I've been working on a new idea, and the math is going to get a little heavy. (So if math ain't your strong suit you're gonna get lost here. I'm half-lost myself. Just a warning, no hard feelings!)

I think it is possible to estimate with great precision the number of bucks on your spread by the sheds you find. The key question in the equation is how many matched sets you find. Let's get to it.

THE THEORY

Let's make two big assumptions to illustrate the equation, then later we'll try to deal with the real world and how those assumptions break down. Our first assumption is that all antlers are equally easy to find, and our second assumption is that the finding of a second antler (i.e. one that completes a matched pair) is an independent event from the finding of the first antler of the pair. remember the definitions; "first antler" is any unmatched shed found for a particular buck, "second antler" is any antler that completes a matched set for a particular buck.

We can find x, where x is the total number of sheds in your search area, with the following equation: # of second antlers/ # of first antlers = Total # of first and second antlers/x. Simply put, the percentage of matches you make from your first antlers represents the percentage of all sheds out there that you are actually finding. For example, suppose I found 9 sheds on my hunting lease this spring, and had two matched pairs among those 9. So, I have 7 first antlers and 2 second antlers. The equation becomes 2/7 = 9/x. Solving for x, we get x=31.5. I would predict that I have 31.5 antlers out on my lease. I found 9, so there's about 21 or 22 more still laying out there somewhere. I could also use this number to project that 15.75 bucks are using my lease in the spring.

THE REAL WORLD (with minimal apologies to MTV)

Of course assumptions break down when we enter the real world. The first assumption is that all antlers are equally easy to find. Of course this is not really true. One antler may be laying out in the open, and another may be stuck in a briar patch. Also, obviously a little forkie horn is going to be much more difficult to see than a BC bruiser shed. The first objection, that some antlers are better hidden than others should not be a concern for our equation, however, because this is a random factor. A big buck's shed is really no more or less likely to be hidden by terrain or vegetation than a small shed. But the second objection carries some weight. This factor is not random, yearling antlers are always going to be tougher to find than sheds off a 5 1/2 y.o. buck. If we lump all antlers together, we're going to get a distorted prediction.

The solution to this dilemma is to compare like to like. Ideally, we should calculate a different equation for each age class - one for 1 1/2 y.o. bucks, another for 2 1/2 y.o. bucks, and so on. Sample size being very imprtant to an accurate estimate, in the real world we would need to keep several years of data on this to attain an accurate picture. (I am starting a shed log book this year.)

The second assumption in my theory is the real sticker. Probably I don't have to argue too much to convince you that finding a second antler is not at all independent of having found the first antler. If you find a big ole shed, what do you do? You stop and start doing a grid search around that first antler, looking for the second antler. Unfortunately, this special effort blows the predictive equation all to hell. For the equation to work, the finding of a second antler must not be in any way the result of having found the first antler. Back to my example, where I found 7 first antlers and 2 second antlers, I should ask myself the following question about the second antlers: Would I have found that second antler even if I had never found the first antler in the the set? (Was the second antler laying fifty feet away smack in the middle of the trail I was following? Did I find it three weeks after I found the first antler? Or was it laying down in a hollow 90 degrees off my intended course, and I only found it because I started a grid search from the first antler?) If the answer is that, yes, I would have found the second antlers regardless, then the equation will hold. But if the answer is no, or some degree of maybe, we've got problems.

I believe this problem can be gotten around. Ask the question of any second antler: How likely is it I would have found this antler even if I had not found the first antler? Answer one of the following: likely, somewhat likely, somewhat unlikely, or unlikely. If the answer is likely score it as a second antler found. If somewhat likely, score it as 2/3 of a second antler. If somewhat unlikely, score it as 1/3 of a second antler. If unlikely, don't count it at all.

In closing, I could take the post by Carll on here titled "Sheds!!" and predict the total number of BC class sheds on his place. He has 5 first antlers and 4 second antlers. If the second antlers were completely independent finds, we would have 4/5 = 9/x, with x = 11.25. Carll would have to reflect on the circumstances of finding each of the four second antlers and tell us to what degree finding them was related to having found the first antler in order for us to fine tune the equation.

Anyhow, I invite any critiques or comments along two lines:

1) Is my math correct?

2)How do I make this equation more relevant to the real world? Can my assumptions be amended? Have I missed a key element?