Normal Distribution Puzzle/Riddle Some time in the future, a lottery takes place, with winning number N. 3 of your friends from the future, John, Joe, and James, provide you with guesses on the number N.
John's guess a is randomly selected from a gaussian distribution centered at N with stdev x;
Joe's guess b is randomly selected from a gaussian distribution centered at N with stdev y;
James' guess c is randomly selected from a gaussian distribution centered at N with stdev z;
Given the values of a, x, b, y, c, z, what would be the best guess of N? Also define what "best" is.

You can calculate and maximize the likelihood of N given a,b,c, with x,y,z being fixed. 
The Likelihood of a value of N (the probability of sampling a,b,c given that the mean is N) is: 
$LL_{a,b,c}(N) = Pr(a | x,N) \cdot Pr(b | y,N) \cdot Pr(c | z,N)$ 
With the distributions being independent and Gaussian, this is 
$LL_{a,b,x}(N) = \frac{1}{x\sqrt{2\pi}} e^{-\frac{(a-N)^2}{2x^2}} \cdot \frac{1}{y\sqrt{2\pi}} e^{-\frac{(b-N)^2}{2y^2}} \cdot 
\frac{1}{z\sqrt{2\pi}} e^{-\frac{(c-N)^2}{2z^2}} = $
$\frac{1}{xyz(\sqrt{2\pi})^3} e^{-\frac{1}{2}(\frac{(a-N)^2}{x^2}  +\frac{(b-N)^2}{y^2}+\frac{(c-N)^2}{z^2})}$ 
And we want to find the N that maximizes this likelihood. To find the maximum, we will search for a point where the derivative of the likelihood equals zero. 
$\frac{d}{dN}LL_{a,b,c}(N) = \frac{1}{xyz(\sqrt{2\pi})^3}\cdot -\frac{1}{2}(\frac{2(a-N)}{x^2} + \frac{2(b-N) }{y^2}+ \frac{2(c-N)}{z^2}) e^{-\frac{1}{2}(\frac{(a-N)^2}{x^2}  +\frac{(b-N)^2}{y^2}+\frac{(c-N)^2}{z^2})}$
This equals zero if and only if 
$\frac{2(a-N)}{x^2} + \frac{2(b-N) }{y^2}+ \frac{2(c-N)}{z^2} = 0$ 
So we get that 
$y^2z^2a - y^2z^2N + x^2 z^2 b- x^2 z^2 N +  x^2 y^2 c-x^2 y^2 N = 0$ 
$N = \frac{y^2z^2a+ x^2z^2b + x^2y^2c}{y^2z^2 + x^2z^2 + x^2y^2}$ 
Is the maximum likelihood estimate. 
