Suppose I play a lottery that has 300 tickets. I can only buy one ticket per draw. Statistically speaking, shouldn't I win once every 300 draws?

Is it more complicated than this?


This question has generated much more response than I had imagined. Thank you all, for your input and your great explanations.

To go into further details: The lottery has 300 tickets. You can buy 1 ticket per draw. Every draw, is all new tickets, so in essence, you can hold 1 out of 300 numbers, at each draw.


Just to give some funny (not really funny) side info. I have now played 1,104 times, and still have not won anything. I suppose I am EXTREMELY unlucky.

  • 10
    $\begingroup$ You would expect to win once every 300 draws. Of course there's no guarantee that you will, though. $\endgroup$
    – Cocopuffs
    Jul 30, 2013 at 6:13
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    $\begingroup$ Simplify: if a lottery has 2 tickets, will you win once every couple of draws? $\endgroup$ Jul 30, 2013 at 7:02
  • 6
    $\begingroup$ @Nicolai: if a dice has 6 sides will it land on 1 every six rolls? $\endgroup$
    – orlp
    Jul 30, 2013 at 10:58
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    $\begingroup$ If you buy 300 tickets on 1 draw, then you will win exactly once :) $\endgroup$ Jul 30, 2013 at 12:08
  • 4
    $\begingroup$ You've implied that every lottery has exactly one winner out of the 300 tickets. To me, that sounds more like a raffle whereas a lottery has no guaranteed winner in a given draw. Scratch-off (aka instant-win) lottery tickets are a different class of problem. You might want to clarify the terms of the drawing. $\endgroup$
    – Patrick M
    Jul 30, 2013 at 21:34

9 Answers 9


Yes it is. The probability of winning $k$ times out of $n$ lotteries is determined from a binomial distribution:

$$P(K=k) = \binom{n}{k} p^k (1-p)^{n-k}$$

where $n=300$, $k=1$, and $p=1/300$. The answer is about $0.368494$, or $36.8\%$. The probability of winning at least once is

$$P(K \ge 1) = 1- P(K=0) = 1-\left( \frac{299}{300} \right )^{300} \approx 0.632735$$

or about $63.3\%$ chance of winning something. Not bad, but not $100\%$.

  • 1
    $\begingroup$ I am unsure if I understand the difference. It's 36.8% chance of winning any given draw? And 63.2% chance that after 300 draws, I will have won once? $\endgroup$
    – Nicolai
    Jul 30, 2013 at 6:14
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    $\begingroup$ "exactly once" means winning just once, not twice or three times or...out of 300 tries. "At least once" covers all those cases. The reasoning there turns out to be equivalent to that in nbubis's answer. $\endgroup$
    – Ron Gordon
    Jul 30, 2013 at 6:16
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    $\begingroup$ @Nicolai When $n$ goes to infinity, the $0.368$ goes to $1/e$, or your chances of winning at least once goes to $1 - (1/e) \simeq 0.63212 = 63.212$%. $\endgroup$
    – Pål GD
    Jul 30, 2013 at 8:40
  • $\begingroup$ @Nicolai: See also Why is the Expected Value different from the number needed for 50% chance of success? $\endgroup$ Jul 30, 2013 at 15:34
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    $\begingroup$ So ... using the same reasoning, if you play 1104 times, the probability of never winning is about 0.025 ... so you are a bit unlucky, but not amazingly so. About 1 person in 40 will be that unlucky. $\endgroup$
    – GEdgar
    Jan 19, 2014 at 17:52

If you flip a coin twice and call it in the air both times, will you necessarily call either flip correctly? Will you necessarily call either flip incorrectly? It isn't that simple.

The very general idea here--by the Law of Large Numbers--is that, if you play this lottery "enough" times, then the fraction of times that you win (the empirical probability) will be "close to" $1/300$ (the theoretical probability). This is a fantastically imprecise notion (hence the quotes), as your tests should make clear. After $1104$ tests, you still haven't won anything, so while $0$ is arguably "close to" $1/300$, you may want your empirical probability to be closer, which means that you haven't played "enough" times, yet.


Not at all.

If your chances of winning are $1/300$ then the chances of you winning over $300$ draws is: $$1-\left(1-\frac{1}{300}\right)^{300}=0.632 \sim 63\%$$ So you still have a $37$ percent chance of not winning, even after playing $300$ draws.


Depends on what you mean by should. Let's do this with 2 tickets and 2 draws. There are two tickets so on the first draw you have a 50% chance. On the second draw, assuming you didn't win the first time you also have 50% chance. However, because the latter case $\textit{assumes}$ that you didn't win the first time, the probability of winning at least once in the first two times is 50% +(50%)(50%)=75%. The first 50 come from your first try while the rest come from the second try.

Hopefully this smaller example helps. You might think that this contradicts what people call the "law of averages" but that's not a real thing. The law of large numbers (mentioned by Cameron Buie) is the closes idea, but it only talks about behavior over a long (approaching infinite) period of time.


If a lottery sells 300 tickets and one buys all 300, one has a 100% chance of winning exactly one lottery, no chance of winning more than one, and no chance of not winning any. If two independent lotteries each sell 300 tickets, and one buys 150 tickets from each lottery, then one gains a chance of winning more than one lottery (precisely two in this case), but must accept in exchange for that a chance to win none. If one divides 300 tickets different ways among various numbers of lotteries, each of which has exactly 300 tickets, the average expectation will be to win exactly one, but one may gain more chances to win more than one; all such chances must be balanced out with chances to win less (i.e. zero).


Strictly speaking, your odds of winning X number of times will follow $$P(K=k) = \binom{n}{k} p^k (1-p)^{n-k}$$

Thus, your odds of winning X number of times will be (Using the calc found at this link)

  • 0- 36.7%
  • 1- 36.8%
  • 2- 18.4%
  • 3- 6.1%
  • 4- 1.5%
  • 5- 0.3%
  • Beyond here, the probabilities get to be very small.

The reason you aren't guaranteed a win with 300 draws is there is a chance you could win more than once. If you took the number of times you won time the probability that you would win that number of times, it would add up to 1 time on average winning, as you would intuitively expect.


Of course, the simplest answer to this is to remember the following:

- Doubling your chances does not halve your odds.

If it did, you would be looking at this:

  • Buying 1 ticket is 1 in 300
  • Buying 2 tickets is 2 in 300, or 1 in 150
  • Buying 3 makes 3 in 300, or 1 in 100
  • Buying 30 tickets is 30 in 300 or 1 in 10 etc

Which is clearly ridiculous.

The odds are 1 in 300 each time you play. If you buy one ticket that doesn't win, the next draw can't know this and reduce your odds to 1 in 299 next time until such time as you win within 300 attempts, it remains 1 in 300


Here's a somewhat intuitive take on the problem:

Let's assume what you said in your question is true. Suppose you've already played 299 days, and you never won. Then you can go and buy ANY ticket, and you'd be sure to win, which doesn't make sense.

It would also mean that you could roll a die 5 times, and if you got no 6, then you'd be able to predict that the next roll must be a 6.

  • 1
    $\begingroup$ It might be better to use the word "assume" instead of "admit." $\endgroup$ Jul 31, 2013 at 13:41

What would happen if there are 300 people, and each buys one of the 300 tickets in each of 300 lotteries (so between them, they buy all the tickets)? After 300 draws, there will have been 300 prices, and all prices went to these 300 people. So you have 300 people and 300 prices.

However, it is very extremely unlikely that each of them one exactly once. There will most likely be some who didn't win at all, some who won once, and some who won more than once.


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