Yes this is easy to calculate

The original dice pool (ODP) of n dice, with each having a probability p of success is a binomial distribution. The probability of exactly k successes is:

Add a die and remove the highest

Adding a dice and removing the highest gives us the following cases to deal with:

1. The ODP has 0 successes. The additional die will have no effect: if it is a success it must be the highest and is therefore removed, if it is a failure then all of the dice are failures (including the highest one that gets removed).
2. The ODP has k(>0) successes. If the additional die is a failure, one success will be removed, if it is a success, it will add one success and one success (from the highest die) will be removed for no net effect.

Ignoring the unusual case of 0 successes in the ODP, the die has a 1-p chance of reducing successes by 1.

Add a die and remove the lowest

Adding a dice and removing the lowest is the reverse of the previous:

1. The ODP has n successes. The additional die will have no effect: if it is a failure it must be the lowest and is therefore removed, if it is a success then all of the dice are successes (including the lowest that gets removed).
2. The ODP has k (<n) successes. If the additional die is a success, one failure will be removed, if it is a failure, it will add one failure and one failure (from the lowest die) will be removed for no net effect.

Ignoring the unusual case of n successes in the ODP, the die has a p chance of increasing successes by 1.

Yes this is easy to calculate

The original dice pool (ODP) of n dice, with each having a probability p of success is a binomial distribution. The probability of exactly k successes is:

$$ f(k;n,p) = Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k} $$

Add a die and remove the highest

Adding a dice and removing the highest gives us the following cases to deal with:

1. The ODP has 0 successes. The additional die will have no effect: if it is a success it must be the highest and is therefore removed, if it is a failure then all of the dice are failures (including the highest one that gets removed).
2. The ODP has k(>0) successes. If the additional die is a failure, one success will be removed, if it is a success, it will add one success and one success (from the highest die) will be removed for no net effect.

Ignoring the unusual case of 0 successes in the ODP, the die has a 1-p chance of reducing successes by 1.

Add a die and remove the lowest

Adding a dice and removing the lowest is the reverse of the previous:

1. The ODP has n successes. The additional die will have no effect: if it is a failure it must be the lowest and is therefore removed, if it is a success then all of the dice are successes (including the lowest that gets removed).
2. The ODP has k (<n) successes. If the additional die is a success, one failure will be removed, if it is a failure, it will add one failure and one failure (from the lowest die) will be removed for no net effect.

Ignoring the unusual case of n successes in the ODP, the die has a p chance of increasing successes by 1.

Yes this is easy to calculate

The original dice pool (ODP) of n dice, with each having a probability p of success is a binomial distribution. The probability of exactly k successes is:

Add a die and remove the highest

Adding a dice and removing the highest gives us the following cases to deal with:

1. The ODP has 0 successes. The additional die will have no effect: if it is a success it must be the highest and is therefore removed, if it is a failure then all of the dice are failures (including the highest one that gets removed).
2. The ODP has k(>0) successes. If the additional die is a failure, one success will be removed, if it is a success, it will add one success and one success (from the highest die) will be removed for no net effect.

Ignoring the unusual case of 0 successes in the ODP, the die has a 1-p chance of reducing successes by 1.

Add a die and remove the lowest

Adding a dice and removing the lowest is the reverse of the previous:

1. The ODP has n successes. The additional die will have no effect: if it is a failure it must be the lowest and is therefore removed, if it is a success then all of the dice are successes (including the lowest that gets removed).
2. The ODP has k (<n) successes. If the additional die is a success, one failure will be removed, if it is a failure, it will add one failure and one failure (from the lowest die) will be removed for no net effect.

Ignoring the unusual case of n successes in the ODP, the die has a p chance of increasing successes by 1.

Yes this is easy to calculate

The original dice pool (ODP) of n dice, with each having a probability p of success is a binomial distribution. The probability of exactly k successes is:

Add a die and remove the highest

Adding a dice and removing the highest gives us the following cases to deal with:

1. The ODP has 0 successes. The additional die will have no effect: if it is a success it must be the highest and is therefore removed, if it is a failure then all of the dice are failures (including the highest one that gets removed).
2. The ODP has k(>0) successes. If the additional die is a failure, one success will be removed, if it is a success, it will add one success and one success (from the highest die) will be removed for no net effect.

Ignoring the unusual case of 0 successes in the ODP, the die has a 1-p chance of reducing successes by 1.

Add a die and remove the lowest

Adding a dice and removing the lowest is the reverse of the previous:

1. The ODP has n successes. The additional die will have no effect: if it is a failure it must be the lowest and is therefore removed, if it is a success then all of the dice are successes (including the lowest that gets removed).
2. The ODP has k (<n) successes. If the additional die is a success, one failure will be removed, if it is a failure, it will add one failure and one failure (from the lowest die) will be removed for no net effect.

Ignoring the unusual case of n successes in the ODP, the die has a p chance of increasing successes by 1.