333Smitty "cactus 2-gun tie"

A Shifflett

New member
Smitty,

On another thread you inquired about the 2-gun tie at the Cactus. This is my understanding of how i won:

Gene Buckys Grands were LV 2682 HV 2551, add these together you get 5233, when you divide the 5233 by 2 you get 2616.5

My grands were LV 2449 HV 2782, add these together you get 5231, when you divide the 5231 by 2 you get 2615.5

On the final 2-gun sheet it had me and Gene both at 2616, why the computer rounded my agg up by 1/2 and not Gene's i have no clue. But i did go to the stat house as soon as it was posted and i did ask them if they were absolutely sure i had won. They told me i had.

As bad as i want to win, i never want to be given a win i didnt deserve and had i thought for one minute that i didnt deserve this one i would have gladly taken second to Gene. Gene is a class act in all aspects of the game and was one of the first persons to walk up, shake my hand and say congratulations. Had it been the other way around i would have done the same for him.

A personal thank you to all the staff at Ben Avery!

And a very special thank you to Ed Adams! He donated $500.00 cash to the 2-gun winner.

Andy
 
Congrats Andy,,,,I hear the wind was pretty stiff,,,sorta like Fairchance !!!,,,,it was NOT a trigger pulling contest,,,,that is a very satisfying victory,,,,you had to work for it,,,
,,,Roger
 
Thanks Andy you beat me to it. I was going to post today, that if you add all 4 aggs and divide by 4 you get:

Andy .26155
Gene .26160

We had it happen a couple of months at one of our other matches. If it's a tie the Bughole program will say tie with a T. But you were right it will round up if it's not a tie.

It was Butch's Reloading/Butch's Guns who donated the $500.00 for the 2-Gun. Butch Fjoser and Ed Adams are partners in Butch's Reloading/Butchs Guns.

Congratulaions again Andy.

Art Clegg
 
I shot Lt32. Lot 6

How do you identify the lot number? I see a "1" moulded into the bottom of the plastic bottle -- does that represent lot number 1? Alternatively, a small label at the upper portion of the back reads "0104132" -- do the first six digits represent the date and the seventh digit represent lot number 2?
 
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hunter

The 8 pounders I have, have a white sticker on the bottom. This sticker has the number 6 on it. I was told that was the lot number.

Andy
 
Mathematical Rounding

I think this will explain why the computer rounded the scores the way it did. The computer was correct.

Rounding, as generally taught, fails to address one very specific condition. We all know that if the fraction is greater than 0.5, the preceding number is "rounded up" and if the fraction is less than 0.5, the fraction is dropped. But, what is often not taught is the condition where the fraction is EXACTLY one-half (0.5).

The "rounding rule" is such that if the fraction is EXACTLY 0.5, rounding should be as follows:
1) if the number immediately preceding the fraction is even, the 0.5 fraction is dropped; or
2) if the number immediately preceding the fraction is odd, the number preceding the 0.5 fraction is rounded UP.

For instance:
13.45 rounded to tenths is 13.4
13.35 rounded to tenths is 13.4
13.55 rounded to tenths is 13.6
13.65 rounded to tenths is 13.6

The reason for this rule is to assure that EXACTLY 50% of all numbers are rounded each way (up or down).

While the computer was correct, I suggest that the program should be modified to move "out" one more decimal place in it's rounding.

Glen O.
 
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This has been argued before, but I will re-introduce it. Instead of going out one more place maybe we should follow the "laws" of statistics and round to the place to which the targets were measured, i.e. thousandths place. I am putting on my fire suit, so fire away. James Mock
 
Rounding of numbers

First let me say I am no mathematician and I don't claim to be:

I have a way of adding yardages and grands that although it would not guarantee we never have another tie, it would certainly make them less likely to occur.
Since the cactus was as close as it was I am going to use that as my example.

Gene's aggs 2120+3243+2760+2341 = 10464
Andy's aggs 2666+2898+2480+2418 = 10462 smallest number wins.....

Also notice if you divide the above totals by 4 you get a clear winner, only when rounding up did it become a tie!

10464/4= 2616
10462/4= 2615.5

To me, the smallest sum of numbers should without question always be the winner.

Andy
 
Right on, Andy

I was logging in to propose just about what you did. What you said earlier is exactly right. The targets are measured to the nearest thousandth and that 3rd decimal digit is the last one in which the person doing the measuring has any confidence. That's another way of saying that any digit(s) following the thousandths are "not significant". The idea of "significant digits" was repeatedly beat into my head in physics and chemistry courses. Summing measurements accurate to plus/minus 0.001" and then dividing to get an "average" group size to plus/minus 0.0001" introduces another digit that is definitely not "significant" -- it is totally meaningless.

I would propose just adding the values and the smallest total wins. That keeps the final sum accurate to 0.001" and does not introduce that extraneous and incorrect 0.0001" value. People are still going to divide it out and come up with their own "average group size"; but, to do it officially can lead to incorrect results.

glen
 
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Andy that is a good idea to add the scores....but not the aggs (which have been rounded). For 100 yards add the MEASURED scores and the smallest would win 100 yards. Next, add the 200 yard MEASURED scores and the smallest would be the 200 yd. winner. Now we must reconcile the 100 and 200. Either the 100 has to be doubled or the 200 halved (round according to the rules posted above). Then the grand total can be ADDED with the smaller number being the winner. I know that this will never happen. Congratulations Andy for some great shooting in some tough conditions. Also, my congratulations to Gene who has been in several of these "barn burners". Good shooting....James
 
James,

How about converting the 100-yd and 200-yd scores to MOA and totalling them? That'd eliminate the "averaging" problem.

glen
 
James,

I have tried for the past year to explain to people that the 100 yard measured score is being doubled when they convert it to an MOA. Which gives the 100 yard agg more weight when figured in the grand. If I beat you by .010 on each of 5 groups at 100 yards, you in turn have to beat me by .020 on each of five groups at 200 in order to tie me in the grand. By adding the MEASURED dimension only, it will eliminate this advantage.

Glen,

As you stated, the final add-up can still be converted to MOA for the purpose of record keeping such, as world records.

Andy
 
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