Re: Math puzzle: Feynman's lesson on bragging

jerry_friedman@xxxxxxxxx wrote:
In /Surely You're Joking, Mr. Feynman/ (pp. 194-5--thank you, Amazon
Book Search), our hero tells how once at Los Alamos he bragged that any
problem that someone could state in 10 seconds, he could solve in a
minute to within 10%. Several co-workers tried to stump him, but he
made all the estimates in time. Then another co-worker named Paul Olum
heard the challenge and unhesitatingly said, "The tangent of 10 to the
100th." As Feynman put it, he was sunk.

So my puzzle is: What would Feynman's best guess have been?
Mathematically, given a fraction x (in this case, x = 0.1), what number
y should you guess in order to have the highest probability that (1 -
x) tan(10^100) < y < (1 + x) tan(10^100)? And while you're at it,
what's the probability that your guess y will be in the allowed range?

Well obviously you should guess y = tan(10^100), whatever that may be :)


If you can imagine calculating tan(10^100) in a reasonable amount of
time, make the exponent big enough so you can't imagine calculating the
answer within the lifetime of the universe.


So let's assume the argument is an unknown number whose value mod pi is distributed uniformly at random on [0,pi)...

If I guess value y, then I'll be correct within faction x if the answer is within the range (y/(1+x),y/(1-x)), since the fraction is measured relative to the answer not my guess y. The objective is then to maximize the span of the inverse tangent of this range:

maximize w.r.t. y: atan(y/(1-x)) - atan(y/(1+x)).

This is straightforward calculus. If I didn't make any mistakes in my hasty computation,

y = sqrt(1-x^2).

For x = 0.1, this gives y = 0.994, or very nearly 1.


(The answer is simple enough that I can imagine it might be possible to
get without calculus, but I have no idea how.)


Nice puzzle.

-Mark
.

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