Weight vs. Odds

[nyjoe14.1]

Ok so here is the scenario:

You are playing 9 ball by the game and you are the weaker player. You usually bet somewhere between $10 and $20 a game. If you typically get the wild 7&8 what kind of odds would you need to play even.

1.5-1?

2-1?

Or am I way off?


Thanks

 

[ironman]

Ok so here is the scenario:

You are playing 9 ball by the game and you are the weaker player. You usually bet somewhere between $10 and $20 a game. If you typically get the wild 7&8 what kind of odds would you need to play even.

1.5-1?

2-1?

Or am I way off?


Thanks

IMO, if you are getting the wild 7 and 8, you would need at least 2-1 on the money. I think you would be much better off with games on the wire playing races than odds on the money by the game. A player able to give up that kind of wieght will usually wear down the other.

 

[RBC]

I think you, the weaker player, are better off with the 7&8. With the ball spot, you get the advantage in every single game. The better player will have the advantage every single game if you get games on the wire. The better player will most likely win many more games because he has the advantage in every single one of them. Once he gets ahead, the spot is gone.

Royce Bunnell
www.obcues.com

 

[Fatboy]

it also depends on the speeds of both players, if the stronger cvan put together 5&6 packs it means nothing to get 2 t o1, if the strong player ant BNR a rack 2 to 1 is strong. So eithout knowing the level of play of BOTH players there is no way to answere, if you alt breaks and stop the big packages then yes a money line can be handicapped alot easier.

 

[nyjoe14.1]

Thanks for the replies.

Just to clarify my original post:

What I was looking for was the general consensus on converting balls given (or as two responses mentioned, games on the wire) in to the appropriate odds on the money. I understand that giving up balls or games on the wire is a great way to handicap a game but it should be assumed that these options have been explored in depth.

Fatboy does raise the interesting point that the relative speed of the two players is a contributing factor. Which is to say that a B player giving odds to a C player would be different from a Pro giving odds to a shortstop. I would guess that someone capable of stringing 4-5 racks together fairly consistently giving up the 7&8 to a player that can also get out regularly would be significantly different than the B & C player scenario. I mean how often does a short stop not get out from the 7 ball?

To extend this hypothetical situation a bit further lets assume that no matter what there will be no weight given, only odds on the money. They will only be playing by the game and there will be no ahead sets or races. Also assume that the weaker player would be generally considered at least a B player and is capable of winning from time to time with the 7&8. So I guess that would make the other player at minimum an A+ more likely a shortstop or something close? Of course that is all contingent on the idea that we all have the same definition of a Pro, short stop, A, B, or C player. I think a Pro is pretty easily definable but there seems to be a far amount of debate about the other levels.

That being said what does everybody think?

Thanks

 

[eyesjr]

Ask fuller question

"I give him the seven, the six, it doesn't matter, I run out every time"-to quote Jose Parica. What we are getting at is the strength of competitors involved. And HOW they play the spot. And how the opponent does at those two things...Be more specific, and youll get an answer.

Thanks for the replies.

Just to clarify my original post:

What I was looking for was the general consensus on converting balls given (or as two responses mentioned, games on the wire) in to the appropriate odds on the money. I understand that giving up balls or games on the wire is a great way to handicap a game but it should be assumed that these options have been explored in depth.

Fatboy does raise the interesting point that the relative speed of the two players is a contributing factor. Which is to say that a B player giving odds to a C player would be different from a Pro giving odds to a shortstop. I would guess that someone capable of stringing 4-5 racks together fairly consistently giving up the 7&8 to a player that can also get out regularly would be significantly different than the B & C player scenario. I mean how often does a short stop not get out from the 7 ball?

To extend this hypothetical situation a bit further lets assume that no matter what there will be no weight given, only odds on the money. They will only be playing by the game and there will be no ahead sets or races. Also assume that the weaker player would be generally considered at least a B player and is capable of winning from time to time with the 7&8. So I guess that would make the other player at minimum an A+ more likely a shortstop or something close? Of course that is all contingent on the idea that we all have the same definition of a Pro, short stop, A, B, or C player. I think a Pro is pretty easily definable but there seems to be a far amount of debate about the other levels.

That being said what does everybody think?

Thanks

 

[Deadon]

Games on the wire are irrelevant when you are playing by the game, as the OP indicated was the situation.

I humbly disagree, to some extent with the above posters whom think the question cannot be answered. While the actual speed of the players might make some very small difference, usually its the relevant speed of the parties that makes the value of the spot. IOW, the difference in their speed. If player A can put together a 6 pack, then if the spot is right, player B must be capable of 2 or 3 racks, and so on. If player B cant, then the spot isn't right to start off.

Assuming that the spot suggested makes an equal game, money odds would usually be the stronger option, ie thats what Jose meant when he made the quote about the balls spotted.

I would rather have 2-1 on the money, than the 7/8, when playing by the game. The better player would have to win 2 out of 3 games to break even. In some kind of set, thats a different story.

 

[androd]

I think you, the weaker player, are better off with the 7&8. With the ball spot, you get the advantage in every single game. The better player will have the advantage every single game if you get games on the wire. The better player will most likely win many more games because he has the advantage in every single one of them. Once he gets ahead, the spot is gone.

Royce Bunnell
www.obcues.com

I'm not sure i understand. the better player has the advantage with games on the wire ? The 7 or 8 may not come into play for long periods of time, but the games on the wire are always there. Sometimes the 7 or 8 rain in in bunches, but many times not. I've had and given both. I'd prefer games on the wire. this isn't relevant playing by the game.
PS; The problem is figuring out how many games on the wire?

 

[Mike Templeton]

I'd say about 3-1 on the money = the wild 7 and 8.

Mike

 

[sjm]

Fatboy makes the legitimate point that the speed of the two players makes a huge difference. The seven and eight aren't worth that much at pro level, but it's a pretty big spot when amateurs, even accomplished ones, are involved.

If we are talking about an A+ player that can give the 7 and 8 to a B player, my guess is that two to one odds would make the B player the favorite. I'll say the B player needs 8/5 odds on every rack for a fair game played by the rack. In other words, at a stake of $50 per rack, in each rack, either the A+ player wins $50 or the B player wins $80.