pic: Alliance Scores Over the 2011 FRC Season

[BHS_STopping]

Originally Posted by IKE

Anyone care to do an alysis on how many teams would have to average net 0 pts. in order for 20% of alliances have a resultant score of 0 pts.?

for example, with dice, if I have 3 dice, the probility of at least 1 of them being a 1 during a roll would be 3*1/6 or 50%. the prob of 2 being 1s would be 3/2*1/36 or 4.5%. The probablility of 3 1s would be 0.5%. At a district event with 80 matches, there would be 160 alliances, and thus I would expect 1,1,1 0.8 times or 80% of events, there would be at least 1 alliance that got 1, 1, 1.

If 0 is assumed as the lower limit, then a 0,0,0 should be difficult to get. If FRC was on 2 vs 2, and 50% of the field could score 1 (or more), and 50% of the field could score 0. I believe you would expect on 25% of alliances to have a score of 0.

For 3 vs. 3, it should (in theory) be significantly more difficult... in theory. I guess my argument is that if "average" robot might correspond with your values, but the "median robot" may perform significantly lower...

Just wanting to correct you slightly on your math here.

If you throw three dice, the probability of at least one of them coming up with a 1 is not simply 3 * 1/6. Using this logic we could then assume that if we throw 6 dice then the number 1 is going to appear every single time (which is false, the actual probability in this case is about 66.5%). When throwing three dice, the probability of throwing at least one 1 is equal to 1 - (5/6)^3. This number turns out to be about 42.1%.

The probability of throwing exactly two 1's is a little bit trickier, but it's not too difficult. There are 216 possible dice rolls for three dice, and 15 of those rolls have exactly two 1's in them. 15/216 is roughly 6.9%. If we include the 1, 1, 1 case (that is, all situations where at least two 1's come up) then our probability is 16/216, or 7.4%.

The probability that all three dice show 1's is 1/216, or .46%, so you were right about that one.