odds on the money per ball as a spot

lll said:

the game is you play each game until all balls are made
and count your score
the stronger players balls count x dollars per ball
the weaker players ball count some percentage of x per ball
example
strong player gives weaker player 2:1 per ball at $20 per ball
end of first rack
strong player makes 10 balls =$200 (10x20)
weak player makes 5 balls =$200 (5x $40)
so they break even
for you guys that like percentages
how can you calculate based on how you play someone by ball count to odds on the cost of each ball
for example
if you can give someone 10-8 what odds on the money per ball could you give

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wincardona said:

Larry, I think there was a thread that addressed that proposition, i'm sure Rodney or someone else can direct you to it.

Lets look at it this way..When playing this prop by placing a value on each ball the way you could accurately determine what the correct value is would be to play the entire rack out with your opponent a number of times. Say you play five racks out completely for a total of 75 balls. Then you would take the number of balls the better player makes and divide by the number of balls that the lesser player made to get a ball value.

Example; In five racks the better player made 45 of the 75 balls and the weaker player made 30 of the 75 balls. You would take the number of balls the better player made which would be 45 and divide it by the number of balls the lesser player made which would be 30. 45 divided by 30 would give you a 1.5 to 1.0 value. 3/2. Which is $1.50 to $1.00

That would be an accurate way to determine what the ball value would be between two players playing out the entire rack (15 balls) Playing out the entire rack places importance on all areas of the game, an ecellent way of playing to pick up on experience playing all parts of the game. However, playing the score now becomes a moot point, this way of playing puts more emphasis on the "risk against reward" value when choosing options. Very interesting.

Dr. Bill

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LSJohn said:

That's the best way, but unlikely to ever be available in the real world. Two guys who want to gamble a fair match against each other aren't likely to play five or ten racks for ice water to find out what's fair, and even if they do, which -- if not both -- are on the lemon? :heh

I think Larry wants to know how to figure out what would be the money-odds equivalent to 10-8 or 9-6 as a pure mathematical question rather than as a real-time test.

IMO, if 10-8 is even, $10 for the weaker player and $8 for the stronger (or $5-$4 or $1.00-$.80) would be close, but would give the better player a small edge from the fact that games will last longer. $9-$6 same way.

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wincardona said:

John, I understand that no one is going to practice playing 5 racks of one pocket all the way through to determine what game would be fare, in terms of applying a money line attached to the ball count but you could start out playing cheap and make the necessary adjustments if you need to as you are playing. Adjustments with the bet or the money line, either way it stimulates action.

Furthermore, if you are familiar with one another's game and have a history with playing one another than based of of the formula I proposed then you can ust that as a way to start the negotiations. 10/8 = $1.25 where 9/6= $1.50. Woudn't that be a good way to start the negotiation from?

Dr. Bill

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