# Basic probability - picking cards

I could really use some help with this homework question.

There are 100 cards in a raffle. 1 of them is the winning card. There are $$i$$ participants, each one buys 1 card. What is the probability the no one wins?

Solution attempt:

I solved it the following way:

The probability the no one wins:

$$\frac{99}{100} \frac{98}{99} \frac{97}{98} \frac{96}{97} \dots =\frac{\frac{99!}{i!}}{\frac{100!}{i!}} = \frac{99!}{100!} = 0.01$$

The problem is that the solution I was given by my teacher is: $$\frac{\binom{99}{i}}{\binom{100}{i}}$$

I seems like the order of choices doesn't matter in this solution. My question is - why? I thought that I should count all the different orders.

Any help would be greatly appreciated!!

Edit: I finally understood my mistake!!

$$\frac{99}{100} \frac{98}{99} \frac{97}{98} \frac{96}{97} \dots =\frac{\frac{99!}{(90-i)!}}{\frac{100!}{(100-i)!}}$$ and this is equivalent..

Thanks for trying to help :)

• Somehow I would have thought the quicker answer was the number of unsold tickets divided by the total number of tickets $\dfrac{100-i}{100}$. Your answer does not seem to depend on $i$ – Henry Apr 5 '19 at 17:34
• Could you explain why? – PhysicsPrincess Apr 5 '19 at 17:35
• There are $100$ tickets or cards. $i$ are sold and $100-i$ are not. Each is equally likely to be the winning ticket – Henry Apr 5 '19 at 17:36
• Okay I understand this solution from this point of view.. But why is the mistake in my first attempt? – PhysicsPrincess Apr 5 '19 at 17:38
• @PhysicsPrincess: Not really, because if the first one buys a winning card, we know for sure that the second won't win. It's like you have $i$ attempts to win. Anyway, your mistake is that the denominators in your expressions should be $(99-i)!$ and $(100-i)!$ – Vasya Apr 5 '19 at 17:47