Players and tickets You are among N players that will play a competition. A lottery is used to determine the placement of each player. You have an advantage. Two tickets with your name are put in a hat, while for each of the other players only one ticket with her/his name is put in the hat. The hat is well shaken and tickets are drawn one by one from the hat. The order of names appearing determines the placement of each player. What is the probability that you will get assigned the $n$th placement for $n = 1, 2, . . . , N$?
The probability that my name is drawn at $k$th attempts is $\frac{2}{N}(\frac{N-2}{N})^{k-1}$ (ie to say $k-1$ failures before the first success at $k$). I know that the solution is $\prod_{k=1}^{n-1}\frac{2}{2+N-n}\frac{N-k}{2+N-k}$.
Let $A_i$ be the event that my name compairs shows up at $i$th attempts. So:

*

*$\mathbb{P}(A_1)=\frac{2}{N+1}$;


*$\mathbb{P}(A_2)=\mathbb{P}(\bar{A_1})\mathbb{P}(A_2|\bar{A_1})=(\frac{N-1}{N+1})(\frac{2}{N})$


*$\mathbb{P}(A_3)=\mathbb{P}(\bar{A_1}\cap \bar{A_2})\mathbb{P}(A_3|\bar{A_1}\cap \bar{A_2})=(\frac{N-1}{N+1})(\frac{N-2}{N})(\frac{2}{N-1})$.
Thus I thought that
$\mathbb{P}(A_n)=(\frac{N-1}{N+1})\cdot (\frac{N-2}{N})\cdot ... \cdot (\frac{N-n-1}{N+1-n})\cdot (\frac{2}{N-n})$
but I can't lead me to the product above. Where am I wrong?
 A: I get (almost) the same result as you, under the assumption "Once a name has been drawn and placed, any future tickets with that name are ignored":
$(\color{green}{\text{green part edited}})$
$$\mathbb{P}(A_n)=\left(\frac{N-1}{N+1}\right)\cdot \left(\frac{N-2}{N}\right)\cdot ... \cdot \color{red}{\left(\frac{N-n+1}{N+3-n}\right)}\cdot \left(\frac{2}{N\color{green}{+2}-n}\right),$$
there are $n-1$ terms dealing with the first $n-1$ unsuccessful draws and the last with the successful draw. The denominators decrease by 1 in each factor as the number of tickets to draw from decreases by 1 after each draw. For the first $n-1$ terms (the unsuccessful draws) the enumerator decreases by 1 as well, so the difference between denominator and enumerator will always be $2$ in those terms (representing your 2 tickets).
That result can be written as
$$\frac{2}{N\color{green}{+2}-n}\prod_{k=1}^{n-1}\frac{N-k}{N+2-k}.$$
Your "known solution" is giving the empty product for $n=1$, so that would mean probability 1, which is obviously not correct. So please check if you copied it correctly.
A: Your solution simplifies extensively.  For some reason, \cancel doesn't work, so I've tried to indicate what I mean by showing the terms that cancel in the same color.  I can only show the first few terms this way, unfortunately.
$$\frac{\color{red}{N-1}}{N+1}\frac{\color{blue}{N-2}}{N}
\frac{\color{green}{N-3}}{\color{red}{N-1}}
\frac{\color{orange}{N-4}}{\color{blue}{N-2}}\dots
$$
Here's a better way to do it.
Let $1\leq n\leq N$ be given.
Let $A$ be the event that your name is on the $n$th ticket.
Let $B$ be the event that you name is not on any of tickets $1,2,\dots,n-1$.  We seek $\Pr(A\cap B)$.
$$\Pr(A\cap B) = \Pr(A)\Pr(B|A) = \boxed{\frac2{N+1}\frac{N+1-n}{N}}$$
The first term is obvious.  For the second, there are $N$ places where the other ticket with your name can be, and $N+1-n$ of them come after the $n$th ticket.
If you carry out the simplification described above, you will come to this simple answer.
A: Maybe I'm not understanding how this probability experiment is being conducted, but I'm getting a different answer than everyone else. Maybe someone can point  out why my reasoning is improper, if any of it is?
Let's assume once a player's name has been chosen, any future tickets with that name are ignored.
The number of ways tickets can be drawn so that your name shows up in the $n^{th}$ slot is ${2 \choose 1} \cdot {{N-1} \choose {N-1}} \cdot (N-1)!$
The number of ways tickets can be drawn in total is ${2 \choose 1}\cdot {{N-1} \choose {N-1}} \cdot N!$
Dividing these two gives the coveted likelihood of $$\frac{ 2 \cdot {{N-1} \choose {N-1}} \cdot (N-1)! }{ 2 \cdot {{N-1} \choose {N-1}} \cdot N! }=\frac{1}{N}$$
To see this we can look at a specific example. Suppose you're person #1 playing in a lottery with $N-1=2$ other people and $ \{1,1^*\} $ represents your two contributing tickets out of the $N+1=4$ tickets in the hat. We will denote the collection of all the tickets in the hat by $\{1,1^*,2,3\}$. We can list all ${2 \choose 1}\cdot {{3-1} \choose {3-1}} \cdot 3!=12$ possible orderings of the names explicitly: $$\{1,2,3\}, \{ 1,3,2\}, \{ 2,1,3\}, \{ 3,1,2\}, \{2,3,1 \},\{3,2,1 \}$$ $$\{1^*,2,3\}, \{ 1^*,3,2\}, \{ 2,1^*,3\}, \{ 3,1^*,2\}, \{2,3,1^* \},\{3,2,1^*\}$$ The probability that you're assigned to the $n^{th}$ placement is $ \frac{1}{N}=\frac{1}{3}$ for $n=1,2,3$ which makes total sense to me.
