Probability of picking bad egg on given day? 
I have a basket with n eggs. Each day, I eat an egg at random from the basket and then I put a fresh on. After a time T spent in the basket an egg goes bad. What is the probability I pick a bad egg a given day? What is the probability I pick bad eggs two days in a row? 

My attempts: Any egg that has been in the basket for T days goes bad, at least that is my interpretation. My initial approach was to pick a couple of values for T and n, and work out what happens, but it gets unruly pretty quickly. I'm wondering if there is a better method using maybe first step analysis, or even markov chains that I'm missing that could be used to solve this...
 A: Imagine you have a fancy egg basket that has labelled slots for the eggs. (Never mind that you could use the labelled slots to keep track of your eggs to keep them from going bad.) When you pick an egg and eat it, you put the replacement in the same labelled slot. That makes it a bit easier to think about this.
The probability that the egg you pick is at least $T$  days old is the probability that you haven’t picked the slot in which it’s sitting for $T$ days, which is $\left(1-\frac1n\right)^T$.
The probability that you pick bad eggs two days in a row is the probablity $1-\frac1n$ that you pick a different slot on the second day, times the probability $\left(1-\frac2n\right)^{T-1}$ that neither of these two slots had been picked for $T-1$ days, times the probablity $1-\frac1n$ that the one you pick on the first day hadn’t been picked the day before those $T-1$ days, for a total probability of $\left(1-\frac1n\right)^2\left(1-\frac2n\right)^{T-1}$.
A: Same result as @joriki by a different approach.  Let's write:
$$P( {\rm egg_1}>T\cap{\rm egg_2}>T)=P( {\rm egg_2}>T|{\rm egg_1}>T)P({\rm egg_1}>T)$$ so that we can recycle @joriki's first result, $P({\rm egg_1}>T)=(1-1/n)^T$. Then, we just need to find $P( {\rm egg_2}>T|{\rm egg_1}>T)$.  Hopefully the notation is clear: $P( {\rm egg}_i>T)$ is the probability that the $i$-th egg we choose is greater than $T$ days old.
Assume $T\ge 1$. Then, in order to both choose a second bad egg and satisfy the ${\rm egg_1}>T$ conditioning criteria, we have to do the following:

*

*The $T-1$ draws preceding ${\rm egg}_1$ can only be taken from the $n-1$ distinct slots not corresponding to ${\rm egg}_1$, by the conditioning criteria.  From these, the draws must also not choose the ${\rm egg}_2$ slot, so the probability for this process is $\left(\frac{n-2}{n-1}\right)^{T-1}$.

*The final ${\rm egg}_2$ draw must correspond to a slot other than the ${\rm egg}_1$ slot, for which we have probability $\frac{n-1}{n}$.

Combining, this gives $$P( {\rm egg_2}>T|{\rm egg_1}>T)=\frac{n-1}{n}\times \left(\frac{n-2}{n-1}\right)^{T-1}=(1-1/n)^{2-T}\times (1-2/n)^{T-1}$$
or, all together:
$$P( {\rm egg_1}>T\cap{\rm egg_2}>T)=(1-2/n)^{T-1}(1-1/n)^{2}$$
