Consider a raffle for which each person may buy any number of tickets. There are many prizes but each each person may only win one prize. What is the probability that person [TEX] k [/TEX] will win a prize?

To make this more concrete, I'll work with some numbers. The following contestants enter the raffle with the indicated quantity:

[TEX] C_1 - 1[/TEX]

[TEX] C_2 - 5[/TEX]

[TEX] C_3 - 10[/TEX]

[TEX] C_4 - 50[/TEX]

[TEX] C_5 - 100[/TEX]

Let [TEX] P(C_k)[/TEX] denote the probability of contestant [TEX] k [/TEX] winning a prize. Let [TEX] E_k[/TEX] denote the number of entries for contestant [TEX] k [/TEX].

If there were 1 prize,

[TEX] P\left(C_k\right) = \frac{E_k}{\sum_{i=1}^n{E_i}}[/TEX]

For example,

[TEX] P\left(C_1\right) = \frac{E_1}{\sum_{i=1}^n{E_i}} = \frac{1}{166}[/TEX]

Now suppose there are 2 prizes. This is where I need help. To update the notation, let [TEX] P(C_k, p)[/TEX] denote the probability of contestant [TEX] k [/TEX] winning prize [TEX] p [/TEX].

A contestant can win the first prize, or, if they lose that, win the second prize. The sum of the entries for the second drawing depends on the winner of the first prize. To find the probability of winning the second prize, sum the probability with each contestant winning the first prize.

[TEX] P\left(C_k, 2\right) = \sum_{j=1, j \ne k}^n{\frac{P\left(C_j, 1\right) \cdot E_k}{\left(\sum_{i=1}^n{E_i}\right)-E_j}}[/TEX]

I'll take a stab with the first contestant. The [TEX] \LaTeX [/TEX] code refused to render the "+" between fractions, so make that expression the

[TEX] P\left(C_1,2\right) = 1 \cdot \left( \frac{P\left(C_2,1\right)}{166-5} - \frac{P\left(C_3,1\right)}{166-10}-\frac{P\left(C_4,1\right)}{166-50}-\frac{P\left(C_5,1\right)}{166-100}\right)[/TEX]

Now what? Is the probability of winning a prize simply the sum of winning the first prize and the second prize? I would prefer an equation that uses counting techniques. Can anyone think of a general formula for any number of prizes?

To make this more concrete, I'll work with some numbers. The following contestants enter the raffle with the indicated quantity:

[TEX] C_1 - 1[/TEX]

[TEX] C_2 - 5[/TEX]

[TEX] C_3 - 10[/TEX]

[TEX] C_4 - 50[/TEX]

[TEX] C_5 - 100[/TEX]

Let [TEX] P(C_k)[/TEX] denote the probability of contestant [TEX] k [/TEX] winning a prize. Let [TEX] E_k[/TEX] denote the number of entries for contestant [TEX] k [/TEX].

If there were 1 prize,

[TEX] P\left(C_k\right) = \frac{E_k}{\sum_{i=1}^n{E_i}}[/TEX]

For example,

[TEX] P\left(C_1\right) = \frac{E_1}{\sum_{i=1}^n{E_i}} = \frac{1}{166}[/TEX]

Now suppose there are 2 prizes. This is where I need help. To update the notation, let [TEX] P(C_k, p)[/TEX] denote the probability of contestant [TEX] k [/TEX] winning prize [TEX] p [/TEX].

A contestant can win the first prize, or, if they lose that, win the second prize. The sum of the entries for the second drawing depends on the winner of the first prize. To find the probability of winning the second prize, sum the probability with each contestant winning the first prize.

[TEX] P\left(C_k, 2\right) = \sum_{j=1, j \ne k}^n{\frac{P\left(C_j, 1\right) \cdot E_k}{\left(\sum_{i=1}^n{E_i}\right)-E_j}}[/TEX]

I'll take a stab with the first contestant. The [TEX] \LaTeX [/TEX] code refused to render the "+" between fractions, so make that expression the

**SUM**of the positive terms.[TEX] P\left(C_1,2\right) = 1 \cdot \left( \frac{P\left(C_2,1\right)}{166-5} - \frac{P\left(C_3,1\right)}{166-10}-\frac{P\left(C_4,1\right)}{166-50}-\frac{P\left(C_5,1\right)}{166-100}\right)[/TEX]

Now what? Is the probability of winning a prize simply the sum of winning the first prize and the second prize? I would prefer an equation that uses counting techniques. Can anyone think of a general formula for any number of prizes?

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