Probability of two people winning a prize, according to the number of tickets each person has

Question

Since I find it hard to abstractly describe what I want to understand, I came up with the following example.

Suppose that 6 people hold each a given number of a very special type of lottery tickets. Person A holds 2 tickets, person B holds 3 tickets, person C holds 1 ticket, person D has 5 tickets, person E has 6 tickets and person F has 3 tickets.

The probability of each person winning a prize is directly proportional to the number of tickets each holds.

So, in the end, N different contenders are going to be chosen winners. What I would like to learn is how can I calculate the probability of A, B, C, D, E and F of being one of the N winners?

We can assume N=2 or whatever, if it makes it easier.

Summary of probabilities: $$\begin{cases}P(A)=\frac{2}{20}\\P(B)=\frac{3}{20}\\P(C)=\frac{1}{20}\\P(D)=\frac{5}{20}\\P(E)=\frac{6}{20}\\P(F)=\frac{3}{20}\end{cases}$$

Answer 1

You haven't specified the method for choosing. I'll suppose that one of the 20 tickets is chosen at random. Then the five people who have no prize join a second similar lottery with their tickets.

Let's figure out the probability that A wins a prize.

Her chances are $2/20$ for the first draw. If she's not chosen (with probability $18/20$) her chance of succeeding the next time depends on who won at first. If it was B (with probability $3/20$) then her three tickets are removed and A's second chance is $2/17$. For C, D, E, and F her chances are $2/19$, $2/15$, $2/14$ and $2/17$ so her chance for on if of the two prizes is $$ \frac{2}{20} + \frac{18}{20}\left( \frac{3}{ 20}\frac{2}{ 17} + \frac{1}{ 20}\frac{2}{19 } + \frac{5}{ 20}\frac{2}{15 } + \frac{6}{ 20}\frac{2}{14 } + \frac{3}{ 20}\frac{2}{17 } \right) . $$

You can work this out for the other folks, and for $N$ participants. One check: you should get probabilty $1$ if $N$ is the number of participants.

I don't know a pretty way to do this.

Answer 2

$$\begin{cases}P(A)=\frac{2}{20}\\P(B)=\frac{3}{20}\\P(C)=\frac{1}{20}\\P(D)=\frac{5}{20}\\P(E)=\frac{6}{20}\\P(F)=\frac{3}{20}\end{cases}$$

You originally asked with regard to two candidates being selected. Let's look at that case. Call the first candidate to be selected $X_1$, the second candidate to be selected $X_2$ and so on... The Probability of $X_1$ being selected is $P(X_1)$.

You want to represent $X_1$ and $X_2$ as being selected. You haven't stated whether they will be selected in one go, or one at a time. Let's assume it's the latter. Time to use some conditional probability, because choosing one affects the probability of choosing the next..!

$$P(X_1 \cap X_2 )=P(X_2|X_1)P(X_1)$$

You choose the first person $X_1\in\{A,B,C,D,E,F\}$, with their probability $P(X_1)$. Easy enough.

Then you select the second person $X_2$, out of the remaining people. If $X_1=A$ then $X_2\in \{B,C,D,E,F \}$. Find $P(X_2|X_1)$ by re-calibrating the likelihood of B,C,D,E & F being chosen now that A is no longer a viable candidate for selection. Then you have the ability to multiply this answer for your chosen $X_2$ and $X_1$ per the conditional probability formula stated above.

You can extend this to the Nth case using further conditional probability. Let N=4: $$P(X_1 \cap X_2 \cap X_3 \cap X_4)=P(X_4|X_1 \cap X_2 \cap X_3)P(X_3|X_1 \cap X_2)P(X_2|X_1)P(X_1)$$

Have fun!