Clarification on fighting technicals

[bob jenkins]

Originally Posted by Back In The Saddle

• Sum each team's "bench leavers", regardless of whether they fought. (NOTE: The summary of penalties may appear to separate fighters and non-fighters into separate penalty groups. But the case play clearly reaches its ruling of no throws and POI based on the equal numbers of bench leavers, irrespective of whether they fought, as illustrated by scenario c.)

This first bullet is wrong. If A1 and A2 leave but do not fight, then B gets two FTs. If, in addition, B1 and B2 leave and fight, A gets four FTs. OFfset where you can (4-2=2) and A will shoot 2 FTs and get the ball.

The rest of your post seemed correct.

[Back In The Saddle]

Originally Posted by bob jenkins

This first bullet is wrong. If A1 and A2 leave but do not fight, then B gets two FTs. If, in addition, B1 and B2 leave and fight, A gets four FTs. OFfset where you can (4-2=2) and A will shoot 2 FTs and get the ball.

The rest of your post seemed correct.

That's what I used to think as well. But that doesn't seem to square with Scrappy's case play ruling, in particular scenario c. In it you have two non-fighters leaving for A, a non-fighter and a fighter leaving for B. It would seem that A would get 4 free throws and B would get 2.

However, the ruling says: "No free throws are awarded for the simultaneous technical fouls as the number of bench personnel leaving the bench and the penalties are the same for both teams."

[bob jenkins]

Originally Posted by Back In The Saddle

That's what I used to think as well. But that doesn't seem to square with Scrappy's case play ruling, in particular scenario c. In it you have two non-fighters leaving for A, a non-fighter and a fighter leaving for B. It would seem that A would get 4 free throws and B would get 2.

However, the ruling says: "No free throws are awarded for the simultaneous technical fouls as the number of bench personnel leaving the bench and the penalties are the same for both teams."

Sigh.

First, offset the number leaving the bench who don't fight. 2 for A is more than 1 for B, so A gets a penalty (one penalty only, here -- even if it was 10 for A and 1 for B, it's only one penalty)

Then, offset the number leaving and fighting. 0 for A, 1 for B == 1 penalty for B (here, it's one per difference -- if it was zero for A and 10 for B, it would be 10 penalties).

Do the same for Players fighting, and coaches leaving.

Then, offset the penalties across the categories.

1 penalty for A, 1 penalty for B == no FTs. (in the parenthetical example, it would be 1 penalty for A, 10 penalties for B == 18 FTs for A)

[slow whistle]

Originally Posted by bob jenkins

Sigh.

First, offset the number leaving the bench who don't fight. 2 for A is more than 1 for B, so A gets a penalty (one penalty only, here -- even if it was 10 for A and 1 for B, it's only one penalty)

Then, offset the number leaving and fighting. 0 for A, 1 for B == 1 penalty for B (here, it's one per difference -- if it was zero for A and 10 for B, it would be 10 penalties).

Do the same for Players fighting, and coaches leaving.

Then, offset the penalties across the categories.

1 penalty for A, 1 penalty for B == no FTs. (in the parenthetical example, it would be 1 penalty for A, 10 penalties for B == 18 FTs for A)

That's what I always thought Bob, but how do you explain casebook play 10.4.5 situation A c)? It says no free throws, go to POI and there are an uneven number of penalties?

[slow whistle]

If you tell me the casebook play is wrong I'll feel much better

[bob jenkins]

Originally Posted by slow whistle

That's what I always thought Bob, but how do you explain casebook play 10.4.5 situation A c)? It says no free throws, go to POI and there are an uneven number of penalties?

I just explained it.

There are an equal number of penalties (meaning FTs) on each side, so they offset.

And, yes, the penalty for being the ONLY team member to leave and not fight is the same as the penalty for being the ONLY team member to leave and fight.

Doesn't seem right, perhaps, and you're not the first person to point that out, but it is the way it is.

[slow whistle]

Originally Posted by bob jenkins

I just explained it.

There are an equal number of penalties (meaning FTs) on each side, so they offset.

And, yes, the penalty for being the ONLY team member to leave and not fight is the same as the penalty for being the ONLY team member to leave and fight.

Doesn't seem right, perhaps, and you're not the first person to point that out, but it is the way it is.

ok I see it now in your note above, thanks!

[slow whistle]

[QUOTE=Back In The Saddle;540700]That's what I used to think as well. But that doesn't seem to square with Scrappy's case play ruling, in particular scenario c. In it you have two non-fighters leaving for A, a non-fighter and a fighter leaving for B. It would seem that A would get 4 free throws and B would get 2.

However, the ruling says: "No free throws are awarded for the simultaneous technical fouls as the number of bench personnel leaving the bench and the penalties are the same for both teams."[/QUOTE

I agree with you , this one confused me as well...in the case play we have no free throws even though it would seem that there is an uneven number of penalties since only one of the four fought. Are they really saying that regardless of whether they fight or not, if you leave the bench area you are getting one flagrant technical and getting tossed, so if you leave AND fight you don't get an additional flagrant technical, still just the one? It seems inconsistent since there is a technical foul penalty listed for each offense (leaving and fighting)....obviously the fact that B7 fought impacts the number if indirect t's going to the coach (two for team B coach and one for team A coach) but you don't shoot ft's based on the number of indirects...not saying the above is correct, just trying to fill in the gaps..appreciate the back and forth to help clarify this one..