Let's say there's a raffle with two prizes. The total number of tickets sold is of 15 (meaning that the probability of winning at least prize, if I buy one ticket, would be $p = 2/15$).
My question would be, what happens if I buy more tickets, for the same raffle? I read some answers for a similar question, and people were calculating the probability of winning the raffle, based on the probability of not winning it.
The formula applied there was:
$p = 1 – (1-x)^{2}$
, if someone buys two tickets. The logic behind it was: what was the probability of actually missing the prize $n$ (or 2, for this example) times in a row. Then subtract that value from 1.
I wanted to know, how can we calculate this probability without actually calculating the failure probability first. For example, if the probability of buying one ticket and winning one of the two prizes is $p=2/15$, why isn't the probability of winning a prize, in case of buying 2 tickets $p = 2 * 2/15$?
I know my assumption is wrong, because if someone would buy 10 tickets, the probability would already be $p=\frac{2 * 10}{15}$, which is already greater than 1.
Why is my logic wrong, in this case?
Best Answer
This is in essence the hypergeometric distribution; from a box with $N$ balls (15 tickets), where $m$ are white (2 winning tickets) and $N-m$ are black (13 non-winning tickets), we pick $n$ balls at random (we buy $n$ tickets). If we denote "$X$" the number of white balls (winning tickets) we pick, the probability that $X=i$ is;
$$P(X=i)=\frac{\binom{m}{i}\cdot \binom{N-m}{n-i}}{\binom{N}{n}}$$
So in your case it would be $P(X=1)+P(X=2)$, with substituting $n$ in the formula accordingly to the number of tickets you buy.