I got tails on the first toss in no way influences the odds of getting tails on the second toss. Those odds remain .5.

That is an independent event.

However, the odds of getting tails on the first toss AND tails on the second toss is .25.

This is analogous to the calculation of surviving all the missions. Surviving the first AND survivng the second And the third......................................

Posted by: E. Avedisian on December 11, 2002 08:50 PMLet me quote myself then:

>>I can see how ex ante a pilot would want to limit the number of flight he would commit himself to, but not necessarily why he would not accept the 50th flight e.g.<<

To clarify, if a pilot has survived 49 flights then I don't see why he should refuse (under primitive assumptions) to fly the 50th flight anymore than the 15th after the 14th.

Now, that obviously assumes he can make a choice about that (and that the probability of being shot down is independent of the number of preceeding flights - no stress or learning by flying e.g.) If, as Jonathan suggests, missions were assigned in blocks, he could only make a commitment to, say, 50 flights, or clean the bathrooms for 3 months instead. And then, it must have been a very hard decision, indeed...

I think we all actually agree.

Posted by: Jean-Philippe Stijns on December 11, 2002 11:47 PMHmmmmm........

JP, You are indeed correct. The odds of surviving any one flight are .96; doesn't matter if its the first, the thirteenth. or the fiftieth.

The odds of surviving any number of flights (N) in a sequence is .96 to the Nth power.

Sorry JP, I did not pay close enough attention to what you were really saying.

Posted by: E. Avedisian on December 12, 2002 08:05 AMPost a comment