Ball spots equating to money line wagers

darmoose

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This is a great thread! My example assumes that it takes 8 balls to win a one-pocket game. When you leave that behind you encounter the variables that I mentioned, and it becomes a crapshoot, but I'm not convinced that it isn't a simple proportion. If it is not, then we need a genuine 24 karat expert. Dr. Dave Alciatore comes to mind, but maybe he's not a math guy. In a practical sense, it's two guys trying to agree on a game, whether they get it clinically correct or not. :unsure:

I think the first assumption people are making, namely that it takes 8 balls to win a OP games is causing difficulty in understanding just what the odds should be and what the % advantage actually is in any given game whare a spot is being offered.

If we are playing 9/7 the person going to 7 has a 22.222% advantage over the other player and the game is a game that takes 9 balls to win.

If we are playing 10/8 the person going to 8 has a 20% advantage and the game is a race to 10, not 8.

There is NO value per ball that remains the same regardless of how you change the game at hand.

This IS just simple math, and there is NO more to it than that. The values put on balls and games by JohnnyTronic and others (which I mean no disrespect towards) simply do not "hold water" as uncle Vinny said ........... :unsure: :rolleyes:

In a game of 10/8 you can say one or both of two things...... the lesser player is advantaged by 20%, or the better player is disadvantaged by 25%, it is math only...
 
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Bob Jewett

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BJ, I imagine the bet was even money between Scott and Dave, the skill difference being addressed in the 17-3 game. You must be asking for a money adjustment for a similar game between equal players - is that correct? I wouldn't even guess, as the correct bet between Joe Blow and John Doe might not be the same for Pete Gomez and Leroy Johnson. Variables, man. At the table, in the poolroom, it's not all arithmetic.
No, it's not all arithmetic, but arithmetic is the best place to start, if you can. It is arithmetic that keeps the casinos and bookies in business, and it is a belief in variables that keeps the mooches coming back.
 

darmoose

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Be careful not to mistake performance, and the variables that attend it, for arithmetic. For example, if two players are equal and one gives the other a 9/8 game, he is taking on a burden of 12.5 % - the value of each ball. Mathematically, he should get a similar adjustment in the bet. On the table, maybe not . . . .

You are correct that a player giving a game of 9/8 is taking on a burden of 12.5%, which is arrived at by dividing his extra burden (1 ball) by the number of balls the opponent needs to win the game (which is 8 balls). 1 divided by 8 = .125 or 12.5%. If the game were 10/8 the player giving the spot would be burdened by 25% or 2 divided by 8 or .250 (25%). In a 9/7 game he is burdened 2 divided by 7 = 28.5%.

The other side of all this is the player getting the spot is "advantaged as such: at 9/8 he has a 1 divided by 9 = 11.1%, at 10/8 he is advantaged thusly, 2 divided by 10 = 20%, and at 9/7 he is advantaged like this, 2 divided by 9 = 22.2% In calculating the advantage the lesser player is given, you get the same answer by dividing the number of balls he must make by the number of balls the opponent must make (8/9=.888 or 11.1%, 8/10= .8 or 20% advantage, and 7/9= .777 or 22.2% advantage)

There is no more to it than simple math. ;)
 

poolboy79

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That's what I'm talking about, you lose the first game and now you are up against a lot of pressure. Also him smashing the stack is actually a smart move, he has a better chance to get lucky, than to win a moving contest with you. Sounds like he might be smarter than people think!
I can't make a ball in the ocean on my good days and I don't have many of them but I'm always going to be the one getting the odds on the money.{ or I'm not playing}
I think Scott Frost gave Dippy Dave 17 to 3 and won.
Now a lot of people I have met think Dave was helpless and they could have beaten Scott with that spot {I told most of them if they would post at least 5 grand I would get in touch with Scott and see if he wanted to come and get the money} I doubt he would have come for that cheap but nobody ever accepted anyway lol} but most of them would have lost playing Dave even, he wasn't a real good player but he wasn't dead money either. I'd say he was a low to mid B shooter but he moved balls to his hole pretty well and defended reasonably. I wonder what everyone thinks the correct money odds should be, for that . {This should be good for some chuckles.}
Billy I, may have already figured this out , Take a stab at it guys , we might learn something valuable.
Sometimes the money line is so high it just simply is off the board.
 

sappo

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I think the first assumption people are making, namely that it takes 8 balls to win a OP games is causing difficulty in understanding just what the odds should be and what the % advantage actually is in any given game whare a spot is being offered.

If we are playing 9/7 the person going to 7 has a 22.222% advantage over the other player and the game is a game that takes 9 balls to win.

If we are playing 10/8 the person going to 8 has a 20% advantage and the game is a race to 10, not 8.

There is NO value per ball that remains the same regardless of how you change the game at hand.

This IS just simple math, and there is NO more to it than that. The values put on balls and games by JohnnyTronic and others (which I mean no disrespect towards) simply do not "hold water" as uncle Vinny said ........... :unsure: :rolleyes:

In a game of 10/8 you can say one or both of two things...... the lesser player is advantaged by 20%, or the better player is disadvantaged by 25%, it is math only...
So what do you think the money line would be if 2 players have been playing 10-5 for months and have been breaking even? Keith
 

darmoose

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So what do you think the money line would be if 2 players have been playing 10-5 for months and have been breaking even? Keith

Keith,

I been thinking about your question for a while and can't come up with a good answer that's not just a guess. The math is simple when simply identifying the advantage one player has or the disadvantage the other player has in a given game. But, converting that to money odds, not so simple. At least I haven't come upon it yet. I will keep thinking about it, you never know............ :unsure:
 

vapros

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If they are breaking even it is an even money bet, for sure. If they change the game, especially a big change, then you would need an expert. Ask a bookie - he will know how to do the math.
 

boingo

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By definition the right ball spot results in an even money bet, each player wins 50% of the time. With no ball spot, 8-8, what is the percentage of games won by the two players? Player A wins p percent of the time and player B wins 1-p percent of the time where 0<p<1. Suppose player A bets X dollars and player B bets Y dollars. The expected value for both players is zero if Y=X*p/(1-p). For instance, suppose p=0.4, then p/(1-p)=0.667 ~2/3 so if player B bets 300 dollars and player A bets 200 dollars neither will make money in the long run, unless they dump the two math professors who are backing them :)
 

unoperro

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By definition the right ball spot results in an even money bet, each player wins 50% of the time. With no ball spot, 8-8, what is the percentage of games won by the two players? Player A wins p percent of the time and player B wins 1-p percent of the time where 0<p<1. Suppose player A bets X dollars and player B bets Y dollars. The expected value for both players is zero if Y=X*p/(1-p). For instance, suppose p=0.4, then p/(1-p)=0.667 ~2/3 so if player B bets 300 dollars and player A bets 200 dollars neither will make money in the long run, unless they dump the two math professors who are backing them :)

Not sure about what you are saying, I looked at it like this-10 to 5 and breaking even means..

1000 games =500 W 500 L each

10 going to 8 is a 20% positive change so he wins 20% of his lost games which is 100 more games

5 going to 8 is a 60 % negative change so he loses 60% of his 500
Which is 300 less games.

So now we are at 900 wins to 100 wins so I say 9 to 1.

I know its not this simple but you have to start somewhere😉
 

BRLongArm

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Scott could easily give that guy 8-1 on the money and never pay a losing bet. Going to eight is one obstacle. Dave never goes to 8. But the other variable is Scott only goes to 8. That's one or two innings usually. I play 10-7 better than Dippy on his best day and I would need about 8-1 to have a reasonable chance against someone of Scott's caliber.
 
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