Probability of winning a prize in a raffle My work is having it's annual Christmas raffle today. 1600 tickets have been sold, and there are 40 prizes to win. I have bought ten tickets. What are the odds I will win a prize?

While an initial estimate of 1/160 is probably within a close enough range to suggest I have little chance of winning, I am curious as to what the precise odds would be. Tickets are not put back in once they have been drawn.
You'll be surprised. The correct probability of winning at least one ticket is around $0.2242$. 

Assuming exactly one prize is given, your answer of $\frac{1}{160}$ is the probability of winning is correct. That is, you go home empty-handed with probability $\frac{159}{160}$. However, $40$ tickets are chosen for prizes, not just one. So even if you miss out on a prize the first time, you could still end up with the second winning ticket; or the third; or the $40^{th}$. What we need to calculate is the chance of winning at least one of those tickets. 



For the moment, assume that the prizes are drawn with replacement. Then in order for you to not get a prize, you need to miss the first time, and the second time, and the third time, and so on, until the $40^{th}$ time. Under our assumption that these are drawn with replacement, all these $40$ events are independent. Therefore, the probability that you miss out on a prize is simply the probability that miss out in any given trial, raised to the power of $40$; i.e., 
$$
\left(\frac{159}{160} \right)^{40} \approx 0.7782.
$$
Hence, the chance that you win a prize is $1 - 0.7782 \approx 0.2218$.



When the prizes are drawn without replacement. Now we are going to compute the exact answer without any assumptions. There are $1600$ tickets, out of which you bought the first ten (say). The judges pick $40$ winners out of the $1600$ tickets; this can be done in $\binom{1600}{40}$ ways. [See binomial coefficients in Wikipedia.] Of these, you will not win a prize if those $40$ tickets are drawn from the $1590$ tickets that you did not buy. That is, there are $\binom{1590}{40}$ possible outcomes in which you will go home empty-handed. That is, you go home empty-handed with probability 
$$
\frac{\binom{1590}{40}}{\binom{1600}{40}}.
$$
Therefore, you will win a prize with the complementary probability
$$
1 - \frac{\binom{1590}{40}}{\binom{1600}{40}} = \frac{1420730930795547} {6335978517846620} \approx 0.2242.
$$
As you can see, that the approximate answer is quite close to the exact one.

[I did these calculations in Wolfram Alpha.]
I was just in a company Christmas raffle and was wondering my odds with the single ticket.

this allows you to change the number of tickets you have, # of prizes and # of remaining tickets after each draw. What this does not cover is the "you must be present to win clause" This assumes all drawn tickets are winners. 

static void Main(string[] args)
        {
            int myTickets = 0;
            int prizes = 0;
            int ticketsRemaining = 1; 

        int.TryParse(args[0], out myTickets);
        int.TryParse(args[1], out prizes);
        int.TryParse(args[2], out ticketsRemaining);

        double numerator = ticketsRemaining - myTickets;

        double odds = 0.000000000000000000000000000000;
        bool first = true;
         for(int i = prizes; i&gt; 0; i--)
         {
             if (first)
             {
                 odds = numerator/ticketsRemaining;
                 first = false;
             }
             else
             {
                 odds = odds*(numerator/ticketsRemaining);
             }
             numerator--;
             ticketsRemaining--;
         }
        odds = 1 - odds;
        Console.WriteLine(odds.ToString(".0000000000000000000000000"));

        Console.Read();
    }