# Probability using combinatorics: probability doesn't sum to 1

In a certain lottery, 10,000 tickets are sold and 10 prizes are awarded. What is the probability of not getting a prize if you buy 2 tickets.

The answer to this question is simple enough: $$\frac {9990 \choose 2}{10000 \choose 2}$$ I understand this answer, since there are 2 ways of picking from the 9990 tickets that don't give you a ticket. And 2 ways of picking some ticket from the 10,000 tickets. But, I wanted to verify this by calculating the probability of getting a ticket.

I used a similar method:

$$\frac{10 \choose 2}{10000 \choose 2}$$

However, when I use a calculator to add them up, they don't sum to 1. I repeated this exercise for buying 10 tickets, in which case the answers for not getting a ticket and getting a ticket were respectively:

$$\frac {9990 \choose 10}{10000 \choose 10}$$ $$\frac{10 \choose 10}{10000 \choose 10}$$

In this case as well, the probability don't sum to 1. I understand that I'm actually decreasing the numerator in the second case. But I can't understand intuitively why this is wrong.

While calculating the probability of not getting a ticket, we divide the number of ways we can pick a ticket that doesn't give you a prize by the number of ways you can pick any ticket. So, while calculuating the probability of getting a ticket, shouldn't we divide the number of ways we can pick a ticket that gives you a prize divided the total number of ways you can pick a ticket?

• These are good questions to ask. What if you buy $20$ tickets; what are the outcomes and their probabilities? – David K Feb 15 at 12:08

## 2 Answers

Note that $$\frac{\binom{10}{2}}{\binom{10000}{2}}$$ is the probability of getting two prizes! Hence, you are missing the case where you get just one price. If you add that, you do indeed get $$1$$:

$$\frac{\binom{9990}{2}}{\binom{10000}{2}} +\frac{\binom{10}{2}}{\binom{10000}{2}} +\frac{\binom{9990}{1}\cdot\binom{10}{1}}{\binom{10000}{2}} = 1.$$

• Shouldn't you also show the correct probability for winning at least one prize? – Wolfgang Kais Feb 15 at 13:13
• Winning at least one prize is the same as winning exactly one or exactly two prices, so the last two summands. – Christoph Feb 15 at 13:51
• Wow, that was really useful! I also used this logic to answer the question posed by @David in the comments, and they summed to 1 as well. Thank you. – WorldGov Feb 15 at 14:00

You can also solve by Binomial as the numbers are large. Hypergeometric distribution used by @Christoph gives the exact probabilities:

Probability that you will get a prize $$= \frac{1}{1000}$$

probability that you will not get a prize $$= \frac{999}{1000}$$

If you buy two tickets, the probability that you will not get a prize is

$$\approx {2\choose0}(\frac{1}{1000})^0(\frac{999}{1000})^2$$ you will get a sum of 1 in this too.

• You are using the symbol "$=$" for an approximation. Please stress the fact that this answer does not give the exact probabilities! – Christoph Feb 15 at 12:08
• @Christoph: Done!! – Satish Ramanathan Feb 15 at 12:10
• Why a downvote, I am introducing another way to look at the problem!! – Satish Ramanathan Feb 15 at 12:27
• I downvoted because this is not an answer to the question "Why do my probabilities do not sum up to $1$?" – Christoph Feb 15 at 12:35
• Although it does not answer the question that the OP asked, it is still a valid answer to the problem posed by OP. I am going to leave it even if it receives further downvotes. – Satish Ramanathan Feb 15 at 12:38