View Full Version : Steve Rourke vs. John V-Neun: 3rd Annual Championship Wager
It's taken some negotiation, but here's the deal this year. Steve wants his signed Canadian $20 back from last year's loss. John prefers to add to his net worth this year. Will John get richer, or will Steve be able to buy a nifty FIRST T-Shirt when John returns the loot from last year???
Steve's position: The championship alliance will have 3 scoring robots. The lowest scoring member of the alliance will score more than 4.5 tubes in total over the two winning matches.
John's position: The championship alliance will most likely have 3 scoring robots. But the lowest scoring member of the alliance will score less than 4.5 tubes in total over the two winning matches.
(Obviously if the winning alliance has a ramp-only 'bot then I'm in the ditch, but I feel it's a necessary risk to get my $20 back. And besides, I took counsel from Karthik on this one -- so I'm feeling confident!)
Jonathan Norris
12-04-2007, 22:54
Hum... this is an interesting one...
I believe that the championship winning alliance will consist of three scoring bots (most likely one being a ramp hybrid). But for all three robots to average over 4.5 tubes (lets say 5) in the finals, where defense will be strong, is going to be very tough. Mainly because that is saying the winning alliance will score at least 15 out of a possible 18 ringers (I believe the max is 18..).
Thats alot... the most ringers we have seen on the rack is 19 (I believe that record from Waterloo is still standing). Say the rack gets filled for the finals on Einstein, that means your alliance puts up 15 ringers and your opponents puts up 9. If they have made it to the finals in Einstein I hope they can put up more then 9 ringers in a match.
Looking at the math I would have to say this is a tough bet, mainly because of how prominent defense will get (not dedicated defense, but the bots that switch from offense to defense quickly), and thats just a tonn o' tubes to be put up by one alliance while playing against another strong alliance.
However I also put trust in Karthik's scouting info...:ahh:
Hum... this is an interesting one...
I believe that the championship winning alliance will consist of three scoring bots (most likely one being a ramp hybrid). But for all three robots to average over 4.5 tubes (lets say 5) in the finals, where defense will be strong, is going to be very tough
No, 4.5 is NOT the average score per match. It is the TOTAL number of ringers scored over the two winning matches. So if Robot C is the lowest scoring 'bot on the winning alliance and scores two in match F1 and three in F2, I win. (I'm counting on attention being drawn by the higher scoring robots, giving Robot C a couple of undefended freebees per match)
-Steve
Thats alot... the most ringers we have seen on the rack is 19 (I believe that record from Waterloo is still standing).
During a practice match today on Galileo with 6 working robots on the field, 21 ringers were scored. I believe it was part two of match 34. It's only a practice match, but I imagine if it happened today, it can happen again.
The lowest scoring member of the alliance will score more than 4.5 tubes in total over the two winning matches.
I'm going to go on this side. I doubt you're going to have many pure-defense robots picked. I'm sure that robots with good ramps will be in ample supply at the 3rd pick, so the dealbreaker will probably be their scoring ability, which means that almost all of the 3rd picks will be scoring robots. At some point the 'lead' robots will be disabled, whether it be by chance, good defense, or needing to deploy a ramp, and that'll be the time for the low-scoring robot to put a few up.
I voted for John because i thought the 4.5 meant the average.. but if thats strung out between 2 matchs then it is a really tough bet, depends on if one is a ramp hybrid or if 1 of them decides to play defense for some odd reason.
Jonathan Norris
12-04-2007, 23:13
No, 4.5 is NOT the average score per match. It is the TOTAL number of ringers scored over the two winning matches. So if Robot C is the lowest scoring 'bot on the winning alliance and scores two in match F1 and three in F2, I win. (I'm counting on attention being drawn by the higher scoring robots, giving Robot C a couple of undefended freebees per match)
-Steve
Ahh ok that clarifies it a bit, now I think you got the better bet :p
Although we were both right in terms of "all-scoring" alliances on Einstein, John was more right than me.....My wallet is lighter this morning, and John's currency collection continues to grow. By my count, a total of 14 tubes were scored by the winning alliance in their two winning matches -- so there's no way that the least scoring 'bot exceeded 4.5 tubes by themselves. Would liked to have seen matches similar to what I saw in many of the division playoffs. I think a couple of alliances that arrived on Einstein were surprised by how things unfolded for them.....
Greg Marra
15-04-2007, 06:39
Although we were both right in terms of "all-scoring" alliances on Einstein, John was more right than me.....My wallet is lighter this morning, and John's currency collection continues to grow. By my count, a total of 14 tubes were scored by the winning alliance in their two winning matches -- so there's no way that the least scoring 'bot exceeded 4.5 tubes by themselves. Would liked to have seen matches similar to what I saw in many of the division playoffs. I think a couple of alliances that arrived on Einstein were surprised by how things unfolded for them.....
In fact, I believe Team 177 scored two tubes total on Einstein. One was scored by their human player, and one by their robot.
Rich Kressly
15-04-2007, 17:10
Steve,
While I feel for you a little, I must say that taking counsel from Kanagarooa..., Kanagarsea..., um .... that guy from 1114, is perhaps not your wisest decision ever :)
and yes, I really can spell "Kanagasabapathy" when I want to
Peter Matteson
15-04-2007, 23:05
That's two years in a row JVN won the bet because of what happened to 177 on Einstein. Happy to help out.
Pete
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