You're in front of a rather odd eight ball pool table with the holes covered up such that the whole table is perfectly flat. It has unusual inner dimensions, 9 and 26, with all balls (1-15) with radius 1 and lighter than usual (except the white ball).
You rack it up and you're just to about to make the first break...
Earlier this day your wealthy tech (he's got a lot of cool technological stuff) friend made a challenge for you.
Friend: "Hey, I've been thinking of an idea for a while. You know that odd pool table I bought with the light balls?"
You: "Yeah? What about it?"
Friend: "Since you're a good puzzler, I've been wanting to make a puzzle for you which I don't really have an answer to. If you manage to do well, you'll get a good prize."
You: "Sure! Let's do this."
Friend: "Alright. I'll go home and make the puzzle, I'll get back to you soon!"
A couple of hours went by and then he finally brought the puzzle.
At most 100 rounds (100 breaks).
Before starting, tell me the number of rounds you want to play and choose only one option A-E, where
option A corresponds to value 20<x≤25,
option B to 15<x≤20,
option C to 10<x≤15,
option D to 5<x≤10, and
option E to 0<x≤5.
Correct value(round 1) gives 1 dollar, correct value(round 2) gives 2 dollars, correct value(round 3) gives 3 dollars,..., correct value(round 100) gives 100 dollars which means maximum winnings are 5050 dollars. The catch is that if you're wrong in any round, you lose all the money you made in previous rounds.
The white ball is not included in the puzzle (except when breaking, of course).
Always ∑ <--->(••) c to c → <---> ?
(Note that my high tech equipment will take care of the above.)
Take your time and when you're ready, tell me the option you've decided and the total number of rounds you want to play.
Finally, after some thinking you feel confident enough to give him an answer.
What option do you choose to get as much money as possible? And how many rounds?
Note: Although mathematics is needed for you to solve the puzzle, it isn't complicated.
Assume the balls have equal probability to end up anywhere on the table (completely random). Oh, and one more thing; not really sum