Odds of Guessing the First 5 NBA Opponents

Question

There are 30 teams. You play each team in your division (5 teams) 4 times, each team in your conference (15) 3 or 4 times and each team in the other conference (15) 2 times. 

 

If there was a contest to "guess the first 5 teams on the Knicks schedule," using only straight probability (and not insider knowledge), what would the odds be?

Answer 1

This question really intrigued me because I'm a big NBA fan. It was tough to figure out, but I'm pretty sure I've figure out how to get the answer. Here are some important notes. I'm going to say the number of teams in the Knicks' division is 4 because I'm not counting the Knicks themselves in the calculations. Because we account for those 4 teams plus the Knicks elsewhere in the calculations, the number of teams left in the East for our calculations is 10. We'll say the West has 15 teams. Also, in reality, each team plays 6 more non-division but in-conference teams 4 times to push us to an 82 game season. To review, here is the important information:

 

29 possible opponents

4 games against 10 teams

3 games against 4 teams

2 games against 15 teams

 

This calculation is tedious because some repetition is allowed, but a certain amount of repetition is not allowed. The person making the selection knows that teams can only be selected a certain number of times, so the number of possible ways to pick the first 5 opponents is not just 29^5. We'll do this in steps.

 

First I'll just do 29^5, which equals 20,511,149. This is the total ways to pick the first 5 opponents with no consideration to how many times you can play an opponent. Now, we have to take out the possibilities that we know cannot happen based on repeated opponents.

 

There is a 1/29^3 chance that you pick an opponent three times in a row. To get the probability of picking a team three in five picks, we multiply this by 5/3. Because this is impossible for 15 of 29 teams (teams in the West), we multiply this by 15.

P3=(1/29^3)*(5/3)*(15)=75/571,787

This is the fraction of our original total the we can exclude because it includes possibilities where we picked out-of-conference teams three times.

 

Next we do the same for picking a team 4 times. The calculation is similar, but we start with 1/29^4, multiply by 5/4, and there are now 19 teams (West plus 4 in-conference teams) for which this type of selection is impossible. The calculation looks like

P4=(1/29^4)*(5/4)*(19)=95/2,829,124

 

Now for picking a team 5 times. We start with 1/29^5, multiply by 5/5 (which we can ignore because it's just 1), and now it's impossible for all 29 teams.

P5=(1/29^5)*29=1/707,281

 

Now we add up P3, P4, and P5 to get the total fraction of possibilities we are going to exclude. Then we subtract this from 1 to get the fraction of possibilities that we will include. Then we multiply this by our original total to get the number of ways you can pick the first 5 teams with all things considered. (I'm just going to work with decimals because the fractions will just have ridiculously large numbers if I add them.)

 

P=P3+P4+P5=0.00016616

1-P=.99983384

Total possibilities=(P-1)*20,511,149=20,504,741

 

There are a total of 20,504,741 ways to pick the first 5 games without repeating teams more than they are allowed.

The probability is 1/20,504,741. You asked for the odds, which can be written as 1:20,504,740.

Your odds are pretty bad, so I wouldn't make bets on this, but this was a fun (yet tedious) little probability experiment. Go Celts!