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.

  • 3
    $\begingroup$ 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 $\frac{10}{40}=\frac{1}{4}$. $\endgroup$ – Henning Makholm Dec 16 '11 at 12:16
  • $\begingroup$ ... which is close to the real value 0.225 . $\endgroup$ – Raskolnikov Dec 16 '11 at 12:21
  • $\begingroup$ 10/1600, forgot to factor in the 40 prizes for that one! $\endgroup$ – Clarkey Dec 16 '11 at 12:31

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.]

  • $\begingroup$ 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. $\endgroup$ – Clarkey Dec 16 '11 at 12:49
  • $\begingroup$ @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. $\endgroup$ – Srivatsan Dec 16 '11 at 12:52
  • $\begingroup$ Can the same person win twice? That means, if two of your tickets get drawn, do you win twice or once? In the case that you can only win once, the whole formula is different, right? The formula you used above is for the scenario that you can win multiple times? $\endgroup$ – Van Du Tran May 11 '16 at 18:41
  • $\begingroup$ I implemented this method but ran into a division by zero when the number of tickets sold is lower than the number of prizes to win (e.g. n choose k where n is lower than k). Example: 2 prizes, but 1 ticket sold. The one ticket has 100% chance to win, but the formula ends up being 1 - ((0 c 2) / (1 n 2) or 1 - 0 / 0. Can I assume that if 0 c 2 is 0, then the win chances is 100%? Or should I look at 1 c 2 being 0? $\endgroup$ – FMCorz Sep 16 '18 at 4:34

$$P(\text{win no prize})=\frac{1600-10\choose 40}{1600\choose 40} \approx 0.776$$ $$\text{Odds}=\frac{1-0.776}{0.776}\approx0.289$$

  • $\begingroup$ 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. $\endgroup$ – Raskolnikov Dec 16 '11 at 12:28
  • $\begingroup$ 1590 choose 40 means that the 40 prizes are chosen from the 1590 tickets that is not bought by the person. What's wrong? $\endgroup$ – FiniteA Dec 16 '11 at 12:36
  • 1
    $\begingroup$ Why are you dividing by .776? Shouldn't the odds of winning a prize just be 1-0.776? $\endgroup$ – Peter Shor Dec 16 '11 at 12:45
  • 3
    $\begingroup$ Odds of an event is $\frac{p}{1-p}$, see: en.wikipedia.org/wiki/Odds $\endgroup$ – FiniteA Dec 16 '11 at 13:10
  • 3
    $\begingroup$ @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?" $\endgroup$ – 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.

  • $\begingroup$ But what if a percent can only win once? And someone hold 100 tickets? $\endgroup$ – Van Du Tran May 11 '16 at 18:45
  • $\begingroup$ I imagine that by a person can only win once you mean that any extra prizes she wins are taken away and perhaps redistributed. Then your probability of winning at least once increases. For other people may at the beginning win multiple prizes, and though you have lost $40$ times in a row, you may get extra chances during the redistribution. But your probability will not change at all if, for example, everybody else only got one ticket. $\endgroup$ – André Nicolas May 11 '16 at 18:55
  • $\begingroup$ In my case, a person can only win "once", so all their tickets are removed from the bucket if they win. That means, I someone own 1000 tickets, and that person get picked first, then on the 2nd run, your odds is 589/599. Does that makes sense? If yes, is there a formulate for calculating this? $\endgroup$ – Van Du Tran May 11 '16 at 19:11
  • $\begingroup$ If you have $40$ tickets as in the problem, your probability of winning will be increased. Your probability of not winning on the next draw is $590/600$, and one continues the calculation as in the various answers. Your problem is quite different from one asked, and is sensitive to the distribution of tickets among ticket buyers. Maybe you can formulate a precise question and ask it. If the question is clear, you will probably get answers quickly. $\endgroup$ – André Nicolas May 11 '16 at 19:22

This is all well and good, but the odds of winning a prize with $10$ tickets in a $1600$ entry raffle with $40$ prizes is $25\%$. One out of every $40$ tickets will be a winner, this is $2,5\%$. Every extra ticket purchased will increase your odds $2,5\%$. $10$ tickets at $2,5\%$ is $25\%$. It is that simple.

  • 2
    $\begingroup$ Wrong. By this logic, if you bought 100 tickets, you would get 250% chance of winning? $\endgroup$ – Van Du Tran May 16 '16 at 17:56

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;
                 odds = odds*(numerator/ticketsRemaining);
        odds = 1 - odds;


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.