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How to determine the probability that a mutation is lost / fixed?

The probability that a neutral mutation gets fixed after an infinite amount of time is equal to its frequency $p$ as you said. Therefore the probability of being lost is $1-p$. This post offers an explanation but there many ways to make the demonstration. You might want to have a look at any good book in population genetics for this demo. Here are book recommendations.

how do you calculate the probability that it exists in 2 copies?

A probability always depends on a priori. What are your a priori? Let's assume that we know that the allele frequency was $\frac{4}{10}$ in the previous time step.

Under the Wright-Fisher model the probability of having 2 copies in the next generation is given by the binomial distribution. Let $N=5$ and therefore $2N=10$ and let the frequency of the allele of interest be $\frac{4}{10}$, the probability of having two alleles in the next generation is ${10 \choose 2} \left(\frac{4}{10}\right)^2 \left(\frac{6}{10}\right)^8 ≈ 0.12$.

Under the Moran model this probability is zero. Moran's model is a birth-death model (Markov model) and therefore transition between time steps can only add or subtract (or make no change) a single allele. You will note that the time step does not mean the same thing for the two models. Loss of heterozygosity is twice as fast under the Wright-Fisher model but this discussion is definitely not what you were asking for!

How to determine the probability that a mutation is lost / fixed?

The probability that a neutral mutation gets fixed after an infinite amount of time is equal to its frequency $p$ as you said. Therefore the probability of being lost is $1-p$. This post offers an explanation but there many ways to make the demonstration. You might want to have a look at any good book in population genetics for this demo. Here are book recommendations.

how do you calculate the probability that it exists in 2 copies?

A probability always depends on a priori. What are your a priori? Let's assume that we know that the allele frequency was $\frac{4}{10}$ in the previous time step.

Under the Wright-Fisher model the probability of having 2 copies in the next generation is given by the binomial distribution. Let $N=5$ and therefore $2N=10$ and let the frequency of the allele of interest be $\frac{4}{10}$, the probability of having two alleles in the next generation is ${10 \choose 2} \left(\frac{4}{10}\right)^2 \left(\frac{6}{10}\right)^8 ≈ 0.12$.

Under the Moran model this probability is zero. Moran's model is a birth-death model (Markov model) and therefore transition between time steps can only add or subtract (or make no change) a single allele. You will note that the time step does not mean the same thing for the two models. Loss of heterozygosity is twice as fast under the Wright-Fisher model but this discussion is definitely not what you were asking for!

How to determine the probability that a mutation is lost / fixed?

The probability that a neutral mutation gets fixed after an infinite amount of time is equal to its frequency $p$ as you said. Therefore the probability of being lost is $1-p$. This post offers an explanation but there many ways to make the demonstration. You might want to have a look at any good book in population genetics for this demo. Here are book recommendations.

how do you calculate the probability that it exists in 2 copies?

A probability always depends on a priori. What are your a priori? Let's assume that we know that the allele frequency was $\frac{4}{10}$ in the previous time step.

Under the Wright-Fisher model the probability of having 2 copies in the next generation is given by the binomial distribution. Let $N=5$ and therefore $2N=10$ and let the freuquency of the allele of interest be $\frac{4}{10}$, the probability of having two alleles in the next generation is ${10 \choose 2} \left(\frac{4}{10}\right)^2 \left(\frac{6}{10}\right)^8 ≈ 0.12$.

Under the Moran model this probability is zero. Moran's model is a birth-death model (Markov model) and therefore transition between time steps can only add or subtract (or make no change) a single allele. You will note that the time step does not mean the same thing for the two models. Loss of heterozygosity is twice as fast under the Wright-Fisher model but this discussion is definitely not what you were asking for!

How to determine the probability that a mutation is lost / fixed?

The probability that a neutral mutation gets fixed after an infinite amount of time is equal to its frequency $p$ as you said. Therefore the probability of being lost is $1-p$. This post offers an explanation but there many ways to make the demonstration. You might want to have a look at any good book in population genetics for this demo. Here are book recommendations.

how do you calculate the probability that it exists in 2 copies?

A probability always depends on a priori. What are your a priori? Let's assume that we know that the allele frequency was $\frac{4}{10}$ in the previous time step.

Under the Wright-Fisher model the probability of having 2 copies in the next generation is given by the binomial distribution. Let $N=5$ and therefore $2N=10$ and let the frequency of the allele of interest be $\frac{4}{10}$, the probability of having two alleles in the next generation is ${10 \choose 2} \left(\frac{4}{10}\right)^2 \left(\frac{6}{10}\right)^8 ≈ 0.12$.

Under the Moran model this probability is zero. Moran's model is a birth-death model (Markov model) and therefore transition between time steps can only add or subtract (or make no change) a single allele. You will note that the time step does not mean the same thing for the two models. Loss of heterozygosity is twice as fast under the Wright-Fisher model but this discussion is definitely not what you were asking for!

How to determine the probability that a mutation is lost / fixed?

The probability that a neutral mutation gets fixed after an infinite amount of time is equal to its frequency $p$ as you said. Therefore the probability of being lost is $1-p$. This post offers an explanation but there many ways to make the demonstration. You might want to have a look at any good book in population genetics for this demo. Here are book recommendations.

how do you calculate the probability that it exists in 2 copies?

A probability always depends on a priori. What are your a priori? Let's assume that we know that the allele frequency was $\frac{4}{5}$ in the previous time step.

Under the Wright-Fisher model the probability of having two copies in the next generation is given by the binomial distribution, that is ${5 \choose 2} \left(\frac{4}{5}\right)^2 \left(\frac{1}{5}\right)^3 ≈ 0.05$.

Under the Moran model this probability is zero. Moran's model is a birth-death model (Markov model) and therefore transition between time steps can only add or subtract (or make no change) a single allele. You will note that the time step does not mean the same thing for the two models. Loss of heterozygosity is twice as fast under the Wright-Fisher model but this discussion is definitely not what you were asking for!

How to determine the probability that a mutation is lost / fixed?

The probability that a neutral mutation gets fixed after an infinite amount of time is equal to its frequency $p$ as you said. Therefore the probability of being lost is $1-p$. This post offers an explanation but there many ways to make the demonstration. You might want to have a look at any good book in population genetics for this demo. Here are book recommendations.

how do you calculate the probability that it exists in 2 copies?

A probability always depends on a priori. What are your a priori? Let's assume that we know that the allele frequency was $\frac{4}{10}$ in the previous time step.

Under the Wright-Fisher model the probability of having 2 copies in the next generation is given by the binomial distribution. Let $N=5$ and therefore $2N=10$ and let the freuquency of the allele of interest be $\frac{4}{10}$, the probability of having two alleles in the next generation is ${10 \choose 2} \left(\frac{4}{10}\right)^2 \left(\frac{6}{10}\right)^8 ≈ 0.12$.

Under the Moran model this probability is zero. Moran's model is a birth-death model (Markov model) and therefore transition between time steps can only add or subtract (or make no change) a single allele. You will note that the time step does not mean the same thing for the two models. Loss of heterozygosity is twice as fast under the Wright-Fisher model but this discussion is definitely not what you were asking for!