Quote:
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...
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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.