Lottery with acceptance chance Problem description:
Suppose I am entering into a lottery. I know that on average there are $n = 500$ people playing. In order to determine the winner, a ranking of all players is created uniformly at random. The player at the top of the ranking gets offered the prize first. However, sometimes players do not show up to accept their prize. There is a p = 0.2 chance that a player does not show up. In this case, the prize is offered to the next player in the ranking and so forth. I would like to know the amount of times $k$ I have to play the lottery to have a chance of $q = 0.5$ to win the lottery (I myself will of course always show up to collect my prize).
I came up with the following equation:
$$\left(1 - \sum_{m=0}^{n-1} \frac{1}{n - m} p^m\right)^k = q$$
For the numbers above, this results in $k = 276.773$, which seems reasonable. However, I would expect that as the probability $p$ that people do not collect their prize converges to 1, the amount of times I need to play also converges to 1. However, this does not appear to be the case. Did I make a mistake in my formula?
 A: Let $X$ denote your ranking by participation of this lottery and let $W$ denote the event that you win.
Then:$$P\left(W\right)=\sum_{m=0}^{n-1}P\left(\left\{ X=m+1\right\} \cap W\right)=\sum_{m=0}^{n-1}P\left(X=m+1\right)P\left(W\mid X=m+1\right)=$$$$\sum_{m=0}^{n-1}\frac{1}{n}p^{m}=\frac{1}{n}\frac{1-p^{n}}{1-p}$$
So to be found is the smallest integer $k$ that satisfies:
$$\left(1-\frac{1}{n}\frac{1-p^{n}}{1-p}\right)^{k}\leq q$$
If $p\uparrow1$ then $\left(1-\frac{1}{n}\frac{1-p^{n}}{1-p}\right)\downarrow0$ so that for large $p$ we get $k=1$.
A: Every individual time you play, you have the same chance of winning. Let us denote that chance as $r$. Your chance of winning (at least once) after $k$ times is equal to $1$ - (your chance of losing $k$ times in a row). Your chance of losing once is $1-r$. Your chance of losing $k$ times is $(1-r)^{k}$. Your chance of not losing $k$ times in a row is $1- (1-r)^{k}$ . We must solve:
$$1- (1-r)^{k} \geq q$$
Now we must compute $r$, that is, your chance of winning after playing once. Your chance of taking the $k'th$ place out of $500$ people is $\frac{1}{500}$, for any given $k\in\{1..500\}$ we select. The chance  of the first $k-1$ people not showing up is $p^{k-1}$. So your chance of winning by taking the $k'th$ place is $\frac{p^{k-1}}{500}$. Your total chance of winning (after one try) is then:
$$r  = \sum_{k=1}^{n}\frac{p^{k-1}}{n} = \frac{1-p^{n}}{(1-p)n}$$ 
We must solve $ 1 - (1-\frac{1-p^{n}}{(1-p)n})^k \geq q$, for $q = 0.5$, $n = 500$, $p = 0.2$. This gives $k≈276.912$, which by rounding is $277$, which is close enough. Now, as $p$ (the probability that people don't show up) goes to $1$, so does your chance of winning.
