Probability of Specific Outcomes

Aseiwell

New member
Joined
Jan 6, 2021
Messages
2
Hi everyone!

This is not a homework question, but rather an actual situation I encountered in life. It has been almost a decade since I took my last prob and stats course in college so I was rusty on the math when trying to work it out myself. Any help would be appreciated!

The necessary information:
I’m in a group with 15 other guys who all collect sports cards. We buy a box of (NFL) football cards and split whatever cards come out of the the box amongst all 16 of us in the group. Since there are 32 teams in the NFL, each person is randomly assigned 2 teams. Each person is only able to keep the cards that come out of the box that are of players which belong to their 2 randomly assigned teams. To determine the random teams, we list each of our names in any arbitrary order from 1-16 and randomize the list of 32 teams. The first name on the list gets teams 1 and 17 on the randomized list of teams; second name on the list gets teams 2 and 18; so on and so forth.

The question:
here are 6 distinct teams that don’t contain any good players that nobody ever wants. If you end up getting one of those 6 teams for both of your randomly assigned teams it’s considered getting “skunked”.

What is the probability (or likelihood?) that any given person gets “skunked”?

Let me know what everyone’s thoughts are on how to break down what the probability is and how to solve it! Thanks I’m advance!

Unnecessary information:
Hobby level boxes of cards are expensive (like >$1000 for a box of 144 cards) and while there is a higher likelihood of a hobby box yielded better cards than a retail box you can buy off the shelves somewhere, there is no guarantee a hobby box will yield cards with a value collective greater than what the box cost. So, in an effort to mitigate risk we split the cost of the boxes up in such a way that we can buy many boxes to minimize risk and maximize reward by allowing each person a shot at getting the best cards in the box.
 
[MATH]P[\text{skunk}] = \dfrac{\dbinom{6}{2}\dbinom{26}{0}}{\dbinom{32}{2}} = \dfrac{15}{496} \approx 3\%[/MATH]
As an aside, have you considered randomly ordering yourselves in a list and choosing cards from the entire batch like you'd choose sides for a pickup sports game? You could choose a new order each round to keep things as fair as possible.
 
[MATH]P[\text{skunk}] = \dfrac{\dbinom{6}{2}\dbinom{26}{0}}{\dbinom{32}{2}} = \dfrac{15}{496} \approx 3\%[/MATH]
As an aside, have you considered randomly ordering yourselves in a list and choosing cards from the entire batch like you'd choose sides for a pickup sports game? You could choose a new order each round to keep things as fair as possible.
Thank you very much for the response! I’ve never thought of doing it like that. That’s definitely an interesting suggestion!
 
Top