# Intuitive way to arrive at the maximizing argument for the binomial probability

The binomial probability term $q^{n}(1-q)^{N-n}$ is maximized when $q=n/N$. This can be easily arrived at by differentiating the given probability term with respect to q. Is there a more intuitive way to arrive at this value of q that maximizes the probability ?

Yes; $N\choose n$$q^{n}(1-q)^{N-n}$ is the probability of obtaining $n$ successes in $N$ independent trials given a probability of success for each trial of $q$. To maximize the probability of obtaining $n$ successes, choose $q$ such that the expected number of successes in $N$ trials is $n$, i.e., $q = n/N$.

• Thank you. When N=2 and n=1,the product is $q(1-q)$.Now, even without looking into the probabilities, we know that(which can be proven in a number of ways) this quadratic function is maximized when both q and 1-q are equal. I was wondering if there is something along that line which would help generalize q=n/N without resorting to differentiation or necessarily invoking the probability. I was looking at it like this. Given two numbers n and N such that N>n, then the pair (a,b) that maximizes $a^{n}b^{N-n}$ will be (n/N,1-n/N).I was wondering if there is something intuitive behind this. Jan 13, 2014 at 6:49
• With regard to the above comment, the maximization of the function is done under the constraint a+b =1 . Jan 13, 2014 at 6:57
• Can't think of anything off the top of my head. The constraint $a + b = 1$ really suggests a probabilistic setup, but there may be some angle along other lines. Jan 13, 2014 at 20:42

This is how I usually explain the situation:

I repeat some experiment (make up one that makes the most sense) 1000 (or any large number) and find that I get a particular result 50 times. What do you think the probability of the result is?

Almost everyone immediately would say 50/1000.

I then use your result to actually show that this makes the most sense.

I carefully avoid terms like a-posterori or maximum likelihood