Contest Winning Probability Example

Let’s say you have entered to win a contest which has 5 possible ways to win, and there are 100 different people who have entered the contest. What are your chances of winning? 5/100, right? Well, let’s take a look and see. First, let’s lay the ground rules for the contest. Each person can only enter the contest once, and each person can only win one of the five possible prizes. Lastly, the 5 winners are chosen at random. There are two basic ways to approach this: 1) using the rules of probability and, specifically, conditional probability and envisioning that the five winners are selected one at a time or 2) using simple probability and treating this situation like all five of the winners form one group that was selected at the same time. As we’ll see, these are both valid ways to envision the winners being selected, and both ways create the same ultimate solution.

Using the Rules of Probability (Unions, Intersections, and Conditional Probability)

In this model, we are going to think of the 5 winners selected one at a time. In this process, we’ll consider the probability that a single person (you) is selected exactly once as one of the 5 winners. Remember, you can’t be selected twice or more, meaning you can’t win more than one prize. The options are that you win zero prizes or exactly one prize. Let’s now define some events. $X_i$represents the event that you were picked on the “i^th” pick. So $X_3$, for instance, means that you were picked on the 3rd pick for the prizes. $Y_i$ will represent that you were not picked on the “i^th” drawing for the prizes. So, for instance, $Y_2$ would represent the event that you were not picked to win the 2nd pick (in other words, someone other than you won the second pick).

So what probability are we looking for? We are looking for the probability that you win EXACTLY 1 of the 5 different drawings. Therefore, that translates to you winning exactly 1 of the drawings, and losing exactly 4 of the drawings.

This means the probability of those events all being true at the same time (the symbol $\cap$ means “and”, meaning all those events are simultaneously true). However, this isn’t the complete picture. This is only one of a few ways you could win 1 of the 5 drawings. This is specifically saying you win the first drawing only and lose the other 4. You could also win one of the prizes by winning the second drawing, but losing the other 4 (the first, the third, the fourth, and the fifth). You could also win one of the prizes by winning the third drawing only, and losing the other four (the first, the second, the fourth, and the fifth). It turns out, there are 5 distinct combinations of winning and losing the drawings that would result in you winning one of the prizes: you could win the first, or the second, or the third, or the fourth, or the fifth. So we can define new events that illustrate that:

Now, what we’re really looking for is the probability of any one of the $W_i$ events to be true. Therefore, we are looking for:

The $\cup$ symbol means “or”. Therefore, we are looking for the probability of winning on the first drawing only OR winning on the second drawing only OR winning on the third drawing only OR winning on the fourth drawing only OR winning on the fifth drawing only. Because these events are mutually exclusive (they can’t happen at the same time) you can find their OR probabilities by just adding them together:

So now, we just need to the find the individual probabilities $P(W_i)$ and add them all together. Let’s look at $P(W_1)$. The probability of winning the first drawing and losing the other 4 can be found as follows.

Let’s first consider the probability that you win on the first prize drawing. Because there are 100 people to choose from and the number of ways that result in you winning is 1, the probability of you winning on the first drawing is:

That’s pretty simple so far. Now, what’s the probability that you win on the first drawing AND lose on the second drawing? Well, a convenient way to determine that is to find the probability of losing on the second drawing GIVEN we won on the first drawing. This called a conditional probability and can be denoted as $P(Y_2|X_1)$. To find this probability, we just need to find the probability of losing the second drawing after the details of you winning the first drawing are in place. After you win the first drawing, the number of ways you can lose the second drawing is the same as the first drawing (one of the 99 other people who are not you whose names have not been drawn yet), but the number of possible outcomes for the second drawing is not the same as the first drawing. Because you have already won the first drawing in this scenario, and therefore, your name has been removed from the second drawing, there are only 99 possible names that can be pulled. Therefore, the probability of you losing the second drawing given that you won the first drawing can be calculated as:

This should make sense since the probability of you losing any of the drawings after you won the first one should be 100%. There is a 100% chance that once you win a drawing, that’s the only one you win. Now, the rules of conditional probability follow that a joint probability (AND probability) can be calculated as follows:

If we continue this way, we can find the probability of $W_1$ as:

So even though there are other events that can happen after winning the first drawing, notably losing the other drawings, because those are the ONLY things that can happen, they don’t affect the probability of you winning once overall.

The calculation for $P(W_2)$ starts out a little a little more complicated, but it simplifies to something very simple. Let’s take a look. Remember, event $W_2$ means you won on the second drawing, but lost on the other four. So let’s start with the first drawing. What’s the probability that you lose this drawing?

Now, what’s the probability that you win on the second drawing given that you lost on the first? Well, there’s only one of you, so that’s one way you can win. Are there still 100 possible winners, though? No. If you lost the first drawing, that means someone else won, which means there name has been removed, and there are only 99 total people to choose from now.

So what’s the joint probability that you win on the second drawing and lose on the first? You can just multiply these probabilities together.

So interestingly, this has the same probability as you winning on the first drawing only. Remember, since after you win on the second drawing, the other 3 drawings are set in stone (you have to lose), their probabilities won’t affect the total probability of the event we’re looking for. Therefore:

By the exact same logic, we can show that all of the probabilities $P(W_i)$ are $\frac{1}{100}$. So, what’s the probability that any one of these events happens (the OR probability of $W_1$, $W_2$, $W_3$, $W_4$, and $W_5$)? Because these are mutually exclusive events (none of these events can happen at the same time), we can just add their probabilities together to get what’s called their union. To reason that they’re mutually exclusive, you would just have to understand that the probability of winning ONLY on the first drawing AND winning ONLY on the second drawing is zero. Those two things can’t both be true. That’s not how the contest is designed. Therefore, as we discussed before:

So after all that work, we were able to show that it is as easy as we thought from the beginning; five different possibilities to win out of one hundred different people. But there’s another good way to prove this. Let’s look at the approach that involves thinking of the winners as grouped together.

Using the Idea of Grouping the Winners Together

Now let’s consider that the person pulling the winners doesn’t pull the names one at a time. Instead, they put the 100 names in a random order and the first five names in that random order are chosen as the 5 winners. Will our ultimate answer change for the probability of you being a winner? No, but the way we analyze the problem will. So let’s see what this probability will be. The question now is how many different ways can the 100 people in the contest be put in unique groups of 5? It turns out there is a very simple calculation that returns that number for us. The calculation is called a combination. To find the number of ways $n$ number of distinct items can be placed in a group of size $k$, we can do the following calculation:

So for our situation, that would be:

What this number represents is the number of ways we can form unique groups of size 5 from a collection of 100 unique things. Now, how many of these unique groupings result in you winning the competition? All the ones that result in you being in the group. Therefore, that would lock in 1 of the people in the group (you). And that would then leave any group of four that could go in the group with you to make a group of 5 with you as one of the winners. Therefore, we’re now interested in how many ways the 99 people that aren’t you can be put into a groups of size 4 so that they could combine with you and form a group of size 5 where you would be one of the winners. We can use that combination calculation again and calculate it as:

So this represents the number of groupings where you would be one of the winners. Let’s divide this (the number of equally likely ways you can win) by the value we calculated earlier (the total number of equally likely groupings of the 100 people into groups of 5) and see what we get:

So here we have a completely different way to think about it, but did we get a different solution? No, our conclusion is the exact same. There is a $\frac{1}{20}$ probability of being a winner in this competition.

Summary

So here we were able to come up with 2 completely different ways to think of this problem that result in the same answer. Are these the only two ways? No, there are potentially dozens of slightly different ways to approach this problem, but these two ways show very different approaches. We can use the rules of probability (intersections of events, unions of events, conditional probability, total probability, etc.), or we can greatly simplify the problem into a grouping -type problem where we just need to consider the total number of groupings possible, and count the number of groupings that result in a specific person winning. The combinations rule we saw earlier is a type of coun