Tuesday, June 26, 2012

Fault, second serve...

Wow, I didn't expect everyone out there to be mathletes, but damn, this was pitiful.  For those of you who entered the contest, I'm giving you a second chance.  If you haven't entered yet, finish reading this, then go here to enter.

The reason I'm giving you a second chance is because nearly everyone completely botched the third question:

How many complete sets will I get?
What card will I get the most of (checklist here)?
How many of that card will I get?
Who will win the Men's Wimbledon tournament this year (draw here)?
Who will win the Women's Wimbledon tournament this year (draw here)?


Maybe it's my fault for leaving you with a math problem, or maybe it's your fault for not recognizing it, but either way, please try again.  Leave a comment on this post with your new answer to question 3.  Or leave it the same and take your chances. By the way...


...here's a gigantic hint:


The minimum number that the answer could possibly be is 24.  That means only 5 of you were even in the realm of possibility.  However, pretty much no one is in the realm of probability.


How did I conclude 24, you say?  Well, given N complete sets, the number of leftover singles is given by


Lc = 1195 - N * 50 


Now, for the card that occurs most often to be at a minimum, we need to calculate the smallest possible number of incomplete (49/50) sets possible.  Therefore, the minimum number of incomplete sets is given by


ISmin = floor ( Lc / 49 )


with any remainder being leftover cards.  Then, we can add together the number of complete sets and the minimum number of incomplete sets.  If the (Lc/49) has a remainder, we add 1, because the most-occurring card must be in that remainder.  So


If rem ( Lc / 49 ) = 0
     MOmin = Lc + ISmin
Else 
     MOmin = Lc + ISmin + 1


Run this algorithm on all possible numbers of sets from 0 to 23, and the answer is either 24 or 25.  For N= 0 to 19 sets, it's 25.  For N= 20 to 23 sets, it's 24.


I told you this would be a gigantic hint.  Here's more hint:


I ran a simulation of 1196 cards 100,000 times and checked to see how often the most-occurring card showed up in each.  Here are some results:




The mean value was 35.5.  Assuming a normal distribution, 99.7% of the time, the value would be between 28 and 43.  The highest answer received so far was 28.   

9 comments:

Josh D. said...

For Q3, I will change to 34, in hopes that their collation is not completely random.

Hackenbush said...

14 sets
Marion Bartoli
37
Djokovic
Sharapova

night owl said...

Zzzzzzzzzzzzz.

Dude, what? I was too busy playing football, hanging out with cheerleaders.

Um ....

I change my number to 42. As you might of gathered, that's a total guess.

Matt Perry said...
This comment has been removed by the author.
Matt Perry said...

Gah, dang it nightowl, beat me by 2 minutes.

Fine. 54 then. That's a good, proper answer for the universe.

Simons44 said...

14 complete sets
Vera Zvonareva
36
Novak Djokovic
Victoria Azarenka

Play at the Plate said...

I change my number to 32...so there.

High-Five Man said...

I'll change mine to 39.

hiflew said...

Man, I took two semesters of statistics in college. I got B's, but I seem to have left the knowledge behind.

I'll change my number to 42, since that is the number that proves I am not a robot this time.