UKC Forums - Rules Question - Split Trees & Dog Running

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-- Rules Question - Split Trees & Dog Running (http://forums.ukcdogs.com/showthread.php?threadid=192139)

Posted by last chance on 02-13-2008 04:15 PM:

Gfults to your last reply they are two different scenarios:Example 5 is the same as origial thread except you have four dogs instead of three dogs.Example 7 is totality different.

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Posted by gfults on 02-13-2008 04:29 PM:

I understand that there are 4 dogs. But I see very little consistency with Allens method of scoring these situations. I can see why dog C shouldnt get moved up in the last scenario. I could even learn to choke down the idea of moving dog D up to 125. Maybe! But the inconsistency lies where they both dont get moved up, or both of them do get moved up. It would be alot simpler to either move them all up all the time, or move none of them up all the time. Think about being in the final 4 at the UKC world hunt and u are handling dog C in the last scenario. You are behind dog D by only 50. How are most folks gonna take seeing dog D get moved up to beat you out of the world title because your dog didnt get moved up. They are both on seperate trees.

Posted by Darrell on 02-13-2008 06:52 PM:

quote:
Originally posted by last chance
In the other scenario we are assuming all dogs are together,but upon arrival dogs A&B are running so dog A gets 125 minus,dog B gets 75 minus,dog C gets 50 and dog D which is on another tree gets 125.

But if you assume all dogs are together, then both should move up because you don't know that C wasn't the one by himself all the time, or neither move up, because again, C could have been alone all the time. You are "assuming" that D was not with A & B, and "assuming" C was with A & B. I just don't see how you can make that assumption since the chances that C was split or D was split are absolutely, 100% equal...

I think it's pretty obvious where the lines of opinions fall on this matter (although moot now). Kinda strange when you think about which people said "nobody moves up", and who they are...

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Posted by gfults on 02-13-2008 07:03 PM:

Darrell,
I'm with ya. Either they both move up or neither one moves up. I can just imagine judging a cast here in southern middle Tennessee where this scenario appears. Go into the tree, A and B are gone. C and D are treed on seperate trees. The judge says he's gonna move dog D up to 125 and leave C at 50. I bet somebody comes completely unhinged! If somebody did that crap to me at a major event, ya'll would see my ass on CNN the next day!!

Posted by Gibbo on 02-13-2008 07:28 PM:

quote:
Originally posted by gfults
Darrell,
If somebody did that crap to me at a major event, ya'll would see my ass on CNN the next day!!

I'm sure we probably will someday anyway

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Posted by GA DAWG on 02-13-2008 08:06 PM:

quote:
Originally posted by Allen / UKC

Example 7 –
Dogs A, B, C and D are declared treed in that order. Upon arriving dogs A and B are gone. C is on one tree and D is on a separate tree.

Tree points:
Dog A = 125-
Dog B = 75-
Dog C = 50
Dog D = 125
(this one will likely bring on the biggest argument, but again, it is staying consistent with the theory of using the root tree. Why? Because it was never obvious that the dogs were split until the cast arrived and “saw” D on a separate tree. We don’t assume A and B were anywhere else other than with C because it wasn’t obvious.)
.

OH ME!!!!!

Posted by CooperCreek on 02-13-2008 08:28 PM:

quote:
Originally posted by Allen / UKC

Tree points:
Dog A = 125-
Dog B = 75-
Dog C = 50
Dog D = 125
(this one will likely bring on the biggest argument, but again, it is staying consistent with the theory of using the root tree. Why? Because it was never obvious that the dogs were split until the cast arrived and “saw” D on a separate tree. We don’t assume A and B were anywhere else other than with C because it wasn’t obvious.)

What if dogs A, B came treed, then dog C gets treed by himself, then dog D went and covered A and B. Split not obvious. Before the cast arrives, dog A and B leave. Now with your logic, we're now awarding first tree points to dog D who treed last, and dog C which got by himself, is getting 3rd tree point. I guess I don't understand why its so obvious that D split from the rest. If we assume D split from the rest once we see it, why can't we assume instead that dog C split from the rest? Oh right, because for a dog to split, it must have treed last, correct?

Posted by GNeal on 02-13-2008 08:46 PM:

Gfults, you almost had me thinking I was a stupid, ignorant, goober, with a learning disability, but since we got the offical ruling I sure hate to think of what it makes you look like. Don't worry though us old big egoed Texans ain't gonna bragg to much.

GNeal

Posted by Tim MACHA on 02-13-2008 08:55 PM:

My last reply on this thread

CooperCreek, I believe Allen made a point of saying that trees will be considered single until proved otherwise split. Therefore we cannot assume splits. I am done here.

Posted by blueticker on 02-13-2008 09:03 PM:

quote:
Originally posted by gfults
Darrell,
I'm with ya. Either they both move up or neither one moves up. I can just imagine judging a cast here in southern middle Tennessee where this scenario appears. Go into the tree, A and B are gone. C and D are treed on seperate trees. The judge says he's gonna move dog D up to 125 and leave C at 50. I bet somebody comes completely unhinged! If somebody did that crap to me at a major event, ya'll would see my ass on CNN the next day!!

Both get moved to 125+ for treeing a coon. If you don't know what tree A & B was on. Dog D was the hound treed with A & B not C. D doesn't get 125. Dog A & B was treed on their own tree, C had his tree and D had his tree. If you don't know for sure score it accordingly when you do know. If you was close enough to tell that ABC was treed together then C doesn't get moved up.

Posted by elvis on 02-13-2008 10:52 PM:

if it is possible to gain access to the old board, i will show you where we got a ruleing from tk in a very similar situation where he said the only way you could move c up is if the judge saw something (in my situation the judge said he saw dog b comming from the direction of dog a's tree) that led him to beleive that dog b had been treed on dog a's tree.otherwise you dont know which tree b left and can not move dog c up.

i had posted the question to todd because of a situation i had at an rqe and felt the rep had not handled it properly.
i know this board is not official and todd may just have been backing his field rep for a miscall,but if thats the case,how official is allens ruleing?

but then,how official is it when it is in the advisor? i still havent forgotten the "official advisor,the final authority and gospel of all rule interpretations" on how to deal with a silent dog. LOL

i dont care what gets changed, i will judge it the way ukc wants it judged. just dont change something and tell us its always been that way.

Posted by blueticker on 02-13-2008 10:54 PM:

Now, back to the orginal question on this post. If A & B leaves the preverbal unknown tree and we see dog C with a coon and dog D with a coon and AB has hauled butt. Both C and D gets the 125+ . Any other way wouldn't even be using common sence that a run over possum should have. Both hounds have treed a coon and we never established a single tree until AB was long gone. All four hounds may have had their own tree if a cast had been standing 40 feet away when called.

GOOD DISCUSSION!!!!!

I always get a kick out of those guys that split tree their dog when the cast is 1/4 away and can't tell anything about what dog is treed where. If dogs ABC and D are all treed, with D called split and when you do get to the area and dog C is split to the left the judge gets a call from the handler of C for split +125 D gets the 125- for treeing with AB.

Posted by Darrell on 02-13-2008 11:52 PM:

My question to Allen is, how do you know you didn't walk into the tree and SEE dog C on a separate tree, and not D? Do I need to call Mrs. Cleo for a ruling on that one? I must be thick as a plank, but I don't know how you can arbritarily say dog D was split and C was not, other than "Allen said so"...

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Posted by Ransom on 02-14-2008 06:04 AM:

Here is to all that cannot understand when there is a split tree that there are 2 sets of treepoints you remind me of a deer running dog that cannot be broke just stuck on stupid.

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Posted by larrypoe on 02-14-2008 06:52 AM:

quote:
Originally posted by blueticker
Both get moved to 125+ for treeing a coon. If you don't know what tree A & B was on. Dog D was the hound treed with A & B not C. D doesn't get 125. Dog A & B was treed on their own tree, C had his tree and D had his tree. If you don't know for sure score it accordingly when you do know. If you was close enough to tell that ABC was treed together then C doesn't get moved up.

Denise,

It cant be scored that way.

When its not possable to determine there is a split tree when the calls are made, when you get to the tree the dog holding the highest position called and still treed is the "root tree" or the tree the calls were taken on. It doesnt matter if that dog is the first dog treed or the last.

If the last dog treed is the only one still there, it doesnt move up at all. If the dogs holding 3rd and 4th tree are the only ones still there, nobody moves up. If the first dog treed is the only dog who has left and all others are there, nobody moves up. If the first and last dog treed are on the same tree and everyone else is gone, nobody moves up.

Why? Because in all those cases there is no split tree evident, so we leave them as called.

When we get to a tree where a split tree becomes evident, we have to use that same scoring method to keep it consistent.

We start with the highest called position still treed, and that is the tree we took the calls on. We move to the next dog who is on a different tree, and move it up to first along with whatever is treed with them moving up behind them. We keep doing that untill all trees are rescored.

If we get in there and have 4 dogs on 4 different trees, we move them all up to 1st tree.

There are cases where somebody is getting the shaft, but no scoring method is going to be perfect. This one at least keeps it consistent in a situation where there is no way of knowing for sure what happened.

Its been scored that way since way before my first hunt, because thats the way I was taught to score it 20 years ago.

I think the confusion has came from the wording of some of those interpretations, IE: if dogs have left nobody moves up, but the root tree scoring has always been the same.

In fact, the scoring situations Allen gave are the same ones I was taught when I was 16 and judging my first NTCH casts as a nonhunting judge.

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Posted by on 02-14-2008 02:03 PM:

If UKC wanted to be consistant and keep it simple, they should just say that anytime a dog is gone from the tree, no dog carded behind the one that moved can move up in tree position,
It may not be perfect but in the long run it is the fairsest, simplest, and most consistant way to handle it.
There is no way I will every remember all of the 7 examples Allen gave us. "Screw that! The MOH can have that headache. I'm just gonna hunt.

Posted by John Wittenborn on 02-14-2008 02:23 PM:

Jim,

I think that is the way it has been explained, except when a split tree is involved.

When was the last time that you seen a tree scored that didn't have 125 plus, 125 minus, or 125 circle attached to the scoring of that tree. Some dog had to of had first tree on every tree, I would think??????????????

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Posted by Darrell on 02-14-2008 11:55 PM:

quote:
Originally posted by larrypoe
Denise,

It cant be scored that way.

When its not possable to determine there is a split tree when the calls are made, when you get to the tree the dog holding the highest position called and still treed is the "root tree" or the tree the calls were taken on. It doesnt matter if that dog is the first dog treed or the last.

If the last dog treed is the only one still there, it doesnt move up at all. If the dogs holding 3rd and 4th tree are the only ones still there, nobody moves up. If the first dog treed is the only dog who has left and all others are there, nobody moves up. If the first and last dog treed are on the same tree and everyone else is gone, nobody moves up.

Why? Because in all those cases there is no split tree evident, so we leave them as called.

When we get to a tree where a split tree becomes evident, we have to use that same scoring method to keep it consistent.

We start with the highest called position still treed, and that is the tree we took the calls on. We move to the next dog who is on a different tree, and move it up to first along with whatever is treed with them moving up behind them. We keep doing that untill all trees are rescored.

If we get in there and have 4 dogs on 4 different trees, we move them all up to 1st tree.

There are cases where somebody is getting the shaft, but no scoring method is going to be perfect. This one at least keeps it consistent in a situation where there is no way of knowing for sure what happened.

Its been scored that way since way before my first hunt, because thats the way I was taught to score it 20 years ago.

I think the confusion has came from the wording of some of those interpretations, IE: if dogs have left nobody moves up, but the root tree scoring has always been the same.

In fact, the scoring situations Allen gave are the same ones I was taught when I was 16 and judging my first NTCH casts as a nonhunting judge.

That has to be the most unfair method of scoring split trees I have ever heard. Assume the last dog treed is by himself? You've just awarded the slowest dog in the cast (remember they were treed so close you couldn't tell they were split, but he still couldn't get treed before 3 other dogs), and moved him up to first tree. I guess I'm the only one seeing the failed logic in that, not to mention inconsistency.

I'm with you Jim (no surprise). All I want to do is judge consistently, and be able to confidently stand behind my call. I'm afraid my willingness to judge has dissipated somewhat...

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Posted by GRAVEDIGGER on 02-15-2008 09:08 AM:

Darrell,
I don't see where the "failed logic" is. If anything it is the most logical interpretation of the rules. The only reason why it seems so complicated is because people are making assumptions, and trying to say what if "......".
Allen breaks it down quite clearly, and tells you that UNLESS it's OBVIOUS you need to score it this way. (1) Because it is not obvious, all dogs go on the card in order. (2) When it's obvious there has been a split, the last dog treed will be decided to have been split per original positioning, and moved up. That is logical.
However, it may not be what actually happened. So the results may end up wrong in some instances--this is the grey area and where UKC interprets what is best for the rules. If you have run the hunts enough, you know that there will be times that the rules will arbitrarily go against you sometimes--the luck factor.
Now, where I see failed logic is when you make assumptions, and believe that because a dog trees last that it probably isn't split because it must be to slow to trail up the other coon. That is a "slippery slope" argument! I guess I should assume that A&B left because they were run off the tree. Even though nothing was heard; so we should scratch both C&D because we don't know which one of them run the other two off. (I guess that is why they don't call it the darrellian method of logic).

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Posted by gfults on 02-15-2008 09:32 AM:

It makes no sense to move one dog up and not both. If UKC wants to stop any confusion they need to either move all of them up on every scenario, or move none of them up on every scenario. The way this is being explained is not consistent. Can u imagine what the handler of dog C will be like when a judge moves dog D up and not dog C?

Posted by last chance on 02-15-2008 01:48 PM:

All dogs tree 50 yards in front of you in this order A,B,C on one tree and D is split.Dogs A,B leave so that means A gets 125 minus,B gets 75 minus,C gets 50 except you want to move him up to 125 because he's the only one left on that tree and D gets 125 because he is split.It has nothing do with last dog treed.

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Posted by Darrell on 02-15-2008 05:07 PM:

quote:
Originally posted by GRAVEDIGGER
Darrell,
I don't see where the "failed logic" is. If anything it is the most logical interpretation of the rules. The only reason why it seems so complicated is because people are making assumptions, and trying to say what if "......".
Allen breaks it down quite clearly, and tells you that UNLESS it's OBVIOUS you need to score it this way. (1) Because it is not obvious, all dogs go on the card in order. (2) When it's obvious there has been a split, the last dog treed will be decided to have been split per original positioning, and moved up. That is logical.
However, it may not be what actually happened. So the results may end up wrong in some instances--this is the grey area and where UKC interprets what is best for the rules. If you have run the hunts enough, you know that there will be times that the rules will arbitrarily go against you sometimes--the luck factor.
Now, where I see failed logic is when you make assumptions, and believe that because a dog trees last that it probably isn't split because it must be to slow to trail up the other coon. That is a "slippery slope" argument! I guess I should assume that A&B left because they were run off the tree. Even though nothing was heard; so we should scratch both C&D because we don't know which one of them run the other two off. (I guess that is why they don't call it the darrellian method of logic).

The assumption is just because D treed last, he was never on the "root tree" (whatever that is), and C was. However, before arriving dog D was assumed to be on the "root tree" also. Wonder what changed? Just because he's by himself? So is dog C?

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Posted by Darrell on 02-15-2008 05:10 PM:

quote:
Originally posted by last chance
All dogs tree 50 yards in front of you in this order A,B,C on one tree and D is split.Dogs A,B leave so that means A gets 125 minus,B gets 75 minus,C gets 50 except you want to move him up to 125 because he's the only one left on that tree and D gets 125 because he is split.It has nothing do with last dog treed.

All dogs tree 50 yards in front of you in this order A,B,D on one tree and C is split. A and B leave. Now what? A scenario of equal probability...

I'll not debate this anymore. All that's left are those siding with UKC, and those that can't see the logic (the dumb ones like Jim, RIP, elvis, gfults, and I). Nobody is changing their mind at this point...

P.S. My friend posed the question to a MOH (or THE MOH) at the World hunt, very similiar to the last question (exactly I think), and was told to split points between 3rd and 1st tree, giving each dog 87.5. Makes more sense to me than any, really. Too bad he couldn't remember his name.

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Posted by Tim MACHA on 02-15-2008 07:16 PM:

quote:

Originally posted by Darrel

P.S. My friend posed the question to a MOH (or THE MOH) at the World hunt, very similiar to the last question (exactly I think), and was told to split points between 3rd and 1st tree, giving each dog 87.5. Makes more sense to me than any, really. Too bad he couldn't remember his name.

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Posted by CooperCreek on 02-15-2008 07:51 PM:

quote:
Originally posted by Darrell

P.S. My friend posed the question to a MOH (or THE MOH) at the World hunt, very similiar to the last question (exactly I think), and was told to split points between 3rd and 1st tree, giving each dog 87.5. Makes more sense to me than any, really. Too bad he couldn't remember his name.

To bad the "Advisor" was a direct result of the lack of MOH's able to make reasonable decisions like the one above. Now we get to follow the blind sheep, follow the thought process of "well, thats the way its always been done," and try to make some sense out of what Todd or Allen had to say in the September 1837 CH Bloodlines on how to score a situation.

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