1. ## Raffle win probability

Could someone please work out the probability of winning a raffle prize under the following circumstances? It's a friendly argument/debate I'm try to settle! Please show workings in any answers.

I have bought one ball in a raffle competition. There are 80 people taking part, so 80 balls in the raffle machine. The thirst three balls drawn out from the raffle win a prize (in descending monetary value with each ball drawn).
I presume the probability of me winning the first prize is 1/80, second prize, 1/79, and third prize 1/78.

What I want to know is firstly, is my above presumption correct? And secondly, what is the probability of me winning any of the three prizes?

Much appreciated!

2. ## Re: Raffle win probability

the probability of you winning any of the prizes is one minus the probability that you won none of the prizes.

The probability that you won none of the prizes is

$\left(1-\dfrac {1}{80}\right)\left(1-\dfrac {1}{79}\right)\left(1-\dfrac {1}{78}\right) = \dfrac {77}{80}$

So the probability that you won any of the prizes is

$1-\dfrac{77}{80} = \dfrac {3}{80}$

3. ## Re: Raffle win probability

Is that correct??

4. ## Re: Raffle win probability

Originally Posted by stakel84
Is that correct??
look at it another way.

3 balls are picked. So if you have any of those 3 balls you win one of the prizes.

What is the probability you have one of those 3 balls? Its $\dfrac 3{80}$

5. ## Re: Raffle win probability

One way:

Total number of ways you can pick 3 unordered balls from 80: 80C3

Total number of ways you can pick 3 ordered balls from 80: 80C3*3!

Total number of ways you can pick 3 ordered balls where your ball is one of them: 1*79*78 + 79*1*78 + 79*78*1 = 3*79*78

Probability of one of the 3 balls picked being your ball = $\dfrac {3*79*78}{^{80}C_3*3!}$ = $\dfrac {3*79*78}{\dfrac{80!*3!}{3!*77!}}$ = $\dfrac {3*79*78}{80*79*78}$ = 3/80

Another way:

Let $W_i$ be the event that your ball is picked in the ith pick. Then the probability that your ball is picked is (Events $W_i$ are mutually exclusive):

$P(W_1) + P(W_2) + P(W_3) =$ [win 1st pick, lose 2nd pick, lose 3rd pick] + [lose 1st pick,win 2nd pick,lose 3rd pick] + [lose 1st and second picks, win 3rd pick]= $\dfrac{1}{80}*\dfrac{79}{79}*\dfrac{78}{78} + \dfrac{79}{80}*\dfrac{1}{79}*\dfrac{78}{78}+\dfrac {79}{80}*\dfrac{78}{79}*\dfrac{1}{78}=\dfrac{1}{80 } +\dfrac{1}{80} + \dfrac{1}{80}=\dfrac{3}{80}$