# draft lottery question

#### 1luffy

##### New member
I'm a lil confused. What are the chances of the 2 seed falling outside of 3rd or not getting a 1st, 2nd and 3rd/ top 3 pick?

tnx.

Last edited:

#### ksdhart2

##### Senior Member
Well, let's think about the question and think about what we know, and see where that leads. You want to know the chance of the 2nd seeded team not getting 1st, 2nd, or 3rd pick. Assuming that each team must be picked at some point (i.e. not being picked isn't an option), that means that your question is the same as asking what's the chance of the 2nd seeded team getting 4th, 5th, 6th, [...] or 14th pick. Looking at your table, we can see that the 2nd seeded team cannot be picked sixth or later (at least I'm making the standard assumption that no listed probability implies that event is impossible; if that's not correct, please reply with what the blank spaces do mean).

So, now we've reduced the problem down to the probability of the 2nd seeded team being picked 4th or 5th. To continue from here, I'll give a few leading questions to hopefully get you thinking in the right direction. Can the 2nd seeded team be picked both 4th and 5th at the same time? What does that tell you about if the events are mutually exclusive? What does that tell you about how you might find the overall probability? If you need a refresher on probabilities, you might try this page from Richland Community College.

#### 1luffy

##### New member
Looking at your table, we can see that the 2nd seeded team cannot be picked sixth or later .
Yes

Can the 2nd seeded team be picked both 4th and 5th at the same time?
Nope. It's either 4th or 5th. Someone told me the answer is .44 but I think it's either .31 or .12?

#### ksdhart2

##### Senior Member
After rounding to two decimal places, I also get 0.44 (44%) as the answer. You say you think the answer is 0.31 or 0.12. Why do think either of those might be the answer? What work did you do to arrive at those answers?

#### 1luffy

##### New member
After rounding to two decimal places, I also get 0.44 (44%) as the answer. You say you think the answer is 0.31 or 0.12. Why do think either of those might be the answer? What work did you do to arrive at those answers?

Cause the 2nd seeded team can't be picked both 4th and 5th at the same time. It's either they're getting 4th or 5th. I'm not sure if my answer is correct though. Is 44% the right answer and not 31% or 12%?

#### ksdhart2

##### Senior Member
Right. The outcome can be one or the other event, but you're looking for the overall probability that it will be either one of the two. Perhaps using two simpler examples will help shine some light on the process, since the linked page didn't help you any. Let's say you flip a fair coin. The probability of it coming up heads is, of course, 50%. The probability of it coming up tails is also 50%. So, what's the probability of it coming up heads or tails? Or, let's say you roll a standard six-sided dice. The probability of rolling a one is 1/6. The probability of rolling a six is also 1/6. So, what's the probability of rolling a one or a six? What process did you use to find these answers? Can you see how that applies to your actual problem? What happens if you use that exact same logic?