If I have two drawings without replacement but add a new name to the second drawing, how do I ensure everyone has equal odds? I have seven names and two raffles. One person is unable to make the first drawing how do I ensure they get equal odds without lowering everyone else?
For instance of on the first drawing the following names are put in along with their probability. Jack has stars since he won the drawing.
*Jack = 1/6*
Craig =1/6
Dillon = 1/6
Billy = 1/6
Gordon = 1/6
Kate= 1/6

And Zoe is left out the first round but we want her to have an equal probability. Since Jack won he is pulled out and obviously his probability will stay at 1/6.
If we put Zoe's name in twice she still has lower odds
Zoe = 1/7
Zoe = 1/7
Craig= 1/7
Dillon = 1/7
Billy = 1/7
Gordon = 1/7
Kate= 1/7

The people who were drawn twice probability
Gordon = 1/6 + 1/7 => 
6/42 + 7/42= 13/42= 0.31

Zoe and Jack
Zoe = 1/7 +1/7 = 2/7 = .29
Jack = 1/6 = .17


Craig = .31
Dillon = .31
Billy = .31
Gordon = .31
Kate= .31
Zoe = .29
Jack = .17

So adding Zoe's name twice won't do the trick. The solution will be to add everyones name multiple times with Zoe's name being the most in the second raffle. How do I figure out the total number of names and how many raffles Zoe gets for the second raffle.
 A: Your approach is reasonable: to have Zoe to have the same chances of possible outcomes as the others, you need to exclude the winner of the first draw from the second (so nobody can win both draws), and you need to give Zoe a higher probability in the second draw to make up for not being in the first. But one of your calculations does not take that into account.
You might start by finding the probability Zoe needs to have of winning in the second draw.  Since there are $7$ people and $2$ prizes, this will have to be $\frac27$.
The easiest way to do this is your suggestion of giving Zoe $2$ tickets in the second draw, which would then have $7$ tickets since you exclude the winner of the first draw. So her probability of winning is indeed $\frac27$.
Everybody else's individual probability of a win is (since they might win the first draw, or they might lose the first and win the second) $\frac16+ \frac56 \times \frac17 =\frac27$, which is the same as Zoe's.
A: I am going to assume that you want a process where no-one can win more than once and everyone has the same probability of winning once.  You do this by removing the winner in the first round from the second draw and adding the new person.  Suppose we have $n>1$ people initially and we set the win-probability for the new person (in the second draw) to $\theta$.  For a person included in the initial round, the probability of winning is:
$$p = \frac{1}{n} + \frac{n-1}{n} \cdot \frac{1-\theta}{n-1} = \frac{1}{n} + \frac{1-\theta}{n} = \frac{2-\theta}{n}.$$
You want to set these to be equal, so your constraint is $\theta = p$.  Solving this equation yields:
$$\theta = \frac{2}{n+1}.$$
In the present instance you have $n=6$ people in people in your initial round so you would set the probability for Zoe winning the second round to $\theta = 2/7$.
