Probability to be the millionth customer (What Would You Do?) I saw this episode
of "What Would You Do?" a few months ago, and I keep wondering what would statistically be the best thing to do in this situation.
Here is the problem formulation:
You are waiting in line at a supermarket, where there is a sign saying that the millionth customer will win an amazing prize. Just before your turn, someone in a hurry asks you if he can cut in line in front of you.
Question: If you let this person cut in front of you, does he have a higher probability of winning the prize than you?
This leads to a more general question which is for me a paradox:  
If you are in line and want to be the millionth customer, should you try to go as soon as possible, or should you wait?   On one hand the more you wait (the more you let people cut in front of you), the closer you get to the millionth. On the other hand the more you wait, the more you risk of someone being the millionth before you.
We can formalize it as follows:
You are customer number $n$. Of course you don't know your number $n$, but you know that you will win if $n=N$ with $N=1000000$. The only thing you know is that $n<=N$ (no one as won the prize yet, but you might). Being the $nth$ customer, your probability to win is then: $Prob(n=N|n<=N)$. But if you let someone cut in front of you, you become the $(n+1)th$ customer and your probability to win becomes $Prob(n+1=N|n<=N)$. 
Now here is my question:
Is there a way to compare $Prob(n=N|n<=N)$ and $Prob(n+1=N|n<=N)$?
 A: Let's say you know that the millionth-customer notice is raised when 999000 customers have passed through.  By whatever means, you form your own subjective probability of you being customer number 999001 (call this $p_{999}$) up to 1000000 (call this $p_0$).  Let $k$ be such that $p_k = \max(p_0,\ldots,p_{999})$.  Then you should let $k$ people get in ahead of you.
In normal circumstances, you have nothing to choose between the possibilities and all your $p$ are equal.  Assume you're at the head of the queue and $p=1/1000$.  If you gave up $j$ positions then you'd lose to one of them with probability $j/1000$ but, if you didn't lose to a queue-jumper, you'd then be in with a chance $1/(1000-j)$ of winning: net, that's a win probability of $(1-j/1000)\times(1/(1000-j)) = 1/1000$ (irrespective of $j$).  In other words, the risk of losing to a queue-jumper and the gain from delaying will, in normal circumstances, exactly cancel.
The practical answer, however, is to ignore the mathematics and get to the checkout as soon as possible, buy any old thing, and return immediately to the queue for another shot at the prize!
