Example #1
This is a 2-way table I made that shows the data of the consensus 2024 Fantasy Football draft in PPR leagues by Fantasypros using ADP (Averaged Draft Position). So for example, for the first top 3 draft picks, you have CMC with an ADP of 1.0, Ceedee Lamb with an ADP of 2.0, and Tyreek Hill with an ADP of 3.2. So then the order would be CMC 1st pick, Lamb 2nd pick, then Tyreek 3rd pick. And then after all 180 players are ordered, I split them into 15 groups of 12 according to their ADP to represent a 12-man league, which means 12 picks per round, with 15 rounds of drafting.
This has 6 columns for each position and 15 rows for each round. Note that there is also a "total" column all the way on the right which indicates the total number of players regardless of position drafted in each round which is 12. And it is always 12. But on the "total" row at the bottom, we see the total number of players for that specific position. In the bottom right, we see the total number of all players drafted from each of the 15 rounds of all types of positions. Basically, this is everyone in our population. We know our table is correct if the values of all the "total" rows, and the "total" columns both lead up to the population number, which is 180.
Note: These are just ADPS and your own fantasy draft, although very similar to this, will be different. But for practical purposes, we can pretend that this was a real draft for a made-up league we had.
Prompt #1: Suppose we randomly select just 1 player from the whole draft without regard to any rounds. What is the probability that we select a Wide Receiver?
Work/Explanation: We can notate this as P (Wide Receiver). To solve this, we have to divide our quantity (total number of WRs) by the total quantity (everyone), in which the boxes will tell us how many since we have a 2-way table.
The blue circle tells us the total number of WRs, and the purple circle tells us the total number of players in the draft without regard to position.
So we simply just divide these two numbers to get our answer.
Answer: Assuming we are selecting randomly from the draft, the probability of us selecting a Wide Receiver, P(Wide Receiver), is 66/180, or 36.67%.
Prompt #2: Suppose we randomly select just 1 player from the whole draft without regard to any rounds. What is the probability that we select a Wide Receiver AND a Runningback?
Work/Explanation: Whenever we see the word "and," we have to stop and think if these two events are mutually exclusive. The prompt says we select just 1 player, meaning that we only have 1 pick, so is it possible to get BOTH a WR and an RB in one pick? Are these two events mutually exclusive?
Answer: The prompt says that we are only selecting 1 player from the draft, making it impossible to get BOTH a WR and an RB in one pick. Therefore, these events are mutually exclusive. The probability of this intersection event, or P (Wide Receiver ∩ Runningback), is 0.
Prompt #3: Suppose we randomly select just 1 player from the whole draft without regard to any rounds. What is the probability that we select a Wide Receiver OR a Runningback?
Work/Explanation: This is now a union scenario, where we apply the addition rule with P (Wide Receiver ∪ Runningback ) = P(Wide Receiver) + P(Runningback) - P (Wide Receiver ∩ Runningback). But we just explained that because getting both a WR and an RB in one pick is impossible, the intersection probability is 0, so we can leave out the last part of our equation.
So now, it's just P (Wide Receiver ∪ Runningback ) = P (Wide Receiver) + P(Runningback).
We know that P (Wide Reciever) is 66/180 from the last problem, now we need to find P (Runningback). Using the same process, by looking at the total row at the bottom, we can find that this would be 53/180.
Finally, we just plug in our numbers: P (Wide Receiver ∪ Runningback)= 66/180 + 53/180
Answer: Assuming we are selecting randomly from the draft, the probability of us selecting a Wide Receiver OR a Runningback, or P (Wide Receiver ∪ Runningback), is 119/180, or 61.11%.
Prompt #4: Suppose we randomly select just 1 player from the whole draft without regard to any rounds. What is the probability that we select a Wide Receiver AND an 8th-rounder?
Work/Explanation: This is an intersection scenario, so we have to figure out if these events are mutually exclusive. Now let's think about this. While it may not be possible to get both a WR and an RB in one pick, it is possible to get a WR and an 8th-rounder in one pick. This is called an 8th-rounder WR. In other words, you can't be both a RB and a WR, but you can be both a WR and an 8th-rounder. Therefore, these two events are not mutually exclusive. So, the probability of this will not be 0.
Now is the next step. We have to recognize that we have a 2-way table. And so, we don't actually have to do any rough calculations. We can just use our eyes to get to the right box where these two events intersect.
This is where these two events intersect, and we get our number: 6. There are 6 8th rounder WRs that came from this draft, and there are 180 players in total. So now to answer the prompt, we just divide these two numbers.
Answer: Assuming we are selecting randomly from the draft, the probability of us selecting a Wide Receiver AND an 8th rounder, or P (Wide Receiver ∩ 8th rounder), is 6/180, or 3.33%
(This is the same 2-way table in the previous exmaple)
Prompt #1: Suppose we drafted at random. What is the probability that our pick was a Quarterback given that the pick was in the 7th round?
Work/Explanation: Because we have a table here, we won't need to resort to our equation. Now, our prompt says GIVEN that the pick was in the 7th round. So, we don't care about any other round here besides the 7th round, so let's block everything else out.
This is what our table looks like now that it's simplified. We are only looking at the 7th round.
And now, we see that in this round that consists of 12 total players, only 1 of them is a QB.
With this information, we can do 1 /12 to get our answer.
Answer: Assuming we randomly selected from the draft, the probability that our pick was a Quarterback given that the pick was in the 7th round, or P ( Quarterback | 7th round), is 1/12 or 8.33%.
Prompt #2: Suppose we drafted at random. What is the probability that our pick was NOT a Quarterback given that the pick was in the 7th round?
Work/Explanation: Our keyword that we know this is a conditional probability is the word "not." With any complementary probability, all we do is subtract our probability value from 1.
Now, we already know from the previous prompt that the probability of our pick being a Quarterback given it was in the 7th round was 1/12. So to find the complementary of this, all we need to do is the calculation 1 - (1/12).
Visually on our 2-way table, it looks like this. Before, we had our red circle on "1" which is there being 1 quarterback. But now, since we don't have a quarterback, our interest is in every other position present.
Answer: Assuming we selected randomly from the draft, the probability that our pick was NOT a Quarterback given that the pick was in the 7th round was 11/12, or 91.67%.