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Chef George and Chef Steve are competing in a cooking competition. There are a dozen eggs in a carton, in which four of the eggs are rotten. The chefs will take turns cracking eggs until a rotten egg is discovered. George will go first. What is the probability that he gets the rotten egg?

Yes, this is homework, but I'm not asking you to do it for me, I just need a push in the right direction. I know on the first crack there's a 4/12 possibility he gets it, and that on his second one there is a 4/10 probability, and so on. What I don't know is how to use this information.

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    Let $p_n$ denote the probability that the first bad egg is in slot $n$. You've computed $p_1$. What is $p_3$? $p_5$? and so on. – 2017-01-19

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You are correct that the chance that George gets a rotten egg on the first try is $\frac 4{12}$. The chance he gets on on his second try is $\frac 4{10}$ assuming he gets a second try. You should compute the chance that he gets a second try at all and multiply by $\frac 4{10}$ to get the chance he finds a rotten one on the second try. Now continue for his other tries, add up the chances, and that is the chance he finds the first rotten egg. If you recognize the geometric series it will ease the computation.

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    Ok. I guess I'm having trouble trying to compute the chance that he gets an nth try. I don't really know where to start. Would that be the chance that no one has found the egg yet? My first instinct to compute that is to add up the probabilities that no one found an egg on their first tries to compute whether or not he gets a second try. Basically, would the chance that he gets a second try be 8/12 + 8/11 (The chance that neither of their first tries yield rotten eggs)? – 2017-01-19
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    For George to get a second try, you need neither chef to find a rotten egg on their first try. The sum you suggest exceeds $1$ so cannot be a probability. Neither sounds like and.... – 2017-01-19
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    Maybe 8/12 * 7/11? I have to ask because regardless of what I do I won't have any way to know if it's right, so I have to be sure before I proceed – 2017-01-19
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    Yes, that is right. – 2017-01-19