Friday, June 30, 2006

More Card Games for Nerds

Actually, what we've come up with is an ubercard game, named Calvincard. The rules are simple: you take a deck of cards, and after that you make up rules as you go along. A new rule can elaborate on an old one, but not contradict it flat-out, like the blessings of fairy godmothers.

You can start out taking turns making new rules, but even that will probably change as the game progresses. You keep going until you are either satisfied with the game, or it has broken out into fisticuffs.

The resulting game this week was rather fun, but I cannot yet report on it because DOB's last rule about scoring grew so complex my brain went on overload and I had to go take a nap. As fun or more fun than playing the game, however, was calculating out the probability of all the different hands to determine how they should be ranked.

OK, maybe we are hopeless nerds.

And on that note, here's a fun little puzzle:

The ants go marching one by one, two by two, all the way up to ten by ten. But the little one is always left out by himself. What's the smallest number of ants that could do this? Bonus points for a particularly simple process to find out.

5 comments:

Anonymous said...

Ok, so I was bored. My guess is 6350401. It's the least common multiple of the squares of 1 thru 10 plus 1. (Finding the least common multiple was done by prime factoring the numbers from 1 thru 10 then making sure those numbers are multiplied the same number of times for your lcm - for example if you're searching for the lcm for 6 and 9, your prime factors are 2x3 and 3x3 respectively. So you need 2x3x3 for your LCM or 18.) You then square this LCM (since your ants were marching quite square) and add the loner. (You could have used the prime factors of the squares to begin with, but that's more numbers to play with at first, hence more key strokes on the calculator, hence more work. I specialize in being a lazy (and hopefully correct) mathematician.

Queen of Carrots said...

The ants were marching square? Is that in the song?

I wouldn't have allowed for that, and thus my answer is much smaller. In other words, I simply picture row upon row of ants, one longer for each verse.

SK: ) said...

My answer is 100 (10x10). From that number you can subtract the number of ants needed for all the other verses of the song.

I first thought you would need 101, to include the "Little One." But in my version of the song the Little One, marching happily with his older brothers, merely stops to tie his shoe. Then they ALL go marching home again.

DJ said...

100 doesn't seem like it would work, because with some of the numbers you end up with more than 1 ant left over. When the ants go marching 8 by 8, for example, you end up with the little one plus three bigger ones marching home together. But of course I'm assuming that only the little one gets left behind.

I also assumed that "two by two" meant two at a time, rather than a square of four ants. So I got my answer (2521) by taking the LCM and adding one.

DJ said...

Wait a minute- if the ants start out marching one by one, can the little one really be said to have been "left out"? He's alone in his rank, but so are all the other ants. Since the ranks are only one ant wide, the little one is part of a full rank, which doesn't seem like it really counts as being left out.

OTOH, I don't think I've actually heard the song for over a decade, so I may be misunderstanding it.