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I am wondering if its possible to model or fit a discrete distribution where there are only three possible outcomes: 0, 1, and 3. These outcomes are numbers which result from a game. I get hung up on the fact that 2 is not an allowed outcome and don't really see how to do it. A sample table is below.

I will also mention that the intention is to determine the average and error on that average, so if there is an alternate method I am oblivious to, please let me know.

 Obs | Counts
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 0   |  35
 1   |  22
 3   |  25

Any help is appreciated.

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    $\begingroup$ what are 0, 1, 3, lables or numbers? In other word where do they come from? $\endgroup$ – utobi Dec 2 '16 at 7:58
  • $\begingroup$ They are numbers which result from a game. I edited my question to add this info. $\endgroup$ – J. Paul Dec 2 '16 at 8:08
  • $\begingroup$ I still don't get it. What kind of number are they? Counts or what? If they are labels then Glen_b's answer should be satisfying. And, is 2 an impossible event or just it never appears by chance? $\endgroup$ – utobi Dec 2 '16 at 11:10
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Wouldn't you just specify a multinomial distribution?

That is, you can just have a $p_0$, $p_1$ and $p_3$ such that $p_0+p_1+p_3=1$ and $p_i>0,i=0,1,3$, and a number of observations $n$, so that you can accommodate any set of probabilities to the three outcomes.

Let the counts for each category be $\mathbf{X} = (X_0,X_1,X_3)^\prime\,,$ where (where $X_0+X_1+X_3=n$)

You observe $(X_0,X_1,X_3)^\prime = (x_0,x_1,x_3)^\prime$ with probability

$\qquad\frac{n!}{x_0! x_1! x_3!} p_0^{x_0} p_1^{x_1} p_3^{x_3} \:$ (with $x_0+x_1+x_3=n$).

The MLEs of the $p$ parameters are just the sample proportions.

You can compute an average just fine, since the category labels are integers (0,1,3), so the expectation is just $n(p_1+3p_3)$. The estimate of the mean is the sample mean $\bar{x}=(x_1+3x_3)/n$ and the variance of the sample mean is $\frac{1}{n^2}[\text{Var}(X_1)+9\text{Var}(X_3)+6\text{Cov}(X_1,X_3)]$.

Note that $\text{Var}(X_i) = np_i(1-p_i)$ and $\text{Cov}(X_i,X_j)=-np_ip_j$ for $i\neq j$.

So we can obtain the standard error of the mean fairly directly.

Or did you want something less general?

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