Likelihood function of $n$ for $X\sim \text{Bin}(n,0.5)$

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I have to find the likelihood function for $X\sim \text{Bin}(n,0.5)$ where we know that $x=14$ and so $n \ge 14$. Several trials were conducted counting successes (14 of which there were) but the number of trials was forgotten, so we need to estimate $n$.

I know that the pmf of a Binomial random variable is $$P(X=x)=\binom{n}{x}\cdot p^x \cdot(1-p)^{n-x}$$ and in this case we have $$P(X=14)=\binom{n}{14}\cdot 0.5^{14} \cdot(1-0.5)^{n-14}=\binom{n}{14}\cdot0.5^{n}\,,$$ however I'm not sure this would be useful since for $n \ge 14$, the function seems to increase at an exponential rate that doesn't appear to have a maximum.

So I'm not sure using the pmf is appropriate - does anyone have any other ideas? Any help would be greatly appreciated.

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I have to find the likelihood function ...(omissis)... Several trials were conducted counting successes (14 of which there were) but the number of trials was forgotten, so we need to estimate n.

No, if you read well your problem you have only to find a way to estimate the $n$ trials with 14 known successes among them

First Idea

using the hint of the text, I would approximate

$$B\Big(n;\frac{1}{2}\Big)\approx N\Big(\frac{n}{2}; \frac{n}{4} \Big)$$

Thus I would maximize

$$f(n)\propto n^{-\frac{1}{2}}e^{-\frac{1}{2n}(2x-n)^2 }$$

Finding

$$\hat{n}=\Bigg\lceil\frac{-1+\sqrt{1+16 x^2}}{2}\Bigg\rceil_{x=14}=28$$


Second idea, Method of Moments

Using a Negative Binomial, in average, how many trials are neeeded to get 14 successes?

$$\hat{n}=\mathbb{E}[N]=\frac{14}{\frac{1}{2}}=28$$


As you can see, the result is 28 in both cases