Computing Maximum Likelihood Estimate of Probability? I am trying to understand Maximum Likelihood Estimation, and am confused by the following problem:
There are 2 slot machines. You can win \$20 with machine A with probability of $\theta$. Machine B has 4 times higher probability of winning, with just one forth of dividend. Suppose you played 10 times with either of machine A or B, and the result was as follows. What is the maximum likelihood estimation of $\theta$?

I tried to approach this problem by finding the likelihood function, which I got as:
$\prod_{i}^{10} P(A)^{n} P(1 - A)^{1 - n}$
$\theta^2 (1-\theta)^0 (4\theta)^3 (1 - 4\theta)^5$
But I don't know how to proceed from here. I also know that the answer is $\frac{1}{8}$, but don't know how to reach it correctly with the steps.
 A: Take the derivative of the function $\theta^5 (1 - 4\theta)^5$, which is $5 (1 - 4 \theta)^5 \theta^4 - 20 (1 - 4 \theta)^4 \theta^5$, and set equal to zero since at the maximum point, the likelihood function has a zero slope. We can verify that $\theta=1/8$ and $\theta=1/4$ gives you zero slope. Also, take the second derivative and verify that it is negative only at $\theta=1/4$. Plot the function to see what is happening.

A: You correctly evaluated the Likelihood Function:
$$ \mathcal{L} \left( x ; \theta \right) = {\theta}^{5} \left( 1 - 4 \theta \right)^{5} $$
Remember the method is called Maximum Likelihood.
So the step you're missing is the maximization part.  
When dealing with Independent events, namely multiplication, is is easiear to work with the Log Likelihood Function (Since $ \log \left( \cdot \right) $ is Monotonic Function):
$$ \hat{\theta} = \arg \max_{\theta} \log \mathcal{L} \left( x ; \theta \right) = \arg \max_{\theta} 5 \log \left( \theta \right) + 5 \log \left( 1 - 4 \theta \right) $$
Taking the derivative yields (Equaling to zero for Stationary Point):
$$ \frac{5}{\theta} - \frac{20}{1 - 4 \theta} = 0 \Rightarrow \hat{\theta} = \frac{1}{8} $$
The result above is after few algebra steps and by the assumption $ \theta \neq 0, \: \theta \neq \frac{1}{4} $ which is clear from the results above.
