Raffle win probability: 80 balls in the raffle machine....

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stakel84

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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?

If you could show your working in the answers you post please?

Much appreciated!
 
stakel84 said:
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.
Click to expand...

Sorry, but as you read in the Read Before Posting thread (you did read it, right? ;)), that's not how this forum works. Showing work is your job. We'll gladly give you hints and help though.

stakel84 said:
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?
Click to expand...

The probabilities you've found are correct. The three balls are picked at random, so every ball has the same chance of being picked. Further, after one person wins, their ball is removed from the pool. Now let's think about the second question can see if we can't puzzle out the answer. In order to win the second prize, a person must first not win the first prize. So, what's the chance of that happening? Then, what's the chance of their ball being picked in the second draw? What does that make the overall probability of winning the second prize? As a hint, remember that the probability of an event happening and the probability of that event not happening must sum to 1 (100%). Then apply similar logic to the third prize, noting that that winner must lose the first and second prizes.
 
ksdhart2 said:
Sorry, but as you read in the Read Before Posting thread (you did read it, right? ;)), that's not how this forum works. Showing work is your job. We'll gladly give you hints and help though.



The probabilities you've found are correct. The three balls are picked at random, so every ball has the same chance of being picked. Further, after one person wins, their ball is removed from the pool. Now let's think about the second question can see if we can't puzzle out the answer. In order to win the second prize, a person must first not win the first prize. So, what's the chance of that happening? Then, what's the chance of their ball being picked in the second draw? What does that make the overall probability of winning the second prize? As a hint, remember that the probability of an event happening and the probability of that event not happening must sum to 1 (100%). Then apply similar logic to the third prize, noting that that winner must lose the first and second prizes.
Click to expand...

So, the probability of not winning on the first draw is 79/80. The probabilty of winning on second draw is 1/79, and not winning is 78/79. The probability of winning on the third draw is 1/78, and of not winning is 77/78.
How then do I work out the probability of winning on any of the three draws?
 
Okay, so if I'm understanding right, you've found the basic probabilities, but you're having trouble synthesizing all the information. The real question here is, given two independent events A and B, what's the probability of A and B happening? To think about that, let's perhaps consider a different, simpler example. Let's say event A is a fair coin coming up heads, and event B is a different fair coin coming up tails. What's the probability of A happening? What's the probability of B happening? What then must be the probability of both A and B happening? (You might try making a probability tree to help you "see" what's going on[/url][/b] here)

Or another, slightly more complicated example, to make sure you understand the underlying concept: Let's say you draw a card from a standard 52-card deck (so no Jokers). Event A is that card being a King. Event B is that card being a Jack. What's the probability of A happening? What's the probability of B happening? What then must be the probability of both A and B happening? In both these examples, which arithmetic operation did you use to find the answers? What does that suggest about how you might solve your raffle example?

Once you've found the individual probabilities of winning each of the three raffle games, you need to find the odds of winning the first, the second, or the third. Here, we can consider a fair six-sided dice. Given that rolling any number is equally likely (1/6), what's the probability of rolling a 1, 2, or 3? Which arithmetic operation did you use to figure that out? What does that suggest about how you might solve your raffle example?

If you need a refresher on probabilities of independent events, you might try this lesson or this lesson from Math Goodies.
 
ksdhart2 said:
Okay, so if I'm understanding right, you've found the basic probabilities, but you're having trouble synthesizing all the information. The real question here is, given two independent events A and B, what's the probability of A and B happening? To think about that, let's perhaps consider a different, simpler example. Let's say event A is a fair coin coming up heads, and event B is a different fair coin coming up tails. What's the probability of A happening? What's the probability of B happening? What then must be the probability of both A and B happening? (You might try making a probability tree to help you "see" what's going on[/url][/b] here)

Or another, slightly more complicated example, to make sure you understand the underlying concept: Let's say you draw a card from a standard 52-card deck (so no Jokers). Event A is that card being a King. Event B is that card being a Jack. What's the probability of A happening? What's the probability of B happening? What then must be the probability of both A and B happening? In both these examples, which arithmetic operation did you use to find the answers? What does that suggest about how you might solve your raffle example?

Once you've found the individual probabilities of winning each of the three raffle games, you need to find the odds of winning the first, the second, or the third. Here, we can consider a fair six-sided dice. Given that rolling any number is equally likely (1/6), what's the probability of rolling a 1, 2, or 3? Which arithmetic operation did you use to figure that out? What does that suggest about how you might solve your raffle example?

If you need a refresher on probabilities of independent events, you might try this lesson or this lesson from Math Goodies.
Click to expand...


Ok, so would the overall probability of winning any of the three raffle prizes be 3/80?
 
That's not quite the answer I get, but it is close. How did you arrive at your answer? Please be as detailed as possible about your thought process and any calculations.
 
stakel84 said:
Ok, so would the overall probability of winning any of the three raffle prizes be 3/80?
Click to expand...
That is the exact answer. Here is a greatly simplified model.
In this holiday season there is club with exactly eighty members. They give themselves a dinner party. As they gather there are two collections of balls numbered one to eighty. One red collection and one blue collection. As each member arrives, she/he retrieves exactly one blue ball; the red balls are put in a forced air selection device as used in national lottery.
At end of the evening the machine selects three red balls at random in some order.
If a member's number on his/hers blue ball matches one of the three drawn red balls, the person is a winner.

\(\displaystyle \large{\dfrac{\dbinom{79}{2}}{\dbinom{80}{3}}= \dfrac {3081}{82160}=\dfrac{3}{80}}\)

See Here. See here. Final calculation.
 
ksdhart2 said:
That's not quite the answer I get, but it is close. How did you arrive at your answer? Please be as detailed as possible about your thought process and any calculations.
Click to expand...

The way I understood your hints is that the probability of somone 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 therefor is

[FONT=MathJax_Size3-Web]([/FONT][FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]−1/[/FONT][FONT=MathJax_Main-Web]80[/FONT] [FONT=MathJax_Size3-Web])[/FONT][FONT=MathJax_Size3-Web]([/FONT][FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]−[/FONT][FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]/79[/FONT] [FONT=MathJax_Size3-Web])[/FONT][FONT=MathJax_Size3-Web]([/FONT][FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]−[/FONT][FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]/78[/FONT] [FONT=MathJax_Size3-Web])[/FONT][FONT=MathJax_Main-Web]=[/FONT][FONT=MathJax_Main-Web]77/[/FONT][FONT=MathJax_Main-Web]80[/FONT]

So the probability that you won any of the prizes is

[FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]−[/FONT][FONT=MathJax_Main-Web]77/[/FONT][FONT=MathJax_Main-Web]80[/FONT] [FONT=MathJax_Main-Web]=[/FONT][FONT=MathJax_Main-Web]3[/FONT][FONT=MathJax_Main-Web]/80[/FONT]



Is that correct?
 
Upon some further reflection, I'm inclined to believe that's probably the correct answer. It's not the process I used, nor the answer I got. However, it does match the answer given by pka, so I can be reasonably certain that it's me who has the wrong answer here. I clearly don't have anywhere near the grasp of even basic probability that I thought I did. Sorry if I led you astray. :oops:
 
ksdhart2 said:
Upon some further reflection, I'm inclined to believe that's probably the correct answer. It's not the process I used, nor the answer I got. However, it does match the answer given by pka, so I can be reasonably certain that it's me who has the wrong answer here. I clearly don't have anywhere near the grasp of even basic probability that I thought I did. Sorry if I led you astray. :oops:
Click to expand...

Could you post the answer you came up with so I can compare?
 
Hm... well, alright. I went to review my solution again, and I must have made an arithmetic error the first time, as now I do get 3/80 as the solution. I'll still post my method though. If we let \(\displaystyle P(W_n)\) be the probability of winning prize #n, and \(\displaystyle P(D_n)\) be the probability of your ball being drawn on the nth draw, then:

\(\displaystyle P(W_1)=P(D_1)=\dfrac{1}{80}\)

\(\displaystyle P(W_2)=P(\neg D_1) \cdot P(D_2) = \dfrac{79}{80} \cdot \dfrac{1}{79} = \dfrac{1}{80}\)

\(\displaystyle P(W_3)=P(\neg D_1) \cdot P(\neg D_2) \cdot P(D_3) = \dfrac{79}{80} \cdot \dfrac{78}{79} \cdot \dfrac{1}{78} = \dfrac{1}{80}\)

The probability of winning any prize is then the sum of these, or 3/80.
 
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