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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.

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3  
Where do you get the estimate $1/160$ from? The probability of any single ticket winning is $\frac{40}{1600}=\frac{1}{40}$, so your first-order estimate of your chance with ten tickets should be $\frac{10}{40}=\frac{1}{4}$. –  Henning Makholm Dec 16 '11 at 12:16
    
... which is close to the real value 0.225 . –  Raskolnikov Dec 16 '11 at 12:21
    
10/1600, forgot to factor in the 40 prizes for that one! –  Clarkey Dec 16 '11 at 12:31

4 Answers 4

up vote 9 down vote accepted

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.]

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Thanks for that. Accepted your answer. Very high quality answer. Just one thing, does your last formula have a small typo? It shows (1590 40) twice. –  Clarkey Dec 16 '11 at 12:49
    
@Clarkey Yes, you're right. It seems I made one typo in that formula while correcting another. :-) If any part of the answer needs more explanation, I can expand or clarify. –  Srivatsan Dec 16 '11 at 12:52

P(win no prize) = $\frac{1600-10\choose 40}{1600\choose 40} \approx 0.776$

$Odd = \frac{1-0.776}{0.776} \approx 0.289$

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Winning no prize when buying 10 tickets means selecting 10 tickets out of the 1560 non-winning tickets. You wrote the formula for selecting 40 tickets out of 1590. –  Raskolnikov Dec 16 '11 at 12:28
    
1590 choose 40 means that the 40 prizes are chosen from the 1590 tickets that is not bought by the person. What's wrong? –  FiniteA Dec 16 '11 at 12:36
1  
Why are you dividing by .776? Shouldn't the odds of winning a prize just be 1-0.776? –  Peter Shor Dec 16 '11 at 12:45
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Odds of an event is $\frac{p}{1-p}$, see: en.wikipedia.org/wiki/Odds –  FiniteA Dec 16 '11 at 13:10
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@PeterShor In fact, FiniteA gave the correct answer to the question which asked "What are the odds that I win a prize?", not "What is the probability that I win a prize?" –  Dilip Sarwate Dec 16 '11 at 13:52

Imagine that the prize numbers are drawn and announced one at a time. If on any draw you do not win, you say "that's too bad," or something more pungent. We find the probability that you say "that's too bad" $40$ times in a row.

The probability that on the first draw, you do not win, is $\frac{1590}{1600}$. Suppose that you do not win on the first draw. Now there are only $1599$ tickets left, of which you hold $10$. So if you lost on the first draw, the probability that you lose on the second draw is $\frac{1589}{1599}$. Thus the probability that you lose on the first draw and on the second draw is $$\frac{1590}{1600}\cdot\frac{1589}{1599}.$$ Suppose that you have not won on the first two draws. Then there are $1598$ tickets left, of which you hold $10$. So if you lost on the first two draws (probability $\frac{1590}{1600}\cdot\frac{1589}{1599}$), the probability that you lose on the third draw is $\frac{1588}{1598}$. Thus the probability that you lose on the first $3$ draws is $$\frac{1590}{1600}\cdot\frac{1589}{1599}\cdot \frac{1588}{1598}.$$ Continue calculating in this way. Each time that you lose, your probability of winning the next time increases a tiny bit, though by a pathetically small amount. We find that the probability of losing $40$ times in a row is $$\frac{1590}{1600}\cdot\frac{1589}{1599}\cdot \frac{1588}{1598}\cdot \cdots \frac{1552}{1562}\cdot\frac{1551}{1561}.$$ We now have an expression for the probability that we lose $40$ times in a row.

Finally, we calculate, or have a piece of software calculate for us. The above product is approximately $0.775768$. Note that this is the probability we lose $40$ times in a row. So the probability that we win at least once is approximately $1-0.775768$, which is about $0.224232$. Pretty good, specially since we may even win more than one prize.

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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> 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();
    } 
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