No account? Create an account
 Nerdsnipe challenge! - Lindsey Kuper — LiveJournal [entries|archive|friends|userinfo]
Lindsey Kuper

 [ website | composition.al ] [ userinfo | livejournal userinfo ] [ archive | journal archive ]

 From: 2012-02-01 03:57 am (UTC) (Link)
It seems like your grouping method (A/D groups with both A/D and D/A) means that you don't have to consider the question and answer cards as distinct: everybody gets two non-duplicating cards and matches anyone with the same two cards. Yes?

There are also 8choose2 = 28 groups, so the average group has 4.86 members.

The most members any group could have, obviously, is 34, which means the fewest groups you could have is 4. The fewest members, then that any group could have, is 0.

How often will any given group be empty? For group AB, 34 players will have an A. In order for AB to be empty, none of them can draw B as their second card. For the first A-holder, there are 7*34 cards that can be drawn, and 6*34 of them are not Bs. For the second, there 7*34 - 1 cards, and 6*34 - 1 non-Bs. So on to the last A-holder, who has a mere (6*34 - 33)/(7*34 - 33) chance. Multiply all these chances together and you have the likelihood of a group being empty.

If I keep writing, it will become very clear that I've taken discrete math and thought plenty about combinatorial math, but have never studied statistics. So I'll stop here.

Edited at 2012-02-01 03:58 am (UTC)