scouts-l Mail Archive for November of 1999: Re: (no subject)
Cheryl Singhal (csinghal@CAPACCESS.ORG
Sat Nov 27 1999 - 17:43:11 CST
On Sat, 27 Nov 1999, Bob Riley wrote:
> Hi James,
> I'm not sure from your post if you've ever been involved in a derby race,
> or at least maybe not in the planning part. No dissing here but, sometimes
> when you're a spectator you're not aware of what went into something.
> For instance, in our format each car raced once in each lane. The only
> way you would have to race 260 times would be if each car raced all
> by itself (65x4=260). Remember, when car A runs so does car B, C
> and D. I believe the minimum # of races, heats or runs you would
> need would be equal to the # of cars in the competition.
Lordy me I wish y'all keep the math questions off-list! (g)
When I first read the question, it sounded like Jim was proposing that of
any 4 cars racing against one another, they ran 4 races as in:
D 4 in Heat 1; then to double check,
D 1 in heat2; and then
D 2 in heat 3; and finally
D 3 in heat four.
So there's 16 races. If you give each car a chance to win against (not
race against) each of the others, then you have to do all 16 races 16 times?
Then the cars that won the most races have to compete against one another
again? Which will run to some other number, the identification of which
is left to the student. (g)
However, if what's going on is that you've got cars A thru let's say Z
and the winners of the first race (who SHOULD all be in the same lane if
one really is faster -g-) then have to race in OTHER lanes in the 2nd
race, then, yes, the first go-through should take 16 races and the next
one 4 races and the last one 1 race, which is only 21. But, by that
fourth one, you're down to only one lane and it'll take 4 because only
one car at a time in each lane?
Sorry, it's been a wierd sort of day ...