# Probability of Winning a Contest

This is my first question so apologies if its unclear/vague.

There exists a contest with me in it, and $5$ others, thus $6$ people in total, along with $5$ prizes. A person can only win one prize, and once they do, they're out of the contest. Winners are chosen at random.

So, what is the chance of me winning at a prize?

My initial thought was: $\frac16$ chance initially, then $\frac15$ if I don't win the first time, $\frac14$ if I don't win the second, ect. as people are removed once they win, resulting in $\frac16+\frac15+\frac14+\frac13+\frac12 = 1.45$ which is $145$%.

This is greater than $100$%, and obviously I am not guaranteed to win as I can be the $1$ loser, so how do you find the correct probability?

Thanks!

Edit: All of you have been extremely helpful. Thank you!

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You have an implicit assumption that the probability of each person winning each prize is equal - this may not be the case for all contests, such as a raffle drawing where people may buy any number of tickets. (That's not a criticism, just a warning that the answers below won't always hold.) – Patrick M Jul 3 '14 at 21:07

Well, if we are choosing $5$ winners at random out of $6$ people, then you have a $\dfrac{5}{6}$ probability of winning.

However, your approach is correct - we can sum the individual probabilities to get the same result, but we have to be cautious. Consider the first two prizes given. As you say, we have a $\dfrac{1}{6}$ probability of winning the first prize, and then a $\dfrac{1}{5}$ chance of winning the second prize if we don't win the first prize. We need to take into account this part - you only have a $\dfrac{5}{6}$ chance to get to the second round in the first place.

Using your method, this would give us a total probability of

$$\frac{1}{6}+\frac{5}{6}\left(\frac{1}{5}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{1}{4}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)\left(\frac{1}{3}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right) \\=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{5}{6}$$

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i understand now! thank you! – user161504 Jul 3 '14 at 17:44
@user161504 Glad to help. – Peter Woolfitt Jul 3 '14 at 17:56

The simplest way to see it is that there is only $1$ person who doesn't win a prize. Your chance of being that person is $\frac 16$, so your chance of winning something is $1-\frac 16=\frac 56$

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so - is the answer just number of prizes divided by number of people always? – user161504 Jul 3 '14 at 18:15
Yes, as long as nobody can win more than one prize. It is called Linearity of expectation. Your expectation of each prize is $\frac 16$, so your total expectation is $\frac 56$ – Ross Millikan Jul 3 '14 at 18:15
@RossMillikan Except we're trying to compute a probability, not an expectation. So you need to add an argument about why the expectation really is a probability because every prize has value 1. – David Richerby Jul 3 '14 at 21:37

Hint: Can you win the second prize if you won the first prize? No. So you need to multiply the probability of winning the second prize by something. Similarly for the other prizes of course.

Alternatively you can do it the other way round and find the probability that you don't win any prize. That one is easier to find.

A third way is as follows: There is symmetry among all people. Nobody has an advantage or disadvantage. And since exactly one of the six people doesn't go home with a prize, we know the probability.

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Calculate it with the probability of the complementary event (not winning a prize), which is: $\frac{5}{6}\cdot \frac{4}{5} \cdot \frac{3}{4}\cdot\frac{2}{3} \cdot\frac{1}{2} = \frac{1}{6}$ , then substract this from 1 to get your probability, which is therefore $1-\frac{1}{6} = \frac{5}{6}$.

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so - is the answer just number of prizes divided by number of people always? – user161504 Jul 3 '14 at 18:14