Normally, lottery tickets are valid for only one drawing. If one wants to play the same numbers for multiple draws, the cost of the ticket is multiplied accordingly. What if one could hold on to it for multiple drawings, without having to pay extra, and choose a particular drawing to cash out? For example, suppose you did not win for four drawings, then got the numbers right for a \$4 prize. However, you chose to hold on to the ticket. Nine drawings later, your numbers are correct to win a \$7 prize, so you chose to cash out. You could have kept the ticket even longer and tried to get the \$100 prize, etc. If you choose not to redeem a ticket immediately after a winning drawing, you lose the opportunity to cash that prize. For example, you were eligible for the \$4 prize, but chose to wait. After losing the next few draws, you would no longer be able to claim the \$4 prize unless your numbers were correct again.
Suppose the lottery had prizes for each drawing with probability $\ p_0, p_1, ... p_n$ and corresponding amounts of $\ q_0, q_1, ... q_n$. Let the zero-indexed prize $q_0 = 0\ $ (not winning) and let the sum of the probabilities $p$ be 1. Then the expected value of one draw is
$$ E = \sum_{i=1}^{n} p_i \cdot q_i$$
Now my knowledge of statistics falls short. If a ticket is valid for $k$ draws, is the expected value of the ticket just $k \cdot E$ ? (Then the value of a ticket increases indefinitely as k increases) If a ticket has $l$ draws remaining and the ticket is eligible to redeem a prize $P$ based on the previous draw, should one redeem it only if $P > l \cdot E$ ? It would also be interesting to see how a jackpot prize, which varies each drawing depending on the prize fund (which itself depends on the outcomes of other players) changes the value.
Suppose the $q_i$ are sorted in increasing order. We'll work backward.
If you have no tries remaining, you should obviously take whatever prize is on offer right now. So the expected value of your ticket is precisely $E$.
If you have $1$ try remaining, you will have the option of taking the prize on offer now or waiting until tomorrow for an expected value of $E$, so you should cash in if you're offered some prize $j$ with $q_j>E$. That means your expected value is $$ E_1=\sum_{i=k_1}^n p_iq_i + E\sum_{i=0}^{k_i-1} p_i $$ where $q_{k_1}$ is the smallest prize with $q_{k_1}>E$. (That is, the expected value of cashing in now, in the cases where you should, plus the expected value of waiting in all other cases.)
So if you have $2$ tries remaining, you should cash in if you're offered a prize $j$ with $q_j>E_1$, which means your expected value is $$ E_2=\sum_{i=k_2}^n p_iq_i + E_1\sum_{i=0}^{k_2-1} p_i $$ and so on.
Obviously this becomes seriously annoying to compute by hand in all but the simplest of cases, but it does give an explicit formula.
As one example of this, let's say there's a \$$1000$ payout that happens with probability $\frac{1}{1000}$ and a \$$1000000$ payout that happens with probability $\frac{1}{1000000}$.
If you have no tries remaining, the expected value of your ticket is $$ E=1000(\frac{1}{1000})+1000000(\frac{1}{1000000})=2 $$
If you have $1$ try remaining, you should take either prize if offered (since $1000>2$). So the expected value is $$ E_1=1000(\frac{1}{1000})+1000000(\frac{1}{1000000}) + 2\left(1-\frac{1}{1000}-\frac{1}{1000000}\right)=4-\epsilon $$ where $\epsilon$ is some smallish number.
So you should still take either prize if offered with $2$ tries remaining, at which point your expected value is
$$ E_2=1000(\frac{1}{1000})+1000000(\frac{1}{1000000}) + E_1\left(1-\frac{1}{1000}-\frac{1}{1000000}\right)\approx 6-2\epsilon $$ and so on.
Eventually, when you have somewhere around $500$ tries remaining, it'll start being correct to forego the \$$1000$-dollar prize. That is, for $k\gg500$, we have $$ E_k=1000000(\frac{1}{1000000})+E_{k-1}\left(1-\frac{1}{1000000}\right)=1+E_{k-1}\left(1-\frac{1}{1000000}\right) $$ As a sanity check, notice that if $E_{k-1}<1000000$ then also $$ E_k<1+(1000000)\left(1-\frac{1}{1000000}\right)=1000000 $$ That is, your expected winnings can never be larger than \$$1000000$ no matter how long you wait — which certainly ought to be true, since you can only claim one prize and the largest one is worth \$$1000000$.